BASIC ALGEBRA
With Applications
Ivan G. Zaigralin
© 2017–2020 Ivan G. Zaigralin
Cover design © 2018 Derrick Santiago
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Edition: 5th
Publisher: Samizdat
Contents
Contents 5
Chapter 1. Concepts 15
1. Expressions, Relations, and Substitution 15
1.1. Expressions versus Relations 15
1.2. Algebraic Substitution 17
1.3. Common Types of Expressions 18
Homework 1.1 21
Homework 1.1 Answers 22
2. Real Number Axioms 23
2.1. Axioms 23
2.2. Applying Axioms 27
2.3. Inverses, Subtraction, Division 28
2.4. Conclusion 30
Homework 1.2 32
Homework 1.2 Answers 33
3. The Number Line and Sets of Numbers 34
3.1. Roster 34
3.2. Integers 34
3.3. Set Builder 36
3.4. Rationals 37
3.5. Reals 38
3.6. Absolute Value 40
3.7. Inequality Relations 42
Homework 1.3 44
Homework 1.3 Answers 46
4. Properties of Real Numbers 47
4.1. Basic Properties 47
4.2. Positive Integer Exponent 50
Homework 1.4 52
Homework 1.4 Answers 53
5. Order of Operations 54
5.1. Evaluating Sums and Products 54
5.2. Like Terms 57
Homework 1.5 59
Homework 1.5 Answers 61
6. Prime Numbers 62
5
CHAPTER 0. CONTENTS
6.1. Integer Division and Prime Numbers 62
6.2. Fractions of Integers in Lowest Terms 64
Homework 1.6 67
Homework 1.6 Answers 69
7. Fractions and Rationals 70
7.1. Fraction Notation 70
7.2. Changing the Denominator 74
Homework 1.7 76
Homework 1.7 Answers 78
8. Fraction Addition 79
8.1. Adding Fractions with a Common Denominator 79
8.2. Least Common Multiple 80
8.3. Adding Fractions of Integers 81
8.4. Mixed Numbers 82
Homework 1.8 84
Homework 1.8 Answers 86
9. Translation 87
9.1. Expressions 87
9.2. Equations 87
9.3. Applications 90
Homework 1.9 91
Homework 1.9 Answers 92
Practice Test 1 93
Practice Test 1 Answers 94
Chapter 2. Linear Equations 95
1. Properties of Equations 95
1.1. Equations and Solutions 95
1.2. Addition Property for Equations 97
1.3. Multiplication Property for Equations 98
1.4. Applications 99
Homework 2.1 101
Homework 2.1 Answers 103
2. Solving Linear Equations 104
2.1. Linear and Non-linear Equations 104
2.2. Isolating The Variable 105
2.3. Applications 110
Homework 2.2 114
Homework 2.2 Answers 116
3. Linear Formulas 117
3.1. Isolating a Variable 117
Homework 2.3 121
Homework 2.3 Answers 122
4. Percent 123
4.1. Percent Units and Notation 123
6
CHAPTER 0. CONTENTS
4.2. Percent Increase and Decrease 125
Homework 2.4 128
Homework 2.4 Answers 129
5. Applications 130
5.1. Distance and Work 130
5.2. Applications of w = r t 132
Homework 2.5 135
Homework 2.5 Answers 137
6. Linear Inequalities 138
6.1. Solution Sets 138
6.2. Interval Notation and Graphs 139
6.3. Addition and Multiplication Properties 141
Homework 2.6 144
Homework 2.6 Answers 145
Practice Test 2 146
Practice Test 2 Answers 147
Chapter 3. Graphing 149
1. Reading and Constructing Graphs 149
1.1. Cartesian Plane 149
Homework 3.1 155
Homework 3.1 Answers 158
2. Graphing Linear Equations 159
2.1. Solution Sets 159
2.2. Graphing Solution Sets 160
2.3. Graphing Lines 162
Homework 3.2 165
Homework 3.2 Answers 166
3. Intercepts 168
3.1. x and y-intercepts 168
3.2. Finding Intercepts 168
Homework 3.3 171
Homework 3.3 Answers 172
4. Slope 174
4.1. Slope of a Line 174
4.2. Using Slope and Intercept to Plot 175
Homework 3.4 177
Homework 3.4 Answers 180
5. Slope-Intercept Form 182
5.1. Slope-Intercept Form 182
5.2. Parallel Lines 186
5.3. Perpendicular Lines 188
Homework 3.5 191
Homework 3.5 Answers 194
6. Point-Slope Form 195
7
CHAPTER 0. CONTENTS
6.1. A Point and a Slope Determine a Line 195
6.2. Two Points Determine a Line 196
6.3. Using Point and Slope to Plot 198
Homework 3.6 199
Homework 3.6 Answers 200
7. Function Notation 202
7.1. Notation 202
7.2. Applications of Linear Functions 204
Homework 3.7 208
Homework 3.7 Answers 209
Practice Test 3 210
Practice Test 3 Answers 211
Chapter 4. Linear Systems 213
1. Graphing Systems of Equations 213
1.1. Solution Sets 213
1.2. Solving Systems by Graphing 215
Homework 4.1 218
Homework 4.1 Answers 219
2. Substitution 221
2.1. Solving Systems using Substitution 221
Homework 4.2 224
Homework 4.2 Answers 225
3. Elimination 226
3.1. Equivalent Systems 226
3.2. Eliminating Variable with Addition 226
Homework 4.3 230
Homework 4.3 Answers 231
4. Applications of Systems 232
4.1. Constant Rate Problems 232
4.2. Mixing Problems 233
Homework 4.4 236
Homework 4.4 Answers 237
5. Multivariate Linear Inequalities 238
5.1. Solution Sets 238
5.2. Solving Inequalities 238
Homework 4.5 244
Homework 4.5 Answers 245
Practice Test 4 247
Practice Test 4 Answers 248
Chapter 5. Polynomial Expressions 249
1. Integer Exponent 249
1.1. Definition 249
1.2. Products and Quotients of Exponents 250
1.3. Exponents of Sums 254
8
CHAPTER 0. CONTENTS
Homework 5.1 255
Homework 5.1 Answers 256
2. Properties of Exponent 257
2.1. Exponent of Exponent 257
2.2. Distributivity 258
Homework 5.2 263
Homework 5.2 Answers 264
3. Introduction to Polynomials 265
3.1. Definitions 265
3.2. Combining Similar Monomial Terms 267
3.3. Evaluating Polynomials in Applications 269
Homework 5.3 271
Homework 5.3 Answers 273
4. Sums of Polynomials 274
Homework 5.4 276
Homework 5.4 Answers 278
5. Products of Polynomials 279
5.1. Monomial Times Monomial 279
5.2. Monomial Times Polynomial 279
5.3. Polynomial Times Polynomial 280
Homework 5.5 282
Homework 5.5 Answers 283
6. Special Products 284
6.1. Difference of Squares 284
6.2. Square of a Binomial 286
Homework 5.6 288
Homework 5.6 Answers 289
7. Quotients of Polynomials 290
7.1. Monomial Divisor 290
7.2. Polynomial Divisor 292
Homework 5.7 298
Homework 5.7 Answers 299
8. Negative Exponent 300
8.1. Negative Integer Exponent 300
8.2. Scientific Notation 304
Homework 5.8 307
Homework 5.8 Answers 309
Practice Test 5 310
Practice Test 5 Answers 311
Chapter 6. Factoring Polynomials 313
1. Greatest Common Factor 313
1.1. Factoring Polynomials 313
1.2. Greatest Common Factor 314
1.3. Monomial GCF 316
9
CHAPTER 0. CONTENTS
Homework 6.1 319
Homework 6.1 Answers 320
2. Grouping 321
2.1. Factoring by Grouping 321
Homework 6.2 323
Homework 6.2 Answers 324
3. Factoring x 2 + bx + c 325
3.1. Guessing the Coefficients 325
3.2. Variations of the Pattern 327
3.3. Basic Factoring Strategy 328
Homework 6.3 330
Homework 6.3 Answers 331
4. Factoring ax 2 + bx + c 332
4.1. Guessing the Coefficients 332
4.2. Factoring Trinomials 333
Homework 6.4 336
Homework 6.4 Answers 337
5. Factoring Special Products 338
5.1. Difference of Squares 338
5.2. Square of a Binomial 340
Homework 6.5 341
Homework 6.5 Answers 342
6. General Factoring Strategy 343
6.1. Irreducible Polynomials 343
6.2. Factoring Strategy 343
Homework 6.6 346
Homework 6.6 Answers 347
7. Solving Equations by Factoring 348
7.1. Zero Product Property 348
7.2. Equation Solving Strategy 349
Homework 6.7 352
Homework 6.7 Answers 353
8. Factoring Applications 354
8.1. Applications to Areas 354
Homework 6.8 358
Homework 6.8 Answers 359
Practice Test 6 360
Practice Test 6 Answers 361
Chapter 7. Rational Expressions 363
1. Simplifying Rational Expressions 363
1.1. Definition 363
1.2. Variable Restrictions 364
1.3. Canceling Common Polynomial Factors 366
Homework 7.1 368
10
CHAPTER 0. CONTENTS
Homework 7.1 Answers 370
2. Products of Rational Expressions 371
2.1. Multiplying Rational Expressions 371
2.2. Dividing Rational Expressions 372
Homework 7.2 375
Homework 7.2 Answers 376
3. LCD for Rational Expressions 377
3.1. LCM for Polynomials 377
3.2. LCD for Rational Expressions 378
Homework 7.3 381
Homework 7.3 Answers 382
4. Sums of Rational Expressions 383
4.1. Sums with a Common Denominator 383
4.2. General Sums 384
Homework 7.4 388
Homework 7.4 Answers 390
5. Complex Fractions 391
5.1. Definition 391
5.2. Simplifying Complex Fractions 391
Homework 7.5 396
Homework 7.5 Answers 398
6. Equations with Rational Expressions 399
6.1. Extraneous Solutions 399
Homework 7.6 404
Homework 7.6 Answers 406
Practice Test 7 407
Practice Test 7 Answers 408
Chapter 8. Radicals 409
1. Introduction to Radicals 409
1.1. Square Root and Radicals 409
1.2. Principal nth Root 411
Homework 8.1 414
Homework 8.1 Answers 415
2. Products with Square Roots 416
2.1. Distributivity Over Multiplication 416
Homework 8.2 420
Homework 8.2 Answers 421
3. Quotients with Square Roots 422
3.1. Distributivity Over Division 422
Homework 8.3 424
Homework 8.3 Answers 426
4. Sums with Square Roots 427
4.1. Radical Like Terms 427
4.2. Rationalizing Sums 429
11
CHAPTER 0. CONTENTS
Homework 8.4 432
Homework 8.4 Answers 433
5. Equations with Square Roots 434
5.1. Squaring Both Sides 434
Homework 8.5 439
Homework 8.5 Answers 440
6. Applications to Right Triangles 441
6.1. Distance Formula 441
Homework 8.6 443
Homework 8.6 Answers 444
7. Rational Exponent 445
7.1. Rational Exponent Definition 445
Homework 8.7 449
Homework 8.7 Answers 451
Practice Test 8 452
Practice Test 8 Answers 453
Chapter 9. Quadratic Equations 455
1. Complex Numbers 455
1.1. Imaginary Unit 455
1.2. The Set of Complex Numbers 456
1.3. Adding and Multiplying Complex Numbers 457
Homework 9.1 459
Homework 9.1 Answers 460
2. Square Root Property 461
2.1. Square Root Property 461
2.2. Solving Equations (nx + k)2 = r 463
2.3. Complex Solutions 464
Homework 9.2 466
Homework 9.2 Answers 467
3. Completing the Square 468
3.1. Completing the Square for x 2 + bx 468
3.2. Solving Quadratic Equations by Completing the Square 470
Homework 9.3 473
Homework 9.3 Answers 474
4. Quadratic Formula 475
4.2. Complex Solutions 477
Homework 9.4 478
Homework 9.4 Answers 479
Practice Final 480
Practice Final Answers 482
Appendix A. To Instructors 483
1. Lecture Notes 483
1.1. Concepts 483
1.2. Linear Equations 485
12
CHAPTER 0. CONTENTS
1.3. Graphing 489
1.4. Systems 490
1.5. Polynomials 492
1.6. Factoring 493
1.7. Rational Expressions 494
1.8. Radicals 495
1.9. Quadratics 497
Appendix B. To Developers 499
1. Contributing 499
1.1. Submitting Contributions 499
1.2. Document Hierarchy 499
2. Document Formatting Guidelines 500
2.1. Mathematical Notions 500
2.2. Sets of Equations and Expressions 501
2.3. Frames 502
2.4. Homework and Answers 504
Homework B.2 505
Homework B.2 Answers 506
3. To Do 507
3.1. To Do Items 507
4. Working with the Source 507
4.1. Getting the Source 507
4.2. Building the Source with LATEX 507
4.3. Branches 508
5. Acknowledgments 508
Index 509
13
CHAPTER 1
Concepts
1. Expressions, Relations, and Substitution
1.1. Expressions versus Relations.
DEFINITION 1.1.1. An algebraic expression or simply expression is a symbolic description
of a single number. We call this number the value of the expression.
DEFINITION 1.1.2. Expressions are equivalent if they describe the same number, even
though they may look very different. For example, expressions 2 · 3 and 5 + 1 are equiv-
alent, because they both evaluate to 6.
BASIC EXAMPLE 1.1.1. Here are some expressions:
expression what it stands for
17 stands for the number 17
10 + 7 is a different expression which also stands for the number 17
(3 + 5) ÷ 2 stands for the number 4
x is a variable expression, and it stands for an unknown number
2x + y stands for a single unknown number; if we knew what expressions x
and y stand for, we would be able to compute the value of 2x + y
DEFINITION 1.1.3. An equation is a statement about the equality of two expressions:
A= B
An equation has an expression on the left, another expression on the right, and an equality
sign in the middle. An equation is true if the value of the expression on the left is the same
as the value of the expression on the right, and false otherwise.
It is important to distinguish between equivalent expressions and equivalent equations. When we
say that equations are equivalent, it means that they have identical solutions (that is, the same
numbers make them true when we substitute them for the variables), even though equations
may look very different. The formal definition of equivalent equations is given later in the text.
15
1. EXPRESSIONS, RELATIONS, AND SUBSTITUTION CHAPTER 1. CONCEPTS
BASIC EXAMPLE 1.1.2. Here are some equations:
equation truth value
4=4 true
17 = 10 + 7 true
3=1+1 false
x +1=7 true if the value of x is 6, false otherwise
x =6 this one is equivalent to the one above: it’s also true if the value of x is
6 and false otherwise
It is crucial to distinguish between expressions and equations. While intimately related to each
other, they are completely different types of objects. For example, we can add the same number
to both sides of any equation and obtain an equivalent equation, which is as true or false as the
original. But if we add an arbitrary non-zero number to an expression, its value will change
and we will end up with a different, unrelated expression.
DEFINITION 1.1.4. Informally speaking, every equation corresponds to a more general
object we call a binary relation, or just relation, which is a true-or-false statement about
two numbers. There are many different types of relations besides equations, and they all
differ from expressions in one key way: expressions denote numbers, whereas relations
denote truth or falsehood.
BASIC EXAMPLE 1.1.3. Besides equations, inequalities are some of the more useful relations
for us:
relation truth value
4<4 false, because 4 is not less than 4
17 > 5 · 2 true, because 17 is greater than 10
x>y true if the value of x is greater than the value of y, false otherwise
16
CHAPTER 1. CONCEPTS 1. EXPRESSIONS, RELATIONS, AND SUBSTITUTION
1.2. Algebraic Substitution. Algebraic substitution is the number one tool in our toolbox.
It can be described as follows: any time and for any reason, we can replace any expression
anywhere by any other expression which has the same value.
BASIC EXAMPLE 1.2.1. We know that expressions 6 · 2, 7 + 5, and 12 all stand for the same
number, so we can always swap one for the other in any expression, even if it is inside an
equation. All of the following are perfectly equivalent statements:
12x = y + 12
(6 · 2)x = y + 12
(6 · 2)x = y + (7 + 5)
As we substitute, it is never a bad idea to guard the value of the inserted expression with
parentheses. With practice, we learn to omit the substitution parentheses in simple cases,
like when we substitute a single positive integer, but more complicated expressions must
be substituted with parentheses so that the correct order of operations is preserved.
THEOREM 1.2.1 (Algebraic Substitution). If the values of expressions A and B are the same,
then we can substitute A by (B) in any expression, and obtain an equivalent expression.
EXAMPLE 1.2.1. Suppose that the value of x is −4. Substitute −4 for x in the expression
(3x − 5) + x.
ANSWER: (3(−4) − 5) + (−4)
EXAMPLE 1.2.2. Suppose that the expression A has the same value as the expression B +5.
Substitute B + 5 for A in the equation
13A = 10 − A
SOLUTION: This example demonstrates why the parentheses are indispensable in some
situations. Without the parentheses, we could make a mistake by writing something like
13B + 5 = 10 − B + 5, which would give us wrong values on both sides.
ANSWER: 13(B + 5) = 10 − (B + 5)
17
1. EXPRESSIONS, RELATIONS, AND SUBSTITUTION CHAPTER 1. CONCEPTS
1.3. Common Types of Expressions. There are four basic types of expressions which are
used heavily in this text: sums, products, terms, and factors.
DEFINITION 1.3.1. A sum is an algebraic expression consisting of expressions being added,
and those expressions are called terms. Because subtraction can be thought of as adding
a negative number, the terms of the sum can be separated by either pluses or minuses.
BASIC EXAMPLE 1.3.1. Here is a sum with 3 terms:
A+ B + C
Here is a sum with 4 terms, and its 3rd term −4 is negative:
a + x − 4 + 7y
Here is a sum with 2 terms (everything inside the parentheses is a single term):
(x + y − 3z) + 14
Incidentally, the first term in the sum above is itself a sum with 3 terms:
x + y − 3z
EXAMPLE 1.3.1. Identify all the terms in the expression
5 y − 3x 2
ANSWER: 2 terms: 5 y, −3x 2
EXAMPLE 1.3.2. Identify all the terms in the expression
3(x + 1) + 5 + x y
SOLUTION: When identifying terms of an expression, we do not look inside the paren-
theses, so 3(x + 1) is a single term, even though another sum with two terms is locked up
inside.
ANSWER: 3 terms: 3(x + 1), 5, x y
18
CHAPTER 1. CONCEPTS 1. EXPRESSIONS, RELATIONS, AND SUBSTITUTION
DEFINITION 1.3.2. A product is an algebraic expression consisting of expressions being
multiplied, and those expressions are called factors.
BASIC EXAMPLE 1.3.2. Here is a product with 2 factors:
4x
Here is a product with 3 factors, and its third factor is itself a sum with 2 terms:
AB(X + 7)
In general, sums and products can go into each other like the Russian dolls:
ABC − 4(A + B)(X − Y) + AB(X + 7)
What we have above is a sum with 3 terms, and the second term −4(A + B)(X −Y ) happens
to be a product with three factors, and its second factor A + B happens to be a sum with
two terms.
2
EXAMPLE 1.3.3. Identify all the factors in the expression: xy
7
SOLUTION: Factors are easier to see if we make the multiplication visible:
2
·x·y
7
We consider the rational number 2/7 as a single factor.
ANSWER: 3 factors: 2/7, x, y
1
EXAMPLE 1.3.4. Identify all the factors in the expression: −10x(x + 1)
z
SOLUTION: Unlike the addition operation, multiplication is often invisible, so it may be
helpful to rewrite the expression with the multiplication dots shown:
1
(−10) · x · (x + 1) ·
z
1
ANSWER: 4 factors: −10, x, x + 1,
z
19
1. EXPRESSIONS, RELATIONS, AND SUBSTITUTION CHAPTER 1. CONCEPTS
DEFINITION 1.3.3. A numerical coefficient or just coefficient is a multiplicative factor in a
product. It usually looks like a numerical constant, but in complicated cases may be any
expression. In the latter case, the variables appearing in the coefficient are often called
parameters, and must be clearly distinguished from the other variables.
Traditionally, the coefficient is written on the left side of the product.
BASIC EXAMPLE 1.3.3.
product coefficient
3x y 2 3
−6a −6
3 3 3
x
4 4
a4 b 1
−x 2 −1
20
CHAPTER 1. CONCEPTS 1. EXPRESSIONS, RELATIONS, AND SUBSTITUTION
Homework 1.1.
Determine the type and the meaning of the 14. Substitute y − 2 for h in the equation
following formal statements. For equations 2h
and inequalities, determine whether they = 5h
3
are true or false. For expressions, deter-
mine their numerical value.
1. 7 = 6 − 1 Identify all the terms in the given expres-
sion.
12 6
2. = 15. 5x − 6
4 2
3. (14 + 2) − 9 16. 1 + x + x y
4. (5 · 3) + 4 17. −(x + 7) − 1 + x
5. 14 < −1 18. (5x − 1) − (4 + x)
6. 5 · 3 = 9 + 6 19. 14x y 2
3+5 20. −(3 + 2x) − A
7.
4
21. (1 + A) − (1 − 2A) + (4A + 3)
8. 5 + 6 > 10
1
22. 3(a + x) − 5(x − 3)(2 y) +
7
Substitute the given expression for the vari-
able. Use the parentheses and do not sim- Identify all the factors in the given expres-
plify the result. sion.
9. Substitute 100 for x in the expression 23. −6x y 5
2x + 1
24. 14x 2 y 2 A6
10. Substitute 4 for x in the expression
25. 2x(u + 1)(u2 − 1)
5 − 4x
26. (A + B)2 (A − B)
11. Substitute −7 for y in the expression
y + y2 27. a2 − b
1 1
12. Substitute −12 for y in the expression 28. (a + 1)
4 x
y 2 (x + y)
1
29. (ab)c
13. Substitute x + 1 for k in the equation 2
k(k − 1) = 4k 30. 11(x y 2 )(x + y)
21
1. EXPRESSIONS, RELATIONS, AND SUBSTITUTION CHAPTER 1. CONCEPTS
Homework 1.1 Answers.
1. A false equation 17. 3 terms: −(x + 7) , −1, x
3. An expression with the value 7 19. 1 term: 14x y 2
5. A false inequality
21. 3 terms: 1 + A, −(1 − 2A), 4A + 3
7. An expression with the value 2
23. 3 factors: −6, x, y 5
9. 2(100) + 1
25. 4 factors: 2, x, u + 1, (u2 − 1)
11. (−7) + (−7)2
27. 1 factor: a2 − b
13. (x + 1)((x + 1) − 1) = 4(x + 1)
1
15. 2 terms: 5x, −6 29. 3 factors: , ab, c
2
22
CHAPTER 1. CONCEPTS 2. REAL NUMBER AXIOMS
2. Real Number Axioms
2.1. Axioms. Axioms are the properties of numbers which are accepted without proof. In
a sense, the axioms serve as a kind of definition for numbers, but unlike traditional definitions,
which define new concepts in terms of old concepts, the axioms define the numbers and the
arithmetic operations by telling us what they can and cannot do. The true nature of a number
is never explained explicitly. We simply agree to call things numbers as long as they obey the
axioms.
The first nine axioms appear almost too simple, and indeed they are. Almost anyone familiar
with basic arithmetic knows these facts by heart.
AXIOM 2.1.1 (Closure of Arithmetic Operations). Adding any two real numbers results in
a real number. Multiplying any two real numbers results in a real number.
BASIC EXAMPLE 2.1.1. Closure guarantees that addition and multiplication are defined for
every pair of real numbers, and the result is a real number. When we substitute 5 for 2+3,
or 6 for 2 · 3, we can say that the substitution is justified by the closure.
AXIOM 2.1.2 (Commutativity of Addition). For any two real numbers X and Y ,
X +Y =Y +X
AXIOM 2.1.3 (Associativity of Addition). For any two real numbers X and Y ,
X + (Y + Z) = (X + Y ) + Z
BASIC EXAMPLE 2.1.2. When considered together, commutativity and associativity of ad-
dition tell us that we can rearrange the terms of any sum, and perform the additions in
any order we consider convenient. Because of that, when we work with expressions like
a+b+c+d
we are free to substitute them with equivalent expressions such as
d+c+b+a
or
(a + c) + (b + d)
AXIOM 2.1.4 (Commutativity of Multiplication). For any two real numbers X and Y ,
XY = YX
23
2. REAL NUMBER AXIOMS CHAPTER 1. CONCEPTS
AXIOM 2.1.5 (Associativity of Multiplication). For any two real numbers X and Y ,
X (Y Z) = (X Y )Z
BASIC EXAMPLE 2.1.3. Just like with addition, commutativity and associativity of multi-
plication allow us to compute values of products by multiplying factors in any order, so a
product with 4 factors such as
3 · x · y · (a − 1)
is equivalent to any other product involving the same factors, regardless of the order of
operations, for example
(a − 1) · (3 · y · x)
or
(3 · y) · (x · (a − 1))
AXIOM 2.1.6 (Additive Identity). There exists a real number 0, called zero, such that for
all real numbers X ,
X +0= X
AXIOM 2.1.7 (Multiplicative Identity). There exists a real number 1, called one or unit,
such that for all real numbers X ,
X ·1= X
While it may seem silly that we have to explicitly demand the existence of numbers such as
zero and one, this is the natural way to proceed from the axioms. Note that other axioms apply
to all numbers, but do not require the existence of any specific number. But once we have the
identities and the closure of operations, we can immediately prove that there exists a number
equal to 1 + 1, and call it 2, and then also a number 2 + 1, which we call 3, and so on.
AXIOM 2.1.8 (Additive Inverse). For each real number X , there exists a real number we call
the additive inverse of X , or the opposite of X , written as −X , with the following property:
X + (−X ) = 0
BASIC EXAMPLE 2.1.4. The opposite of 5 is −5 because 5 + (−5) = 0.
At the same time, the opposite of −5 is 5 because (−5) + 5 = 0, so we can say 5 and −5
are opposites of each other. In general, it can be proven that every number has a unique
opposite.
The opposite of zero is zero itself, because 0 + 0 = 0.
24
CHAPTER 1. CONCEPTS 2. REAL NUMBER AXIOMS
AXIOM 2.1.9 (Multiplicative Inverse). For each non-zero real number X , there exists a real
1
number we call the multiplicative inverse of X , or the reciprocal of X , written as , 1/X ,
X
or X −1 , with the following property:
1
X· =1
X
Note that zero is very special. It is the only real number which is its own opposite, and the
only real number with no reciprocal: there is no real number that would give us one when
multiplied by zero.
BASIC EXAMPLE 2.1.5. The reciprocal of 2 is 1/2, which can also be written as 0.5. To
check, simply multiply them, and make sure to get 1 as the result. By the same token, the
reciprocal of 1/2 is 2.
In practice, obtaining the reciprocal amounts to writing the number as a fraction, and
turning that fraction “upside-down”. Recall that we can multiply fractions by multiplying
numerators and multiplying denominators:
A X AX
· =
B Y BY
Recall also that we can write numbers like 2 as 2/1. It should be easy to see now that the
reciprocal of any fraction A/B is the fraction B/A, as long as neither A nor B is zero. Here
are some more reciprocal pairs:
2 1 29
X 1 −1 4 − −
3 17 31
1 1 3 31
1 −1 −17 −
X 4 2 29
The last axiom deserves special attention. It is far less intuitive, and it takes a lot of practice to
apply it correctly, or even to notice that it can be applied.
AXIOM 2.1.10 (Distributivity). For any three real numbers X , Y , and Z,
X (Y + X ) = X Y + X Z
25
2. REAL NUMBER AXIOMS CHAPTER 1. CONCEPTS
EXAMPLE 2.1.1. Use the distributivity to rewrite the expression without parentheses:
4(x + 7)
SOLUTION:
4(x + 7) = 4x + 4 · 7
= 4x + 28
ANSWER: 4x + 28
EXAMPLE 2.1.2. Use the distributivity to rewrite the expression without parentheses:
2(x + 2 y + 3z)
SOLUTION: The distributive property can be shown to work for sums with more than two
terms, so that each term gets multiplied by the number from the outside of the parentheses.
2(x + 2 y + 3z) = 2 · x + 2 · 2 y + 2 · 3z
= 2x + 4 y + 6z
ANSWER: 2x + 4 y + 6z
EXAMPLE 2.1.3. Use the distributivity to rewrite the expression as a product with two
factors:
4x + 4
SOLUTION: The greatest common factor for these two terms is 4, so a single application
of the distributive property gives a product of two factors: 4 and x + 1.
4x + 4 = 4x + 4 · 1 multiplicative identity
= 4(x + 1) distributivity
ANSWER: 4(x + 1)
26
CHAPTER 1. CONCEPTS 2. REAL NUMBER AXIOMS
EXAMPLE 2.1.4. Use the distributivity to rewrite the expression as a product with two
factors:
ab + 3b
SOLUTION: The greatest common factor for these two terms is b, so a single application
of the distributive property gives a product of two factors: a + 3 and b.
ab + 3b = (a + 3)b
ANSWER: (a + 3)b
2.2. Applying Axioms. Unlike the easy axioms, the distributivity of multiplication over
addition will play a major role in our study, and we will refer to it explicitly throughout the
text in order to justify algebraic substitutions. The rest of the axioms will still be in heavy use,
but they will be applied under the hood, so to speak.
Before we leave this section, let us have some fun and see how the familiar algebraic concepts
can be broken down into steps so tiny, that each step is justified by an axiom.
EXAMPLE 2.2.1. Simplify the expression by appealing to the axioms:
(x + 5) + (−x)
SOLUTION:
equivalent expressions axiom used for substitution
(x + 5) + (−x)
(5 + x) + (−x) By commutativity of addition, x + 5 = 5 + x, so we were able
to substitute (5 + x) for (x + 5) without a change in value.
5 + (x + (−x)) The whole expression was substituted by associativity of addi-
tion.
5+0 x and −x are opposites and they must add up to zero, so we
were able to substitute zero for x + (−x).
5 zero is the additive identity, so we were able to substitute 5 for
5 + 0.
ANSWER: 5
27
2. REAL NUMBER AXIOMS CHAPTER 1. CONCEPTS
EXAMPLE 2.2.2. Rewrite the expression without parentheses and then simplify by appeal-
ing to the axioms:
3(2 + x) + (−6)
SOLUTION: This is just like the previous example, but we will try to justify the steps in a
concise manner, rather than describing them in with sentences. Since we are going strictly
one substitution at a time, the substitutions we make should be obvious.
equivalent expressions axiom used for substitution
3(2 + x) + (−6)
(3 · 2 + 3x) + (−6) distributivity
(6 + 3x) + (−6) closure
(3x + 6) + (−6) commutativity of +
3x + (6 + (−6)) associativity of +
3x + (0) additive inverse
3x additive identity
ANSWER: 3x
2.3. Inverses, Subtraction, Division.
THEOREM 2.3.1. Each number has a unique opposite, and zero is the only number which
is its own opposite. An opposite of an opposite of a number is the number itself. Formally,
for any real number x,
−(−x) = x
PROOF. We appeal directly to the axioms:
−(−x) = −(−x) + 0 identity of +
= −(−x) + (−x + x) additive inverse
= (−(−x) + (−x)) + x associativity of +
= 0+ x additive inverse
= x identity of +
28
CHAPTER 1. CONCEPTS 2. REAL NUMBER AXIOMS
THEOREM 2.3.2. Each non-zero number has a unique reciprocal. Only 1 and −1 are equal
to their respective reciprocals:
1 1
1= −1=
1 −1
A reciprocal of a reciprocal of a number is the number itself. Formally, for any non-zero
real number x,
1
=x
(1/x)
PROOF. The proof is very similar to the previous one:
1/(1/x) = 1/(1/x) · 1 identity of ·
= 1/(1/x) · (1/x · x) multiplicative inverse
= (1/(1/x) · 1/x) · x associativity of ·
= 1· x multiplicative inverse
= x identity of ·
Axioms do not mention subtraction or division, but we can define them in terms of addition,
multiplication, and inverses.
DEFINITION 2.3.1. Subtraction is an operation defined in terms of the addition and the
opposite. For any two real numbers X and Y ,
X − Y = X + (−Y )
In other words, subtracting a number amounts to adding its opposite.
DEFINITION 2.3.2. Division is an operation defined in terms of the multiplication and the
reciprocal. For any two real numbers X and Y ,
1
X ÷Y =X ·
Y
In other words, dividing by a number amounts to multiplying by its reciprocal.
X
Other notations commonly used for X ÷ Y are and X /Y .
Y
Now that we finally have all of the arithmetic operations, let us go back for a moment to the
definitions of sum, product, term, and factor. Since subtraction is essentially addition, we are
fully justified in calling the following expression a sum with 3 terms:
4x − 5 − Y
29
2. REAL NUMBER AXIOMS CHAPTER 1. CONCEPTS
because it is equivalent to a sum of 4x, the opposite of 5, and the opposite of Y
4x + (−5) + (−Y )
The same kind of reasoning applies to the division and the products. We can get away with
calling this a product with 5 factors:
1 1
x · y ·z· ·
a b+c
even though we usually write down an equivalent fraction:
x yz
a(b + c)
x+y
EXAMPLE 2.3.1 (Fraction Notation). Identify all the factors in the expression
x−y
SOLUTION: The fraction notation is a division with a gotcha: unlike the regular division ÷
which behaves similarly to the multiplication · in the order of operations, the fraction bar
waits for the values of its numerator and denominator, making it the very last operation to
be performed within the fraction. One way to make sense of this is by enclosing numerator
and denominator in parentheses, so that the order of operations becomes unambiguous:
x+y (x + y)
=
x−y (x − y)
It may be helpful to remember that every fraction comes equipped with these parentheses,
but most of the time they are invisible, so to speak. Now we can see that this expression,
when thought of as a product, has 2 factors: (x + y) and 1/(x − y), which is the reciprocal
of (x − y).
1
ANSWER: 2 factors: (x + y),
(x − y)
2.4. Conclusion. Two concepts stand out in this section, and will be alluded to throughout
the text. The distributive property is both indispensable and hard to grasp, and it is the only
axiom we will keep calling by name. The way we defined subtraction and especially division
will help us not to get lost while working with fractions and other complicated expressions.
The rest of the axioms are about to fade into shadows, but there is one last thing we want
to stress, as it is the kind of insight that is simple yet profound at the same time. Not only
the axioms can be used to break up everything we do into tiny, elementary, and fully justified
steps; we also must be able to justify everything we do with axioms alone. Every single algebraic
argument in this text is justifiable from the axioms and the logical principles, and nothing else
would be recognized as valid. The rigorous study of the properties of numbers which arise
directly from the axioms is called abstract algebra, and it is a fascinating, albeit a very difficult
30
CHAPTER 1. CONCEPTS 2. REAL NUMBER AXIOMS
subject, consisting almost entirely of formal proofs, the simplest ones being similar to the step-
by-step simplification examples above.
31
2. REAL NUMBER AXIOMS CHAPTER 1. CONCEPTS
Homework 1.2.
Use the distributivity to rewrite the expres- 13. 6x + 12 + 18 y
sion without parentheses:
14. 15a + 45b + 30
1. 3(x + 16)
15. 28p + 7t + 7
2. 4(x + 6)
16. 24b + b + ab
3. 5(2 + 2a)
4. 8(1 + 2 y)
Simplify the expression by appealing to the
5. a(10 + x) axioms:
6. b(x + y) 17. 2(x + 5) + 3
7. 5(6x + 9 + 4g) 18. 5 + 3(4 + a)
8. 6(2t + 3 + 4s) 19. (6 + x) + (−6)
20. g + (2 + (−g))
Use the distributivity to rewrite the expres-
1
sion as a product with two factors:
21. x 7 + + 8, given that x 6= 0
x
9. 3a + 3b
1
22. 12 + a 4 + , given that a 6= 0
10. 6 y + 6z a
11. 33x + 11 23. 5x + (6 + 3x)
12. 21a + 7 24. (2a + 9) + 8a
32
CHAPTER 1. CONCEPTS 2. REAL NUMBER AXIOMS
Homework 1.2 Answers.
1. 3x + 48 19. expression axiom
(6 + x) + (−6)
3. 10 + 10a (x + 6) + (−6) commutativity of +
x + (6 + (−6)) associativity of +
5. 10a + a x
x + (0) additive inverse
x additive identity
7. 30x + 45 + 20g
21. expression axiom
9. 3(a + b)
1
x 7+ +8
x
11. 11(3x + 1) 1
x ·7+ x · +8 distributivity
x
13. 6(x + 2 + 3 y) (7x + (1)) + 8 multiplicative inverse
7x + (1 + 8) associativity of +
15. 7(4p + t + 1) 7x + 9 closure
17. expression axiom 23. expression axiom
2(x + 5) 5x + (6 + 3x)
(2 · x + 2 · 5) + 3 distributivity 5x + (3x + 6) commutativity of +
(2x + 10) + 3 closure (5x + 3x) + 6) associativity of +
2x + (10 + 3) associativity of + (5 + 3)x + 6 distributivity
2x + 13 closure 8x + 6 closure
33
3. THE NUMBER LINE AND SETS OF NUMBERS CHAPTER 1. CONCEPTS
3. The Number Line and Sets of Numbers
3.1. Roster.
DEFINITION 3.1.1. A set of numbers is any collection of numbers. Any number that belongs
to the set is called an element or a member of that set.
DEFINITION 3.1.2 (Roster Notation). A set can be written down in the roster notation,
which is a comma-separated list of its elements inside the curly brackets. A set with
elements 1, 2, and 3 can be written as {1, 2, 3}. The order in which elements are listed is
not important, so {1, 2, 3} is the same set as {3, 1, 2}. A particular number either belongs
to a set or it does not, just like a particular cat can be either alive or dead, so writing {7, 7}
is just as redundant as saying “this cat is alive and alive”.
BASIC EXAMPLE 3.1.1. Very large and infinite sets can be stated in roster notation using
the ellipsis. . . For example, the infinite set of positive odd numbers can be stated as
{1, 3, 5, 7, 9, 11, 13, . . .}
It is important to list enough elements for the pattern to become clear to the reader, and
so the roster notation is too ambiguous for many interesting sets, like this one:
{6, 28, 496, 8128, . . .}
What is the next number in this sequence? If you cannot tell, then perhaps the roster
notation is not the best way to describe this set.
DEFINITION 3.1.3. A set with zero elements exists, and it is called the empty set. The
traditional notation for the empty set is ∅, which is why it is probably not a good idea to
strike through your zeroes: ;. Roster notation {} can also be used.
3.2. Integers.
DEFINITION 3.2.1. The set of positive integers is
{1, 2, 3, 4, 5, . . .}
DEFINITION 3.2.2. The set of non-negative integers is
{0, 1, 2, 3, 4, 5, . . .}
which is exactly the same as the set of positive integers with one additional element: zero,
which is neither positive nor negative. The standard notation for this set is N, although
some texts will use the same notation for positive integers.
Many texts use “whole numbers” and “natural numbers” to refer to either positive or non-
negative integers, but we will not use either term in order to avoid confusion.
34
CHAPTER 1. CONCEPTS 3. THE NUMBER LINE AND SETS OF NUMBERS
DEFINITION 3.2.3. The set of integers is
{0, 1, −1, 2, −2, 3, −3, 4, −4, 5, −5, . . .}
which can also be written as
{. . . , −5, −4, −3, −2, −1, 0, 1, 2, 3, 4, 5, . . .}
One great way to visualize the integers is the number line, which is a graphical representation
of the collection of real numbers:
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
The number line extends infinitely far to the right and to the left, but we can only draw a finite
segment of it. Every point on the number line corresponds to a real number, and the picture
above highlights the integers.
The number line provides a convenient way to explain the addition of integers. Because addi-
tion is associative and commutative, the order in which numbers are added does not affect the
result. To compute a sum, we start at zero on the number line. For each positive term of the
sum we go that many units to the right on the number line, and for each negative term of the
sum we go that many units to the left. The point where we end up is the value of the sum.
EXAMPLE 3.2.1. Find 5 + (−7)
SOLUTION: 5 units to the right from zero take us to 5, and 7 units left take us to −2, so
the value is −2.
ANSWER: −2
EXAMPLE 3.2.2. Find −10 − (−7)
SOLUTION: By definition of subtraction, this expression is equivalent to
−10 + (−(−7))
But the opposite of an opposite is the number itself: −(−7) = 7, so this is equivalent to
−10 + 7
Starting at zero, we move 10 units to the left to reach −10, and then 7 units to the right,
and end up at −3.
ANSWER: −3
35
3. THE NUMBER LINE AND SETS OF NUMBERS CHAPTER 1. CONCEPTS
EXAMPLE 3.2.3. Find −3 + 2 − 4 − 5
SOLUTION: By definition of subtraction, this expression is equivalent to
(−3) + 2 + (−4) + (−5)
Starting from 0 on the number line,
• 3 units to the left take us to −3
• 2 units to the right take us to −1
• 4 units to the left take us to −5
• 5 units to the left take us to −10
ANSWER: −10
3.3. Set Builder.
DEFINITION 3.3.1 (Set-builder Notation). A set can be defined by the set-builder notation,
which looks like this:
expression relation
Recall that relation is basically a statement which can be true or false. Defined as above,
the set will contain values of all expressions for which the relation is true.
EXAMPLE 3.3.1. Write the set of even integers in set-builder notation.
SOLUTION: In roster notation, the set of even integers looks like this:
{. . . , −4, −2, 0, 2, 4, 6, 8, . . .}
In set-builder, we can write
x x is even
which reads “the set of numbers x such that x is even”. This may look like an arcane way
to rewrite a simple English statement, but the real usefulness of set-builder is made clear
when we start using nontrivial expressions. For example, and with a lot more rigor, we
can express the same set as
2k k is an integer
This reads “the set of all values of the expression 2k, where k is an integer”. So this set
has 2, since 2 = 2 · 1; it has 4 since 4 = 2 · 2, and so on. Defined this way, the set has every
even integer, and none of the odd ones.
ANSWER: 2k k is an integer
36
CHAPTER 1. CONCEPTS 3. THE NUMBER LINE AND SETS OF NUMBERS
EXAMPLE 3.3.2. List a few members of the set 3x x is a negative integer
SOLUTION: Plug in a few negative integers into the expression to the left of the bar | to
get elements such as 3(−1) = −3, 3(−2) = −6 and so on. State the answer in the roster
notation.
ANSWER: {−3, −6, −9, −12, −15, . . .}
1
§ ª
EXAMPLE 3.3.3. List a few members of the set k is a positive integer
k
SOLUTION: Plug in a few positive integers into the expression to the left of the | to get
elements such as 1/1 = 1, 1/2, 1/3 and so on. State the answer in the roster notation.
1 1 1 1
§ ª
ANSWER: 1, , , , , ...
2 3 4 5
3.4. Rationals.
DEFINITION 3.4.1. The set of rational numbers is the collection of all numbers which can
be represented by a fraction a/b where a is an integer and b is a positive integer:
1 −1 17
§ ª
, , , ...
2 3 13
Entries above are listed in no particular order. It is possible, but not easy to come up with
a pattern that would allow to list all rational numbers. The set-builder way of defining
rationals completely removes the ambiguity:
na o
a is an integer and b is a positive integer
b
The standard notation for the set of all rational numbers is Q.
A number is rational as long as there is some way to write it as a fraction of integers. So zero
is a rational number because 0 = 0/1, and 0.5 is also a rational number because 0.5 = 1/2.
Every rational number corresponds to a point on the number line. For example, 1/2 is right in
between 0 and 1, and 1/3 is one third of the way from 0 to 1:
0 1 1 2 1
3 2 3
37
3. THE NUMBER LINE AND SETS OF NUMBERS CHAPTER 1. CONCEPTS
Rational numbers are what topologists call dense on the real line, meaning that infinitely many
can be found in every little interval. However, not all points on the number line correspond to
rational numbers.
3.5. Reals.
DEFINITION 3.5.1. The set of real numbers is the collection of all numbers with signs
(positive or negative) and decimal representations. A decimal representation is a finite
list of digits from 0 to 9 (the integer part), then a decimal point, then an infinite list of
digits (the fractional part). It is traditional to omit an infinite tail of zeroes, and to avoid
an infinite tail of nines:
2.100000 . . . = 2.1
17.99999 . . . = 18
The standard notation for the set of all real numbers is R.
THEOREM 3.5.1. Each real number corresponds to a unique point on the real number line,
and each point on the real number line corresponds to a unique real number.
All rational numbers are real. 1/2 is a real number because it has a decimal representation
0.5, and 1/3 is a real number because it has a decimal representation 0.333333 . . . = 0.3.
A decimal representation of any integer fraction can be produced with long division. Some
divisions (like 1/3) seem to go on forever, but it can be shown that all rational numbers end
in simple repeating patterns.
Rewriting decimal representations as fractions of integers is straightforward when possible.
EXAMPLE 3.5.1. Write the number 0.125 as a fraction of integers.
SOLUTION: The last digit 5 is in the third position after the decimal point, which corre-
sponds to thousandths, so we can represent this number as a fraction with denominator
1000.
125
ANSWER:
1000
EXAMPLE 3.5.2. Write the number 1075.56 as a fraction of integers.
SOLUTION: This number ends with hundredths, so it is equivalent to
38
CHAPTER 1. CONCEPTS 3. THE NUMBER LINE AND SETS OF NUMBERS
107556
ANSWER:
100
p
Probably the oldest known example of a real number that is not rational isp 2, which is the
length of the diagonal of a square with side of length 1. It can be written as 2p= 1.41421 . . .,
but unlike with rational numbers, the digits of the decimal representation of 2 do not have
a simple repeating pattern.
p
2
1
1
Another famous example of a real number that is not rational is the ratio of a circle’s circum-
ference to its diameter, π = 3.14159 . . ., which is exactly the circumference (the length of the
border) of a circle with diameter 1.
π
1
Real
p numbers are drawn on the number line in according to their decimal representations, so
2 is between 1 and 2 while π is shortly after 3:
p
2 π
0 1 2 3 4
DEFINITION 3.5.2. Real numbers which are not rational are also known as irrational num-
bers.
EXAMPLE 3.5.3. Identify all the integers in the set
1 9
§ ª
−4, 0, , , 27.7
2 3
39
3. THE NUMBER LINE AND SETS OF NUMBERS CHAPTER 1. CONCEPTS
SOLUTION:
• −4 is an integer Ø
• 0 is an integer Ø
• 1/2 is rational, but not an integer
• 9/3 is equal to 3, so it is an integer Ø
• 27.7 is not an integer
We will state the answer as a subset consisting of integers, and note that we can put 3
instead of 9/3 because sets contain numbers, not the expressions which happen to define
the numbers.
ANSWER: {−4, 0, 3}
EXAMPLE 3.5.4. Identify all the non-negative integers in the set
5 −10 0
§ ª
5, 5.5, , , , −12, −0.17
5 2 −6
SOLUTION:
• 5 is a non-negative integer Ø
• 5.5 is not an integer
• 5/5 is equal to 1, which is a non-negative integer Ø
• −10/2 is equal to −5, which is not non-negative
• 0/(−6) is equal to 0, which is a non-negative integer Ø
• −12 is not non-negative
• −0.17 is neither non-negative nor an integer
ANSWER: {5, 1, 0}
3.6. Absolute Value.
DEFINITION 3.6.1. The absolute value of a real number x, written |x|, is the distance from
the number to zero on the real number line. Alternatively, the absolute value of a non-
negative number is the number itself, and the absolute value of a negative number is its
opposite.
Absolute value bars alter the order of operations like parentheses, but the distributive property
does not work on them. In general, we must know the value of the expression (that is, reduce
it down to a single number) before removing the bars and the negative sign. It is a grave but
40
CHAPTER 1. CONCEPTS 3. THE NUMBER LINE AND SETS OF NUMBERS
common mistake to change pluses to minuses inside the absolute value bars before the value
of the expression is known.
BASIC EXAMPLE 3.6.1.
| − 17| = 17
−|17| = −17
|4 − 13| = | − 9| = 9
|2(−5)(−3)| = | − 30| = 30
2 − | − 5 − 3| = 2 − | − 8| = 2 − 8 = −6
EXAMPLE 3.6.1. Find the value of the expression if x = 5
|x − 7| + 2x
SOLUTION:
|x − 7| + 2x = |(5) − 7| + 2(5)
= | − 2| + 10 replaced 5 − 7 by its value −2
= 2 + 10 because | − 2| = 2
= 12
ANSWER: 12
41
3. THE NUMBER LINE AND SETS OF NUMBERS CHAPTER 1. CONCEPTS
3.7. Inequality Relations. The real number line provides an ordering of the set of real
numbers. If the number line is oriented so that 1 is to the right of 0, then given any two
distinct real numbers, the one on the right is greater than the one on the left.
DEFINITION 3.7.1. We will use five inequality relations in this text:
a < b a is (strictly) less than b a b
a > b a is (strictly) greater than b b a
a ≤ b a is less than or equal to b
a ≥ b a is greater than or equal to b
a 6= b a is not equal to b
< and > are called strict inequalities, while ≤ and ≥ are called non-strict inequalities.
Other ways to read a ≤ b are “a is at most b” and “b is at least a”.
EXAMPLE 3.7.1. Determine whether the inequality is true or false:
17 ≥ 17
SOLUTION: 17 = 17, so the non-strict inequality holds.
ANSWER: true
EXAMPLE 3.7.2. Determine whether the inequality is true or false:
−6 < −10
SOLUTION: −6 is to the right of −10 on the real number line, so it is greater than −10,
and the inequality is false.
ANSWER: false
EXAMPLE 3.7.3. Determine whether the inequality is true or false:
3.1315 ≥ 3.14
42
CHAPTER 1. CONCEPTS 3. THE NUMBER LINE AND SETS OF NUMBERS
SOLUTION: The number 3.1315 is just a tad to the right of 3.1400 on the number line, so
it is greater, and the inequality holds.
ANSWER: true
EXAMPLE 3.7.4. Determine whether the inequality is true or false:
2 1
<
6 3
SOLUTION: To compare these fractions, we need to rewrite them with the same denomi-
nator somehow. If we multiply 1/3 by 2/2, for example, its value will not change, but the
denominator will:
1 1 2 1·2 2
= · = =
3 3 2 3·2 6
So the inequality is equivalent to
2 2
<
6 6
which is false.
ANSWER: false
43
3. THE NUMBER LINE AND SETS OF NUMBERS CHAPTER 1. CONCEPTS
Homework 1.3.
Find the value of the expression: if z = −16
1. 4 − 9 20. Find the value of the expression
−k + 4 − 10
2. 17 − 24
if k = −4
3. −2 − 6
4. −14 − 18 List at least five elements of the given set
using the roster notation:
5. 5 − 7 − 13
21. 2k + 100 k is a non-negative integer
6. 6 + 8 − 33
7. 7 − 7 22. 3(m − 1) m is a negative integer
x +1
§ ª
8. (−7) − (−5) 23. x is a positive integer
x
9. (−8) − (−5) − 2
24. x 2 x is a non-negative integer
10. (−2) + (−5) + 5
11. 4 − 9 − (−4)
25. Identify all the negative integers in the
set
12. (−12) − (−12)
−15
§ ª
−7, 4, −7.5, −1, 0, , 100
13. 8 + (−6) + 2 3
14. 10 + (−12) − 3 26. Identify all the non-negative integers
in the set
15. −2 + 20 − 6 − 17 16 1
§ ª
−2, 0, , −100,
4 17
16. 4 + (−3) + (−2) + 1
27. Identify all the integers in the set
17. Find the value of the expression §
4 −12
ª
12, , −10.1, , π
−12 + x + 3 3 2
if x = −9 28. Identify all the positive integers in the
set
18. Find the value of the expression §
4 60 −10 p 0
ª
, −4, , , 2,
−10 + 14 + y 6 6 −10 1
if y = −5
19. Find the value of the expression
−4 − z − 1
44
CHAPTER 1. CONCEPTS 3. THE NUMBER LINE AND SETS OF NUMBERS
Write the real number as a fraction of inte- Determine whether the inequality is true or
gers. false:
29. 0.4 35. 7 ≤ 7
30. 0.3 36. 9 > 5
31. −0.17 37. −10 < −1
32. −0.87 38. −5 ≥ 4
33. 3.1415 39. 0.1 < 0.1
34. −2.718 40. −9 ≥ −9
1 1
41. >
8 4
42. −6 ≤ −10
43. π > 3
p
44. 2 ≤ 2
45
3. THE NUMBER LINE AND SETS OF NUMBERS CHAPTER 1. CONCEPTS
Homework 1.3 Answers.
1. −5 25. {−7, −1, −5}
3. −8 27. {12, −6}
5. −15 4
29.
10
7. 0
−17
9. −5 31.
100
11. −1
31415
33.
13. 4 10000
15. −5 35. true
17. −18 37. true
19. 11 39. false
21. {100, 102, 104, 106, 108, . . .}
41. false
3 4 5 6
§ ª
23. 2, , , , , . . .
2 3 4 5 43. true
46
CHAPTER 1. CONCEPTS 4. PROPERTIES OF REAL NUMBERS
4. Properties of Real Numbers
4.1. Basic Properties. The following properties of real numbers are well-known, but we
are going to highlight and prove them here, just to demonstrate the axiomatic approach we are
taking. As you go through the proofs, you can have some fun by covering up the comments,
and trying to name the axioms responsible for each step in the proof.
THEOREM 4.1.1. For any real number A,
0·A=0
PROOF. We appeal directly to the axioms:
0A = 0A + 0 identity of +
= 0A + (1A − 1A) additive inverse
= (0A + 1A) − 1A associativity of +
= (0 + 1)A − 1A distributivity
= 1A − 1A identity of +
= 0 additive inverse
THEOREM 4.1.2. For all real numbers A and B,
(−A)B = A(−B) = −(AB)
and
(−A)(−B) = AB
As a consequence, multiplying numbers with opposite signs produces a negative result,
while multiplying numbers with the same signs produces a positive result.
PROOF. To show that (−A)B = −(AB), we appeal to the axioms, as well as the theorem
above:
(−A)B = (−A)B + 0 identity of +
= (−A)B + (AB − AB) additive inverse
= ((−A)B + AB) − AB associativity of +
= (−A + A)B − AB distributivity
= 0B − AB identity of +
= 0 − AB theorem 4.1.1
= −AB identity of +
47
4. PROPERTIES OF REAL NUMBERS CHAPTER 1. CONCEPTS
Showing A(−B) = −(AB) is very similar. To show the last identity, we appeal to the statements
we’ve just proven:
(−A)(−B) = −(A(−B))
= −(−(AB))
= AB theorem 2.3.1
EXAMPLE 4.1.1. Find (−5)(−7)
SOLUTION: (−5)(−7) = 5 · 7 = 35
ANSWER: 35
EXAMPLE 4.1.2. Simplify (−4)(7x)
SOLUTION: (−4)(7x) = −(4 · 7x) = −28x
ANSWER: −28x
EXAMPLE 4.1.3. Find (−2)(−3)(−5)
SOLUTION: (−2)(−3)(−5) = (6)(−5) = −30
ANSWER: −30
THEOREM 4.1.3. For any real number A,
(−1)A = −A
This is a striking result so many of us take for granted, telling us that taking an opposite
of a number is the same as multiplying it by the opposite of the multiplicative identity.
48
CHAPTER 1. CONCEPTS 4. PROPERTIES OF REAL NUMBERS
PROOF. We appeal to the axioms, as well as the facts we have proven earlier:
(−1)A = (−1)A + 0 additive identity
= (−1)A + (1A − (1A)) additive inverse
= ((−1)A + 1A) − (1A) associativity of +
= ((−1) + 1)A − (1A) distributivity
= 0A − (1A) additive inverse
= 0 − (1A) theorem 4.1.1
= −(1A) additive identity
= −A multiplicative identity
EXAMPLE 4.1.4. Remove the parentheses and simplify:
−(x − y + 2)
SOLUTION: We will rewrite the opposite as a multiplication by −1 and apply the distribu-
tive property. Recall also that subtracting a number is the same as adding its opposite.
−(x − y + 2) = (−1)(x + (− y) + 2)
= (−1)x + (−1)(− y) + (−1)2 distributivity
= −x + y + (−2) (−1)(− y) = 1 y = y
= −x + y − 2
ANSWER: −x + y − 2
Multiplying by −1 and then distributing may seem tedious, so in practice we simplify opposites
of sums using an even nicer property.
THEOREM 4.1.4. For any collection of real numbers A, B, C, . . ., the opposite of their sum
is a sum of their opposites:
−(A + B + C + . . .) = (−A) + (−B) + (−C) + . . .
EXAMPLE 4.1.5. Remove the parentheses and simplify:
−(x − 3 y + 5z − 1 − a)
49
4. PROPERTIES OF REAL NUMBERS CHAPTER 1. CONCEPTS
SOLUTION: As we remove these parentheses, we replace each term of the sum by its
opposite:
−(x − 3 y + 5z − 1 − a) = −x + 3 y − 5z + 1 + a
ANSWER: −x + 3 y − 5z + 1 + a
EXAMPLE 4.1.6. Remove the parentheses and simplify:
−(−a + 2 − (5 − b))
SOLUTION: We can remove the outer parentheses, and then the inner ones:
−(−a + 2 − (5 − b)) = a − 2 + (5 − b)
= a−2+5− b −2 + 5 = 3
= a+3− b
Alternatively, we can remove the inner parentheses first, but then of course we will get
the same result:
−(−a + 2 − (5 − b)) = −(−a + 2 − 5 + b)
= a−2+5− b
= a+3− b
ANSWER: a + 3 − b
4.2. Positive Integer Exponent.
DEFINITION 4.2.1. For any real number b and any positive integer n, the nth power of b
is the product of n numbers b, written as
b n = |b · b · {z
b · . . . · b}
n times
n
Within the expression b , b is called base and n is called exponent.
BASIC EXAMPLE 4.2.1. Base −6, exponent 2:
(−6)2 = (−6)(−6) = 36
BASIC EXAMPLE 4.2.2. Base 5, exponent 3:
53 = 5 · 5 · 5 = 125
50
CHAPTER 1. CONCEPTS 4. PROPERTIES OF REAL NUMBERS
BASIC EXAMPLE 4.2.3.
(−3)4 = (−3)(−3)(−3)(−3) = 81
BASIC EXAMPLE 4.2.4.
(−2)5 = (−2)(−2)(−2)(−2)(−2) = −32
Looking carefully at the examples above allows us to make a general statement about products
of negative numbers:
THEOREM 4.2.1. A product of even number of negative factors is positive, while a product
of odd number of of negative factors is negative.
EXAMPLE 4.2.1. Evaluate (−1)1000
SOLUTION:
(−1)1000 = (−1)(−1)(−1) . . . (−1)
| {z }
1000 times
This is too long to write out by definition, but since (−1)(−1) = 1 and the signs cancel
in pairs, it is possible to prove that (−1)n = 1 when n is even, and (−1)n = −1 when n is
odd.
ANSWER: 1
BASIC EXAMPLE 4.2.5. The following is an example of a notational convention: the expo-
nent always supersedes the sign in the order of operations.
−72 = −(72 ) = −(7 · 7) = −49
Compare it with
(−7)2 = (−7)(−7) = 49
BASIC EXAMPLE 4.2.6. Unlike multiplication, the exponent does not distribute over addi-
tion. You can easily check that
(1 + 2)2 = 33 = 9
is completely different from
12 + 22 = 1 + 4 = 5
51
4. PROPERTIES OF REAL NUMBERS CHAPTER 1. CONCEPTS
Homework 1.4.
Simplify the expression:
1. (−7)(−3)
Find the value of the expression:
2. (−1)(−17)
13. (−4)2
3. (−2a)(−4b)
14. (−5)2
4. (−6c)(−3d)
15. 25
5. (−5)(−6)(−7)
16. (−4)3
6. (−10)(−3)(−1)
17. (−7)3
Remove the parentheses and simplify: 18. (−1)6
7. −(a − 2b − 6 + c) 19. −34
8. −(−4 − 3a + 7b) 20. (0.5)2
9. −(x + 5) − (5 − 3 y) 21. (−0.6)2
10. −(6 − x) − (x + 20) 22. −0.42
11. −(10 − (−4 + 2x)) 23. (−1)17
12. −(−7 − (19x − 5)) 24. (−1)54
52
CHAPTER 1. CONCEPTS 4. PROPERTIES OF REAL NUMBERS
Homework 1.4 Answers.
1. 21 13. 16
3. 8a b 15. 32
5. −210 17. −343
7. −a + 2b + 6 − c 19. −81
9. −x − 10 + 3 y 21. 0.36
11. −14 + 2x 23. −1
53
5. ORDER OF OPERATIONS CHAPTER 1. CONCEPTS
5. Order of Operations
5.1. Evaluating Sums and Products. The order of operations makes arithmetic expres-
sions unambiguous. We end up evaluating from inside out, so to speak, starting with the
innermost parentheses and absolute values.
Every expression in this text can be thought of as a sum of terms. Some of these terms will
be products of factors, and some of these factors will have exponents. We think of subtraction
as of adding the opposite, so our sums may have subtractions. We also think of division as of
multiplying by the reciprocal, so our products may have divisions.
DEFINITION 5.1.1 (Order of Operations). To evaluate a sum of terms:
(1) Find the value of each term.
(2) Find the value of the sum by applying additions and subtractions from left to
right.
Some of the terms will be products of several factors. To evaluate a product of factors
with exponents:
(1) Find the value of each exponential expression by using a single factor on the left
of the exponent as the base.
(2) Find the value of the product by applying multiplications and divisions from left
to right.
At any point in this process we are free to remove or to insert parentheses by using the
distributive property or any other axiom.
EXAMPLE 5.1.1. Evaluate the expression
5 + 4x − 2x 2
if x = 3
SOLUTION: This is a sum of 3 terms.
5 + 4x − 2x 2 = 5 + 4(3) − 2(3)2
= 5 + 4(3) − 2(9) evaluate the exponent
= 5 + 12 − 18 all terms are evaluated, time to add
= 17 − 18
= −1
ANSWER: −1
54
CHAPTER 1. CONCEPTS 5. ORDER OF OPERATIONS
EXAMPLE 5.1.2. Evaluate the expression
(3 − 5)(−8 + 14)
SOLUTION: This is a product with 2 factors.
(3 − 5)(−8 + 14) = (−2)(6) evaluate each factor
= −12
ANSWER: −12
EXAMPLE 5.1.3. Evaluate the expression
x 2 ÷ (x + 8)(x + 3)
if x = −6
SOLUTION: This is a product with 3 factors, the second factor being the reciprocal of
(x + 8). Recall that multiplications and divisions are to be done left to right.
x 2 ÷ (x + 8)(x + 3) = (−6)2 ÷ ((−6) + 8)((−6) + 3) substitute −6 for x
= (36) ÷ (2)(−3) all factors are evaluated
= (18)(−3) 36/2 = 18
= −54
ANSWER: −54
EXAMPLE 5.1.4. Evaluate the expression
a − a2
5+a
if a = −3
SOLUTION: It is important to remember that the fraction notation (Example 2.3.1) comes
with invisible parentheses, and the division of numerator by denominator is last in the
order of operations:
a − a2
= (a − a2 ) ÷ (5 + a)
5+a
55
5. ORDER OF OPERATIONS CHAPTER 1. CONCEPTS
So we can think of this fraction as of a product of 2 factors, and we have to find the value
of each factor before we can divide:
a − a2 (−3) − (−3)2
= substitute −3 for a
5+a 5 + (−3)
−3 − (9)
=
2
−12
=
2
= −6
ANSWER: −6
EXAMPLE 5.1.5. Evaluate the expression
| − 6 · 4| ÷ (−5 + 7)3 (−52 )
SOLUTION: This can be seen as a product with 3 factors. We have to evaluate each factor
before we can divide or multiply them. Note that we have to evaluate the base (−5 + 7)
before we can exponentiate it.
| − 6 · 4| ÷ (−5 + 7)3 · (−52 ) = | − 24| ÷ (2)3 · (−25)
= (24) ÷ 23 · (−25)
= 24 ÷ 8 · (−25) all factors are evaluated
= (3)(−25)
= −75
ANSWER: −75
EXAMPLE 5.1.6. Evaluate the expression given that x = 2 and y = −1
|x − y|
−(x + y)2
SOLUTION: We view this fraction as a product of 2 factors:
|x − y|
= |x − y| ÷ (−(x + y)2 )
−(x + y)2
56
CHAPTER 1. CONCEPTS 5. ORDER OF OPERATIONS
We have to evaluate the numerator and the denominator before we can divide.
|x − y| |2 − (−1)|
=
−(x + y)2 −(2 + (−1))2
|3|
=
−(1)2
3
=
−1
= −3
ANSWER: −3
5.2. Like Terms.
DEFINITION 5.2.1. Like terms, sometime called similar terms, are the terms which have all
the same variable factors with the same corresponding exponents.
EXAMPLE 5.2.1. Simplify the expression 3x y − 6x y
SOLUTION: Both terms have the same variable factors x y, so we can factor them out
using the distributivity:
3x y − 6x y = (3 − 6)x y
= (−3)x y
= −3x y
ANSWER: −3x y
EXAMPLE 5.2.2. Simplify the expression 5a − 2(a2 − 3a)
SOLUTION: We begin by applying the distributive property to remove the parentheses:
5a − 2(a2 − 3a) = 5a − 2a2 + 2 · 3a
= 5a − 2a2 + 6a
57
5. ORDER OF OPERATIONS CHAPTER 1. CONCEPTS
Here 5a and 6a are like terms, while −2a2 is different because it has exponent 2 instead
of 1.
5a − 2a2 + 6a = 5a + 6a − 2a2 associativity of +
= (5 + 6)a − 2a 2
distributivity
= 11a − 2a 2
ANSWER: 11a − 2a2
EXAMPLE 5.2.3. Simplify the expression 5x − 1 − (3x − 1 + 10 y)
SOLUTION: We begin by removing parentheses, using the fact that the opposite of a sum
is the sum of opposites. After that we use commutativity and associativity to change the
order of addition and put like terms next to each other, and then simplify them using
distributivity.
5x − 1 − (3x − 1 + 10 y) = 5x − 1 − 3x + 1 − 10 y
= 5x − 3x + 1 − 1 − 10 y properties of +
= (5 − 3)x + (1 − 1) − 10 y distributivity
= 2x + 0 − 10 y
= 2x − 10 y
ANSWER: 2x − 10 y
58
CHAPTER 1. CONCEPTS 5. ORDER OF OPERATIONS
Homework 1.5.
Evaluate the expression: 17. Evaluate the expression
−75 ÷ x 2
1. 3 + 8 ÷ | − 4|
if x = −5
2. 8 ÷ 4 · 2
18. Evaluate the expression
3. (−6 ÷ 6)3
45 ÷ 32 y( y − 1)
4. 5(−5 + 6) · 62 if y = 3
5. 7 − 15 ÷ 3 + 6 19. Evaluate the expression
6. (−9 − (2 − 5)) ÷ (−6) 6x ÷ 12x 3
if x = −2
7. | − 7 − 5| ÷ (−2 − 2 − (−6))
20. Evaluate the expression
8. 4 − 2 · |32 − 16|
−30 ÷ a(a + 4)2
−10 − 6 if a = −6
9. −5
(−2)2
2 + 4 · |7 + 22 |
10. Simplify the expression by combining like
4·2+5·3
terms:
−18
11. (6 · 2 + 2 − (−6)) −5 + 21. r − 9 + 10
6
−13 − 2 22. −4x + 2 − 4
12.
2 − (−1) + (−6) − (−1 − (−3))
3
23. 4b + 6 + 1 + 7b
|3 − (6 + 11)| 24 24. −7x − 2 − 2x
13. +
7 (1 − 3)2
25. −8px + 5px
−52 + (−5)2
14. 2
|4 − 25 | − 2 · 3 26. 2r 2 − r + 6r 2
−9 · 2 − (3 − 6) 27. t − 2t 2 − t 2 − 7t
15.
1 − (−2 + 1) − (−3)
28. −5x + y + 12x − y
5 + 3 − 24 ÷ 6 · 2
2
16.
(5 + 3(22 − 5)) + |22 − 5|2
Simplify the expression by removing paren-
theses and combining like terms:
29. 9(b + 10) + 5b
59
5. ORDER OF OPERATIONS CHAPTER 1. CONCEPTS
30. 4v − 7(1 − 8v) 37. a(x + 3) − 4ax + 2a
31. −3(1 − 4x) − 4x 38. 9x − 11x y − x(1 + 2 y)
32. −8x + 9(−9x + 9) 39. 5(1 − 6k) + 10(k − 8)
33. −10 − 4(n − 5) 40. −7(4x − 6) + 2(10x − 10)
34. −6(5 − m) + 3m 41. b(m + y) − m(b − y)
35. (8n2 − 3n) − (5 + 4n2 ) 42. (2b − 8) − 2(b2 + 5)
36. (7x 2 − 3) − (5x 2 + 6x)
60
CHAPTER 1. CONCEPTS 5. ORDER OF OPERATIONS
Homework 1.5 Answers.
1. 5 23. 11b + 7
3. −1 25. −3px
5. 8
27. −6t − 3t 2
7. 6
29. 14b + 90
9. −9
31. −3 + 8x
11. −40
33. 10 − 4n
13. 6
35. 4n2 − 3n − 5
15. −3
37. −3ax + 5a
17. −3
19. 8 39. −20k − 75
21. r + 1 41. b y + m y
61
6. PRIME NUMBERS CHAPTER 1. CONCEPTS
6. Prime Numbers
6.1. Integer Division and Prime Numbers.
DEFINITION 6.1.1. A positive integer m is an integer divisor or just divisor of a positive
integer n if dividing n by m leaves no remainder.
BASIC EXAMPLE 6.1.1.
• 4 is a divisor of 12 because 12/4 = 3
• 5 is a divisor of 5 because 5/5 = 1
• 3 is not a divisor of 5 because 5/3 leaves a remainder of 2
• 2.5 is not a divisor of 5 because 2.5 is not an integer
DEFINITION 6.1.2. We call a positive integer prime when it has exactly two positive integer
divisors: one and itself.
BASIC EXAMPLE 6.1.2.
• 1 is not prime because it only has a single divisor, itself
• 2 is prime because it has exactly two divisors: 1 and 2
• 3 is also prime
• 4 is not prime because it has three divisors: 1, 2, and 4
• 5 is prime
• 6 is not prime because it has four divisors: 1, 2, 3, and 6
THEOREM 6.1.1. The set of prime numbers is infinite, and it starts like this:
{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, . . .}
DEFINITION 6.1.3. A positive integer with three or more positive integer divisors is called
composite. Incidentally, 1 is neither prime nor composite.
THEOREM 6.1.2 (Fundamental Theorem of Arithmetic). Every positive integer greater
than 1 is either prime or is a product of prime factors, and this product is unique, up to
the order of the factors.
EXAMPLE 6.1.1. Write 10 as a product of prime factors.
ANSWER: 2 · 5
62
CHAPTER 1. CONCEPTS 6. PRIME NUMBERS
EXAMPLE 6.1.2. Write 60 as a product of prime numbers.
SOLUTION: A quick and greedy way to success is to try to express the number as a product
of smaller numbers, and keep doing that until nothing is left but primes.
60 = 6 · 10 = (2 · 3) · (2 · 5)
The order is not important within the product, but a traditional way to state the answer
is by using the exponential notation, and listing lower bases first.
ANSWER: 22 · 3 · 5
EXAMPLE 6.1.3. Write 231 as a product of prime numbers.
SOLUTION: When it is not immediately clear how to factor, we try dividing by every prime,
starting with the lowest one, as many times as possible. 231/2 leaves a remainder of 1,
so 2 is not a factor. 231/3 = 77, so 3 is a factor, and then
231 = 3 · 77 = 3 · 7 · 11
ANSWER: 3 · 7 · 11
EXAMPLE 6.1.4. Write 450 as a product of prime numbers.
SOLUTION:
450 = 10 · 45 = (2 · 5)(5 · 9) = 2 · 5 · 5 · (3 · 3)
ANSWER: 2 · 32 · 52
63
6. PRIME NUMBERS CHAPTER 1. CONCEPTS
EXAMPLE 6.1.5. Write 448 as a product of prime numbers.
SOLUTION: Every even number has 2 as a factor, so we will keep dividing by 2 until we
encounter an odd quotient:
448 = 2 · 224
= 2 · 2 · 112
= 2 · 2 · 2 · 56
= 2 · 2 · 2 · (8 · 7)
= 2 · 2 · 2 · (2 · 2 · 2) · 7
ANSWER: 26 · 7
6.2. Fractions of Integers in Lowest Terms.
DEFINITION 6.2.1. A rational number is written in lowest terms if numerator and denom-
inator have no common prime factors.
When numerator and denominator do have common factors, we can reduce the fraction to
lowest terms by canceling them.
THEOREM 6.2.1. Common factors cancel in fractions. For any real number A and any two
non-zero real numbers D and C,
AC A
=
DC D
EXAMPLE 6.2.1. Write 12/15 in lowest terms.
SOLUTION: If we can detect a common factor, we can simplify by canceling it.
12 3 · 4 4
= =
15 3 · 5 5
4
ANSWER:
5
EXAMPLE 6.2.2. Write 60/315 in lowest terms.
64
CHAPTER 1. CONCEPTS 6. PRIME NUMBERS
SOLUTION: Sometimes it is hard to identify common factors right away, but we can always
completely factor both numerator and denominator, and then all of the common factors
will become apparent.
60 = 6 · 10 = (2 · 3) · (2 · 5) = 2 · 2 · 3 · 5
315 = 5 · 63 = 5 · (7 · 9) = 5 · 7 · (3 · 3) = 3 · 3 · 5 · 7
The common factors are 3 and 5:
60 2·2·3·5 2·2 4
= = =
315 3 · 3 · 5 · 7 3 · 7 21
4
ANSWER:
21
EXAMPLE 6.2.3. Simplify the fraction assuming all variables are non-zero:
4abx
4b y
SOLUTION: Common factors 4 and b cancel.
ax
ANSWER:
y
EXAMPLE 6.2.4. Simplify the fraction assuming all variables are non-zero:
6 y3
18 y 2
SOLUTION:
6 y3 y3
= cancel common factor 6
18 y 2 3 y2
y
= cancel common factor y 2
3
y
ANSWER:
3
65
6. PRIME NUMBERS CHAPTER 1. CONCEPTS
EXAMPLE 6.2.5. Simplify the fraction assuming all variables are non-zero:
60x 2
40x
SOLUTION:
60x 2 60 · x
= cancel common variable factor x
40x 40
3 · 20 · x
= 20 is a common numerical factor
2 · 20
3x
= rewrite in lowest terms
2
3
ANSWER: x
2
66
CHAPTER 1. CONCEPTS 6. PRIME NUMBERS
Homework 1.6.
Write the number as a product of prime fac- Rewrite the rational number in lowest
tors with exponents. terms.
1. 12 21
23.
28
2. 14
21
24.
3. 16 35
16
4. 18 25.
26
5. 19 12
26.
6. 20 48
12
7. 22 27.
40
8. 24 75
28.
80
9. 26
52
10. 27 29.
13
11. 30 110
30.
5
12. 36
4
31.
13. 40 19
14. 42 46
32.
48
15. 45
150
33.
16. 48 450
17. 160 72
34.
45
18. 180
27
35.
19. 625 150
8
20. 225 36.
100
21. 196 24
37.
22. 1000 64
67
6. PRIME NUMBERS CHAPTER 1. CONCEPTS
25 Simplify the fraction assuming all variables
38.
65 are non-zero:
90 18x 3
39. 45.
225 6x
253 7u2
40. 46.
11 28u
160 −6ab2
41. 47.
180 −10b
625 −8x y
42. 48.
225 12x y
196
43. −x
1000 49.
5x 3
230
44. 20
460 50.
−60b
48x y 2
51.
30x 2 y
46x 2 y 2
52.
−26x 2
68
CHAPTER 1. CONCEPTS 6. PRIME NUMBERS
Homework 1.6 Answers.
1. 22 · 3 4
31.
19
3. 24
1
33.
5. 19 3
7. 2 · 11 9
35.
50
9. 2 · 13
3
11. 2 · 3 · 5 37.
8
13. 23 · 5 2
39.
5
15. 32 · 5
8
17. 25 · 5 41.
9
19. 54 49
43.
250
21. 22 · 72
3 45. 3x 2
23.
4 3ab
47.
8 5
25.
13
−1
49.
3 5x 2
27.
10
8y
29. 4 51.
5x
69
7. FRACTIONS AND RATIONALS CHAPTER 1. CONCEPTS
7. Fractions and Rationals
7.1. Fraction Notation. Recall that for any non-zero real number B there is a reciprocal,
which we denote by 1/B, such that B · (1/B) = 1. Recall also that division is multiplication by
the reciprocal:
A 1
A÷ B = = A·
B B
THEOREM 7.1.1. For any two non-zero real numbers A and B, the reciprocal of a product
is the product of reciprocals:
1 1 1
= ·
AB A B
BASIC EXAMPLE 7.1.1. This theorem generalizes to larger products as well:
1 1 1 1 1
· · = =
7 10 −2 7 · 10 · (−2) −140
This also also gives us a justification for multiplying fractions the way we do, by computing
the product of numerators over the product of denominators:
3 2 1 1
· = 3· · 2· by definition of ÷
7 5 7 5
1 1
= (3 · 2) · · reorder multiplication
7 5
1
= (3 · 2) · apply theorem
7·5
1
= 6· simplify
35
6
= by definition of ÷
35
EXAMPLE 7.1.1. Identify all the factors in the expression by rewriting the fraction as a
product:
3b
xy
SOLUTION: Formally, this fraction is a product of (3b) and the reciprocal of (x y), which
we can now separate into individual factors using the axioms and the basic properties:
3b 1 1 1
= (3b) · = 3· b· ·
xy (x y) x y
70
CHAPTER 1. CONCEPTS 7. FRACTIONS AND RATIONALS
1 1
ANSWER: 4 factors: 3 · x · ·
x y
EXAMPLE 7.1.2. Identify all the factors in the expression by rewriting the fraction as a
product:
−5A(B − 1)
3X
SOLUTION: It may be useful to rewrite the expression with multiplications made visible
so that factors stand out more:
(−5) · A · (B − 1)
3·X
1 1
ANSWER: 5 factors: (−5) · A · (B − 1) · ·
3 X
THEOREM 7.1.2. To multiply fractions, multiply numerators to obtain the new numerator,
and multiply denominators to obtain the new denominator. For any two real numbers A
and B and any two non-zero real numbers C and D,
A B AB
· =
C D CD
EXAMPLE 7.1.3. Multiply fractions:
x 4
·
3 5
SOLUTION:
x 4 x · 4 4x
· = =
3 5 3·5 15
4x
ANSWER:
15
71
7. FRACTIONS AND RATIONALS CHAPTER 1. CONCEPTS
7
EXAMPLE 7.1.4. Multiply and simplify the result: 8·
12
SOLUTION: When multiplying (or dividing) a fraction by a number, it may be useful to
represent that number as a fraction first:
7 8 7
8· = ·
12 1 12
8·7
=
1 · 12
4·2·7
= 4 is a common factor
4·3
2·7
=
3
14
=
3
14
ANSWER:
3
EXAMPLE 7.1.5. Multiply and state the answer as a single fraction:
3 x +1
·
2 5
SOLUTION: Recall that the fraction notation (Example 2.3.1) comes with invisible paren-
theses, and so the sum x + 1 in the numerator becomes a single factor when we multiply
numerators:
3 x +1 3 (x + 1)
· = ·
2 5 2 5
3(x + 1)
=
2·5
3(x + 1)
=
10
3(x + 1)
ANSWER:
10
72
CHAPTER 1. CONCEPTS 7. FRACTIONS AND RATIONALS
EXAMPLE 7.1.6. Find the area of a rectangular piece of fabric 5/4 feet long and 2/7 feet
wide.
SOLUTION: The area of a rectangle is the product of its length and width:
5 2 5·2
· = multiply the dimensions
4 7 4·7
5
= cancel common factor 2
2·7
5
=
14
5
ANSWER: square feet
14
THEOREM 7.1.3. To divide a fraction by a fraction, multiply it by the reciprocal of the
divisor. For any real number A and any three non-zero real numbers B, C and D,
A B A D AD
÷ = · =
C D C B CB
EXAMPLE 7.1.7. Divide fractions:
6 2
÷
5 5
SOLUTION:
6 2 6 5
÷ = ·
5 5 5 2
6·5
=
5·2
6
=
2
= 3
ANSWER: 3
73
7. FRACTIONS AND RATIONALS CHAPTER 1. CONCEPTS
8 a+b
EXAMPLE 7.1.8. Divide fractions ÷ and state the answer as a single fraction.
a−1 2
SOLUTION: Make the fraction notation parentheses visible in order to multiply sums
correctly:
8 a+b 8 (a + b)
÷ = ÷ fraction parentheses
a−1 2 (a − 1) 2
8 2
= · multiplication by reciprocal
(a − 1) (a + b)
8·2
=
(a − 1)(a + b)
16
=
(a − 1)(a + b)
16
ANSWER:
(a − 1)(a + b)
7.2. Changing the Denominator. Before we start adding fractions, we need to figure out
a way to change a denominator without affecting the value of a fraction.
EXAMPLE 7.2.1. Rewrite the fraction 7/10 so that it has the denominator 40.
SOLUTION: The challenge is in keeping the value of the fraction the same, while changing
its appearance. If we multiply a fraction by 1, its value will not change. So when we
multiply it by 4/4, both numerator and denominator change, but the value of the fraction
stays the same:
7 7 4 7·4 28
= · = =
10 10 4 10 · 4 40
28
ANSWER:
40
74
CHAPTER 1. CONCEPTS 7. FRACTIONS AND RATIONALS
EXAMPLE 7.2.2. Rewrite the fraction −11/3 so that it has the denominator 18.
SOLUTION:
11 11 6 11 · 6 66
− =− · =− =−
3 3 6 3·6 18
66
ANSWER: −
18
75
7. FRACTIONS AND RATIONALS CHAPTER 1. CONCEPTS
Homework 1.7.
Identify all the factors in the expression by −12 −9
14. ÷
rewriting the fraction as a product: 7 5
3x 14
1. 15. ÷2
5 5
6 −13 −15
2. 16. ÷
7a 8 8
x +2 5 5
3. 17. ÷
2x(1 + a) 3 6
4 yz −4 −13
4. 18. ·
a+b 5 8
−4x(x + 3) 1 3
5. 19. ÷
a b(5 + c) 10 2
5 5
12u 20. ·
6. 3 3
−5(x + 1)(x + 2)
21. Find the area of a square with the side
length of 3/4 feet.
Perform fraction multiplication or division,
simplify the result, and state the answer as 22. Find the area of a rectangular strip
a single fraction: of land 21/5 meters long and 2/3 meters
1 2 wide.
7. ·
3 5
1 3
8. ·
8 8
−2
9. 9 ·
9
5
10. 2 ·
6
6 11
11. ·
5 8
−2 3
12. ·
3 4
3
13. −2 ÷
4
76
CHAPTER 1. CONCEPTS 7. FRACTIONS AND RATIONALS
Rewrite the fraction so that it has the given 13
27. with denominator 21
denominator: 7
2 9
23. with denominator 12 28. with denominator 16
3 2
−1 −11
24. with denominator 8 29. with denominator 600
4 6
−3 −5
25. with denominator 28 30. with denominator 99
14 9
7
26. with denominator 20
5
77
7. FRACTIONS AND RATIONALS CHAPTER 1. CONCEPTS
Homework 1.7 Answers.
1 17. 2
1. 3 factors: 3 · x ·
5
1 1 1 1
3. 4 factors: (x + 2) · · · 19.
2 x (1 + a) 15
1 1 1 21. 9/16 square feet
5. 6 factors: (−4) · x · (x + 3) · · ·
a b (5 + c)
2 8
7. 23.
15 12
9. −2 −6
25.
33 28
11.
20
39
−8 27.
13. 21
3
7 −1100
15. 29.
5 600
78
CHAPTER 1. CONCEPTS 8. FRACTION ADDITION
8. Fraction Addition
8.1. Adding Fractions with a Common Denominator.
THEOREM 8.1.1. A sum of fractions with a common denominator is the sum of numerators
over the same denominator. For all real numbers A, B, and C 6= 0
A B A+ B
+ =
C C C
A B A− B
− =
C C C
PROOF. Express fractions as products, and then apply the distributive property to the com-
mon denominator:
A B 1 1
+ = A· + B ·
C C C C
1
= (A + B)
C
A+ B
=
C
The subtraction case is similar.
3 21
EXAMPLE 8.1.1. Simplify the expression +
8 8
SOLUTION:
3 21 3 + 21 24
+ = = =3
8 8 8 8
ANSWER: 3
4 9
EXAMPLE 8.1.2. Simplify the expression −
15 15
SOLUTION:
4 9 4 − 9 −5 −1
− = = =
15 15 15 15 3
1
ANSWER: −
3
79
8. FRACTION ADDITION CHAPTER 1. CONCEPTS
8.2. Least Common Multiple.
DEFINITION 8.2.1 (Least Common Multiple). The least common multiple for a collection
of positive integers a, b, c, . . . (LCM for short) is the smallest positive integer divisible
by each of the a, b, c, . . . without a remainder. To find the LCM, we factor each integer
completely, and then take the product with each prime factor to the highest degree we
found.
BASIC EXAMPLE 8.2.1.
integers LCM
4, 12 12
4, 6 12
8, 10 40
2, 3, 5 30
4, 6, 8 24
EXAMPLE 8.2.1. Find the LCM for 16 and 20.
SOLUTION: When it’s too hard to guess the LCM, we can factor each integer to expose the
prime factors:
16 = 24
20 = 22 · 5
The highest power of 2 is 4, and the highest power of 5 is 1, so the LCM is
24 · 5 = 16 · 5 = 80
ANSWER: 80
EXAMPLE 8.2.2. Find the LCM for 34 and 40.
SOLUTION: Factor each integer:
34 = 2 · 17
40 = 23 · 5
So the LCM is 23 · 5 · 17 = 680
ANSWER: 680
80
CHAPTER 1. CONCEPTS 8. FRACTION ADDITION
EXAMPLE 8.2.3. Find the LCM for 121, 66, 8.
SOLUTION: Factor each integer:
121 = 112
66 = 2 · 3 · 11
8 = 23
So the LCM is 23 · 3 · 112 = 2904
ANSWER: 2904
8.3. Adding Fractions of Integers.
DEFINITION 8.3.1. The lowest common denominator for a collection of rational numbers
(LCD for short) is the LCM of their denominators.
THEOREM 8.3.1. Fractions with different denominators can be added by rewriting each
fraction in an equivalent form using the LCD, and then adding as usual.
EXAMPLE 8.3.1. Simplify and state as a fraction in lowest terms:
5 3
+
2 4
SOLUTION: The LCD is 4, so we rewrite the first fraction:
5 5 2 10
= · =
2 2 2 4
and then we can add:
5 3 10 3
+ = +
2 4 4 4
10 + 3
=
4
13
=
4
13
ANSWER:
4
81
8. FRACTION ADDITION CHAPTER 1. CONCEPTS
EXAMPLE 8.3.2. Simplify and state as a fraction in lowest terms:
5 2
−
3 5
SOLUTION: The LCD is 15, so we rewrite both fractions:
5 5 5 25
= · =
3 3 5 15
2 2 3 6
= · =
5 5 3 15
and then we can subtract:
5 2 25 6
− = −
3 5 15 15
25 − 6
=
15
19
=
15
19
ANSWER:
15
8.4. Mixed Numbers.
DEFINITION 8.4.1. A rational number is written as a proper fraction when the absolute
value of the numerator is less than the absolute value of the denominator. In other cases
it is called an improper fraction.
BASIC EXAMPLE 8.4.1. Some proper fractions: 17/19, −1/2, 10/20.
Some improper fractions: 3/2, 7/7, −101/100.
DEFINITION 8.4.2. A fraction is expressed as a mixed number when it is a sum of an integer
and a proper fraction. Traditionally, the plus sign is not shown. For example, the number
3.25, which is equal to
1
3+
4
looks like this in the mixed number notation:
3 14
82
CHAPTER 1. CONCEPTS 8. FRACTION ADDITION
EXAMPLE 8.4.1. Rewrite 60/13 as a mixed number.
SOLUTION: When we divide 60 by 13, the integer quotient is 4 and the remainder is 8, so
60 8
=4+
13 13
8
ANSWER: 4 13
EXAMPLE 8.4.2. Rewrite 7 45 as an improper fraction.
SOLUTION:
4 35 4 35 + 4 39
7 45 = 7 + = + = =
5 5 5 5 5
39
ANSWER:
5
83
8. FRACTION ADDITION CHAPTER 1. CONCEPTS
Homework 1.8.
Perform the summation and simplify. 2 5
17. −
5 4
1 3
1. + 2
8 8 18. −1 −
3
4 5
2. + −5 15
15 15 19. −
7 8
14 24
3. − 3 9
5 5 20. +
2 7
12 19
4. − 5
11
7 7 21. − −
3 6
17 10
5. +
6π 6π 13 4
22. −
18 9
10 22
6. −
3 3 1 11
23. −
2 6
1
Find the LCM for given integers. 24. −1 −
3
7. 3, 6
11 1
25. −
8. 5, 15 8 2
4 2
9. 6, 9 26. − + −
3 15
10. 10, 25
5
27. −6 −
11. 2, 9, 10 3
12. 3, 8, 15 15 5
28. − +
8 3
13. 48, 84
3 7 1
29. −
14. 28, 42 2 9 9
2 6 9
Perform the operations, simplify, state the 30. +
3 5 5
answer as a single fraction in lowest terms:
5 7 13
15 31. + ÷
15. 2 + 9 3 6
8
7 11 5
3 5 32. − ÷
16. + 8 4 4
5 4
84
CHAPTER 1. CONCEPTS 8. FRACTION ADDITION
Simplify the expression by combining like Rewrite the mixed number as an improper
terms: fraction:
5 2 45. 1 32
33. x+ x
9 3
11 46. 2 27
34. a − 3a
2 5
47. −10 11
5y 5y
35. − 48. −9 49
7 21
13b 11b
36. +
15 45
3 2 1
37. 5a − − a
4 3 3 Rewrite the fraction as a mixed number:
5 1 9
22
38. b + + b 49.
3 6 10 5
1 3 50
39. (6a + 5b) + (4a − 9b) 50.
2 2 7
2 1 110
40. (9x − 4 y) − (3a − 2 y) 51. −
3 3 21
41. Evaluate the expression 82
52. −
3 13
6a +
2
if a = −7/5
42. Evaluate the expression
4 53. Find the perimeter of the shown shape
3b + if side lengths are given in inches:
15
if b = −4/7
43. Evaluate the expression 9/14
3 1
x+ 5/14
5 x 2/7
if x = 2/3
3/7
44. Evaluate the expression
3 3 1
y− · 54. Find the perimeter of a rectangular
10 4 y strip of land 21/5 meters long and 2/3 me-
if y = 3/2 ters wide.
85
8. FRACTION ADDITION CHAPTER 1. CONCEPTS
Homework 1.8 Answers.
1 29. 1
1.
2
4
3. −2 31.
3
9
5. 11
2π 33. x
9
7. 6
10 y
35.
9. 18 21
11. 90 21 1
37. a−
4 2
13. 336
39. 9a − 11b
31
15.
8 −69
41.
10
−17
17.
20 13
43.
−145 10
19.
56 5
45.
7 3
21.
2
115
47. −
4 11
23. −
3
49. 4 52
7
25.
8 5
51. −5 21
23
27. − 53. 12/7 inches
3
86
CHAPTER 1. CONCEPTS 9. TRANSLATION
9. Translation
9.1. Expressions. The order of operations in mathematical expressions makes them un-
ambiguous, and there are ways to carry some of that certainty into plain English. Here are
some of the common ways to describe arithmetic expressions:
a+b the sum of a and b
a−b the difference of a and b
ab the product of a and b
a/b the quotient of a and b
a2 the square of a
To see the difficulty with translation, consider the phrase
the sum of a and b times c
It could mean a + bc, or it could mean (a + b)c, and these expressions are not equivalent.
Sometimes the ambiguity can be removed by finding a different phrasing:
the sum of a and the product of b and c
definitely means a + bc, while
the sum of a and b, times c
will probably be interpreted as (a + b)c, because of where the comma is placed in the sentence.
9.2. Equations. Many English sentences can be translated into equations almost word for
word. Here are some of the most notorious and useful patterns:
BASIC EXAMPLE 9.2.1.
a is 2 units greater than b
a = b + 2
BASIC EXAMPLE 9.2.2.
a is 5 units less than b
a = b − 5
BASIC EXAMPLE 9.2.3.
a is 6 times greater than b
a = 6 · b
87
9. TRANSLATION CHAPTER 1. CONCEPTS
BASIC EXAMPLE 9.2.4.
a is 7 times less than b
a = b / 7
EXAMPLE 9.2.1. Describe the variables used to represent the given quantities, and write
the statement as an algebraic relation:
One coffee costs as much as three donuts.
SOLUTION: Whenever it is not quite clear what to write, it may be useful to rephrase a
statement to make it sound like one of the patterns above. The price of one coffee is three
times the price of a donut. If c is the price of one coffee in dollars and d is the price of
one coffee in dollars, then c = 3d. We will state the answer by defining the variables and
their units, and then stating the relation between them.
ANSWER:
c is the price of one coffee in dollars
d is the price of one donut in dollars
c = 3d
EXAMPLE 9.2.2. Describe the variables used to represent the given quantities, and write
the statement as an algebraic relation:
Alice is biking 2.5 miles per hour faster than Bob.
SOLUTION: This statement compares speeds, so let a be Alice’s speed, and let b be Bob’s
speed, and then the sentence can be rephrased as
Alice’s speed is 2.5 miles per hour greater than Bob’s speed.
ANSWER:
a is Alice’s speed in mph
b is Bob’s speed in mph
a = b + 2.5
88
CHAPTER 1. CONCEPTS 9. TRANSLATION
EXAMPLE 9.2.3. Describe the variables used to represent the given quantities, and write
the statement as an algebraic relation:
The width of a FIBA regulation basketball court is 13 meters shorter than its length.
ANSWER:
w is the width of a court in meters
l is the length of a court in meters
w = l − 13
EXAMPLE 9.2.4. Describe the variables used to represent the given quantities, and write
the statement as an algebraic relation:
A tiger weighs 3.1 times less than a giraffe.
ANSWER:
t is the weight of a tiger in kg
g is the weight of a giraffe in kg
g
t=
3.1
89
9. TRANSLATION CHAPTER 1. CONCEPTS
9.3. Applications. We will solve applications by describing the variables used to represent
the quantities of interest, and stating appropriate equations. As you construct your own solu-
tions to applications, you may be using different variable names, and your equations may also
be somewhat different, depending on how you translate English sentences. The final numerical
answers, however, should always come out the same, regardless of translation.
EXAMPLE 9.3.1. A brownie recipe is asking for 350 grams of sugar, and a pound cake recipe
requires 270 more grams of sugar than a brownie recipe. How much sugar is needed for
the pound cake?
SOLUTION: Let b stand for the weight of sugar needed for the brownies, and p for the
weight of sugar needed for the pound cake, both measured in grams. Then the phrase
“pound cake recipe requires 270 more grams of sugar than a brownie recipe” translates as
“p is 270 grams greater than b”, or
p = b + 270
Since b = 350, we get
p = 350 + 270 = 620
ANSWER:
b and p are sugar amounts for brownies and pound cake respectively, in grams,
equation: p = b + 270
solution: p = 620
EXAMPLE 9.3.2. The size of a compressed file is 1.74 MiB, while the size of the original
uncompressed file is 5.5 times greater. What is the size of the uncompressed file?
SOLUTION: We can restate the given informationas follows: the size of the uncompressed
file is 5.5 times greater than the size of the compressed file:
u = 5.5c
ANSWER:
u and c are the sizes of the uncompressed and the compressed file respectively, in MiB,
equation: u = 5.5c
solution: u = 9.57
90
CHAPTER 1. CONCEPTS 9. TRANSLATION
Homework 1.9.
Describe the variables used to represent the 15. The upload speed in bytes per second
given quantities, and write the statements is 10 times slower than twice the download
as algebraic relations. speed.
1. Suzy is 5 years older than Ron. 16. Nancy’s wage in dollars per hour is 1.2
times greater than three times Phil’s wage.
2. A buffalo is 4 times heavier than a tiger.
3. The radius of Jupiter is 11 times greater
than the radius of the Earth. Solve applications by describing the vari-
ables used to represent the quantities of in-
4. A puppy is 6 pounds heavier than a kit-
terest, and stating appropriate equations.
ten.
17. Margo is 5 years old and Ivan is 34
5. The air is 30 ◦ F cooler than the water.
years older than Margo. How old is Ivan?
6. A donut is 2 dollars cheaper than a tea.
18. The temperature in Boston is −3 ◦ C
and the temperature at the South Pole is 57
7. The highway is twice as wide as the al- ◦
C lower than the temperature in Boston.
ley.
What is the South Pole temperature?
8. In winter, the day is twice as short as the
19. Alice runs 1.9 times faster than Bob,
night.
who runs at 4.7 mph. How fast does Alice
9. Abby is 7 inches shorter than Kurt. run?
10. A bicycle is 4 times faster than a run- 20. In February 2017 the US House of Rep-
ning person. resentatives had 362 congressmen. There
were 286 fewer women in the House. How
11. Jeff’s pool is one and a half times many congresswomen were there?
longer than Kate’s pool.
21. Tao’s first car was 3 times cheaper than
12. The postage this year is 2 cents more his second car, and he bought his second
expensive than it was the last year. car for 7899 dollars. How much was Tao’s
first car?
13. The perimeter of a rectangle is 2 cm
greater than 4 times the length of the 22. The distance by car from Sacramento
shorter side. to San Francisco is 88 miles. The distance
by car from Sacramento to Cool is 2.2 times
14. The price of a nail is 1.5 dollars shorter than the distance to San Francisco.
cheaper than one tenth of the price of a How long is the trip from Sacramento to
hammer. Cool?
91
9. TRANSLATION CHAPTER 1. CONCEPTS
Homework 1.9 Answers.
1. s is Suzy’s age in years, r is Ron’s age in 15. u is the upload speed in bps, d is the
years, s = r + 5 download speed in bps, u = (2u)/10
3. J is the radius of Jupiter in meters, E is
the radius of the Earth in meters, J = 11E 17.
M and I are Margo’s and Ivan’s ages respec-
5. a is the temperature of the air in ◦ F, w tively, in years,
is the temperature of the water in ◦ F, a = equation: I = M + 34
w − 30 solution: I = 39
7. h is the width of the highway in meters,
a is the width of the alley in meters, h = 2a 19.
A and B are Alice’s and Bob’s speeds respec-
9. a is Abby’s height in inches, k is Kurt’s tively, in mph,
height in inches, a = k − 7 equation: A = 1.9B
solution: A = 8.93
11. J is the length of Jeff’s pool in feet, K is
the length of Kate’s pool in units of length,
J = 1.5K 21.
f and s are the prices of the first and the
13. P is the perimeter of the rectangle in second cars respectively, in dollars,
cm, x is the length of the shorter side in equation: f = s/3
cm, P = 4x + 2 solution: f = 2633
92
CHAPTER 1. CONCEPTS 9. PRACTICE TEST 1
Practice Test 1
1. Identify the terms in the expression 11. Simplify the expression assuming that
3 + 4x − 5(x + y) all variables are non-zero:
8x 6x
2. Identify the factors in the expression ÷
ya 5y
−3x( y − 1)( y + 1)
12. Find the LCM for 10, 16.
3. Use the distributivity to rewrite the ex-
13. Simplify the expression:
pression without parentheses:
−3( y − 4x + 1) 3 5 3
+ ·
4 6 2
4. Use the distributivity to rewrite the ex-
14. Evaluate the expression if
pression as a product with 2 factors:
16x − 12 y − 2 a = −1 and b = 7
a5
b
5. List at least four elements of the given ÷ 4−
set using the roster notation: 3 3
3 − 2x x is a negative integer 15. Simplify the expression by combining
like terms:
6. Write the number 36 as a product of
2
1
prime factors. − 3x − y + (20x + 7 y)
5 5
7. Rewrite the fraction 20/32 in lowest 16. Describe the variables used to represent
terms. the given quantities, and write the state-
ments as algebraic relations.
8. Simplify the expression by combining
like terms: The width of a rectangular room is 4 feet
2(8 − y) − 3( y − 5) shorter than its length.
9. Simplify the expression: 17. Solve the application by describing the
14 − |5 − 7| variables used to represent the quantities of
30 − 3 · 23 interest, and stating appropriate equations.
10. Evaluate the expression if x = −4 Alex runs 1.1 times faster than Beth, who
x −5 runs at 2.8 miles per hour. Find how fast
1 − 2x Alex runs.
93
9. PRACTICE TEST 1 CHAPTER 1. CONCEPTS
Practice Test 1 Answers.
1. 3 terms: 3, 4x, −5(x + y) 12. 80
2. 4 factors: −3, x, ( y − 1), ( y + 1)
13. 2
3. −3 y + 12x − 3
14. −1/5
4. 2(8x − 6 y − 1)
9
5. {5, 7, 9, 11, . . .} 15. x + y
5
6. 22 · 32
16. w and l and width and length respec-
7. 5/8 tively, in feet.
w= l −4
8. 31 − 5 y
9. 2 17. A and B are the respective speeds in
miles per hour.
10. −1
A = 1.1 · B
20
11. A = 3.08
3a
94
CHAPTER 2
Linear Equations
1. Properties of Equations
1.1. Equations and Solutions. Recall that an equation is a statement about equality of
two expressions, and it has to be either true or false. When an equation has no variables, we
can simplify the expressions and compare the numbers to find out whether the equations holds.
3(4 − 5) = 5−8
12 − 15 = −3
−3 = −3
So the equation above was true all along.
When an equation contains a variable, it will typically be true for some values of the variable,
and false for some others. Any value for which the equation is true will be known as its solution.
DEFINITION 1.1.1. A number is called a solution for the equation involving one variable if
substituting this number for the variable makes the equation true. The set of all solutions
for the equation is called its solution set. To solve the equation means to find its solution
set.
EXAMPLE 1.1.1. Determine if −7 is a solution for the equation
3 − x = 5x
SOLUTION: Substitute (−7) for x everywhere in the equation and simplify both sides:
3− x = 5x
3 − (−7) = 5(−7)
3+7 = −35
10 = −35
The equation is false, so −7 is not a solution.
ANSWER: No
95
1. PROPERTIES OF EQUATIONS CHAPTER 2. LINEAR EQUATIONS
EXAMPLE 1.1.2. Determine if 6 is a solution for the equation
1
w−4= w
3
SOLUTION: Substitute 6 for w everywhere in the equation and simplify both sides:
1
w−4 = w
3
1
(6) − 4 = (6)
3
2 = 2
The equation is true, so 6 is a solution.
ANSWER: Yes
DEFINITION 1.1.2. Equations are equivalent if they have the same solutions sets.
We will typically solve equations by finding equivalent equations with known solution sets.
For example, simplifying an expression cannot change its value, so any algebraic simplification
produces an equivalent equation.
EXAMPLE 1.1.3. Solve the equation x = 2(3 − 5)
SOLUTION: The following equations are equivalent:
x = 2(3 − 5)
x = 2(−2)
x = −4
The last equation is true when x is −4, and false otherwise. We will state the answer as a
set with one element, using the roster notation.
ANSWER: {−4}
96
CHAPTER 2. LINEAR EQUATIONS 1. PROPERTIES OF EQUATIONS
1.2. Addition Property for Equations.
THEOREM 1.2.1. Adding the same number or expression to both sides of an equation
produces an equivalent equation. Formally, for all expressions A, B, and C the following
two equations are equivalent:
A = B
A+ C = B+C
Since every number has an opposite, and subtracting a number means adding its oppo-
site, we can also subtract the same number on both sides of an equation, and obtain an
equivalent equation:
A− C = B − C
EXAMPLE 1.2.1. Solve the equation x − 3 = 10
SOLUTION: We try to isolate x on the left side by canceling −3:
x −3 = 10
x −3+3 = 10 + 3 added 3 to both sides
x = 13 combined like terms
ANSWER: {13}
EXAMPLE 1.2.2. Solve the equation 4x = 3x
SOLUTION: We don’t know what number 3x stands for, but we know it is a number, and
every number has an opposite, so we can add −3x to both sides, which is the same as
subtracting 3x.
4x = 3x
4x − 3x = 3x − 3x added −3x to both sides
x = 0 combined like terms
ANSWER: {0}
97
1. PROPERTIES OF EQUATIONS CHAPTER 2. LINEAR EQUATIONS
1.3. Multiplication Property for Equations.
THEOREM 1.3.1. If C is an expression denoting a non-zero real number, then multiplying
both sides of an equation by C produces an equivalent equation. Formally, for all expres-
sions A and B, and all non-zero numbers C the following two equations are equivalent:
A = B
AC = BC
Since every non-zero number has a reciprocal, and dividing by a number means multi-
plying by its reciprocal, we can also divide both sides by the same non-zero number, and
obtain an equivalent equation:
A B
=
C C
Multiplying both sides of an equation by zero almost guarantees a non-equivalent result. For
example, the equation x = 3 has a single solution, while the equation 0 · x = 0 · 3 is true for
every real number x, and so has infinitely many solutions.
EXAMPLE 1.3.1. Solve the equation 5x = 75
SOLUTION: We would like to get rid of 5 on the left so that x is isolated, so we divide
both sides by 5, which is the same as multiplying both sides by 1/5:
5x = 75
1 1
(5x) = (75)
5 5
x = 15
ANSWER: {15}
3 6
EXAMPLE 1.3.2. Solve the equation − x =
7 5
SOLUTION: We would like to isolate x on the left side, so we need to cancel the coefficient
−3/7. Multiply both sides by −7/3, which is the reciprocal of −3/7:
3 6
− x =
7 5
7 3 7 6
− − x = −
3 7 3 5
98
CHAPTER 2. LINEAR EQUATIONS 1. PROPERTIES OF EQUATIONS
On the left, reciprocals cancel. On the right we reduce the fraction to lowest terms by
canceling the common factor 3.
7 · (2 · 3)
x = −
3·5
14
x = −
5
14
§ ª
ANSWER: −
5
EXAMPLE 1.3.3. Solve the equation 6x − 4 = 20
SOLUTION: Addition and multiplication properties may be exploited any number of times
and in any order. Here we can isolate x on the left side if we add 4 first, and then divide
both sides by 6.
6x − 4 = 20
6x − 4 + 4 = 20 + 4 addition property
6x = 24 combined like terms
6x 24
= multiplication property
6 6
x = 4
ANSWER: {4}
1.4. Applications. We will solve applications by describing the variables used to represent
the quantities of interest, and stating appropriate equations. As you construct your own solu-
tions to applications, you may be using different variable names, and your equations may also
be somewhat different, depending on how you translate English sentences. The final numerical
answers, however, should always come out the same, regardless of translation.
EXAMPLE 1.4.1. The profit of ACME Corporation in the second quarter was 8.84 billion
dollars, which was 1.3 times greater than the profit in the first quarter. What was ACME’s
profit in the first quarter?
99
1. PROPERTIES OF EQUATIONS CHAPTER 2. LINEAR EQUATIONS
SOLUTION: Let f and s be the profits in the first and second quarter respectively, measured
in billions of dollars. Then we can translate the statement above as “s is 1.3 times greater
than f ” or
s = 1.3 f
Since s = 8.84, we can substitute it and solve the resulting equation for f :
8.84 = 1.3 f
8.84 1.3 f
=
1.3 1.3
6.8 = f
ANSWER:
f and s are the first and the second quarter profits respectively, in billions of dollars,
equation: s = 1.3 f
solution: f = 6.8
EXAMPLE 1.4.2. The width of a rectangle is 125 meters shorter than the length. Find the
length if the width is 205 meters.
SOLUTION: Let w and l be the width and the length respectively, measured in meters. Then
the statement “The width of a rectangle is 125 meters shorter than the length” translates
as “w is 125 meters less than l” or
w = l − 125
Since w = 205, we can substitute it and solve the resulting equation for l:
205 = l − 125
205 + 125 = l − 125 + 125
330 = l
ANSWER:
w and l are the width and the length respectively, in meters,
equation: w = l − 125
solution: l = 330
100
CHAPTER 2. LINEAR EQUATIONS 1. PROPERTIES OF EQUATIONS
Homework 2.1.
Determine whether the given set is a possi- 1 2
21. a − =−
ble solution set for the given equation. 6 3
1. {4}, 5x + 7 = 29 x 2
22. − =
6 9
2. {8}, 6 − x = −2
2 5
23. x − =−
3. {−2}, x +7=3− x 3 6
n 2 3
4. {4}, −3 = 5 − 24. − + y = −
2 3 4
5. {17}, 5x − 10 = 5(x − 2) 3
25. x = 18
4
6. {−100}, 5x − 10 = 5(x − 2)
26. −8.2x = 20.5
7. {0}, x − 4x = 3x + 1
2
27. −6 = y + 25
8. {0}, (1 − x) = x + 1
2
28. −6 = x + 9
9. {−1, 9}, |x − 4| = 5
29. x − 4 = −19
10. {−3}, |3x| = −9
30. t − 7.4 = −12.9
31. −x = 28
Solve equations and check solutions.
32. −t = −8
11. 7x = 56
33. 12 = −7 + y
12. 12x = 72
1 8
34. x + =
13. x + 21 = 10 3 3
14. x + 12 = −7 3 3
35. − y = −
5 5
15. 4.5 + x = −3.1
2 4
36. − x = −
16. −15x = −20 5 15
37. m − 2.8 = 6.3
17. 400 = −x
38. y − 5.3 = 8.7
18. −2x = −300
4
19. −7x = 49 39. x = 16
5
y
20. = 11
8
101
1. PROPERTIES OF EQUATIONS CHAPTER 2. LINEAR EQUATIONS
3 number of US astronauts. Find the num-
40. x = 27
4 ber of ISS crew members from the United
States.
41. 2x + 3 = 13
51. According to NASA, the global land-
42. 3x − 1 = 26
ocean temperature index (average surface
temperature of Earth) in 2016 is 1.26 ◦ C
43. 2t + 9 = 43
greater than it was in 1920. Find the aver-
44. −5 y + 7 = −18 age temperature in 1920 if it was 0.99 ◦ C
in 2016.
45. 84 = 7x − 7
52. When Kyle went to Europe, his ticket
46. 50 = 9t + 1 to Berlin was 1.25 times cheaper than the
return ticket to New York. How much was
the return ticket if the ticket to Berlin cost
$480?
Solve applications by describing the vari-
ables used to represent the quantities of in- 53. Damian’s age is one year less than
terest, and stating appropriate equations. twice the Colin’s age. How old is Colin, if
Damian is 17 years old?
47. The payout for playing a single num-
ber on a roulette table is 35 times greater 54. Mary spends 14 minutes on her morn-
than the bet. Cindy receives the payout of ing commute, which is 5 minutes shorter
$3850. How much was Cindy’s bet? than three times the duration of her
evening commute. Find how much time
48. A croissant has 53 more calories than she spends on her evening commute.
a donut. Find the amount of calories in one
donut if each croissant has 445 calories. 55. Bob says to Alice: if you give me 3 ap-
ples and then take half of my apples away,
49. A history textbook costs $90, which is then I will be left with 13 apples. How
five times cheaper than an anatomy text- many apples do I have now?
book. Find the price of the anatomy text.
56. Charlie says to Diane: if I give you 5 of
50. As of July 28, 2017 Russia contributed my apples and then also half of the remain-
39 crew members for the International ing amount, then I will have 10 apples left.
Space Station, which is 12 fewer than the How many apples do I have now?
102
CHAPTER 2. LINEAR EQUATIONS 1. PROPERTIES OF EQUATIONS
Homework 2.1 Answers.
1. No 39. {20}
3. Yes 41. {5}
5. Yes 43. {17}
7. No 45. {13}
9. Yes 47.
b and p are the bet and the payout amounts
11. {8} respectively, in dollars,
equation: p = 35b
13. {−11} solution: b = 110
15. {−7.6} 49.
h and a are the prices of the history text
17. {−400} and the anatomy text respectively, in dol-
lars,
19. {−7} a
equation: h =
5
§ ª
1 solution: a = 450
21. −
2
51.
1 x and y are the 1920 and the 2016 tem-
§ ª
23. − peratures respectively, in ◦ C,
6
equation: y = x + 1.26
25. {24} solution: x = −0.27
27. {−31} 53.
D and C are the ages of Damian and Colin
29. {−15} respectively, in years,
equation: D = (2C) − 1
31. {−28} solution: C = 9
33. {19} 55.
b and a are the amounts of apples Bob has
35. {1} before and after the exchange respectively,
equation: (b + 3)/2 = a
37. {9.1} solution: b = 23
103
2. SOLVING LINEAR EQUATIONS CHAPTER 2. LINEAR EQUATIONS
2. Solving Linear Equations
2.1. Linear and Non-linear Equations.
DEFINITION 2.1.1. A linear equation is an equation in which each term is either a numerical
constant or the product of a numerical constant and the first power of a single variable.
BASIC EXAMPLE 2.1.1. Here are some linear and some non-linear equations.
A linear equation in one variable x:
3 + x = 4x − 5
A linear equation in two variables x and y:
y = 4x − 17
A linear equation in three variables x, y, and z:
1
x − 5 y + z = 14
2
The following equations are non-linear because they have non-linear terms, highlighted
below. This one has a second power of a variable, when only first power is allowed:
1 − x2 = 4x
This one has a term with multiple variable factors, when at most one is allowed:
5z = 3xy + 7x
This one has an absolute value term, which is neither a numerical constant nor a variable
product:
|x − 4| = 2x + 7
This equation is non-linear, but a few algebraic operations can show it is equivalent to a
linear equation:
2(4 − x) = 3(x + 5)
8 − 2x = 3x + 15 distributed
This equation is also non-linear, but amazingly it is also equivalent to a linear equation:
5x2 + 1 = 2x(1 + 2x) + x2
5x2 + 1 = 2x + 4x2 + x2 distributed
5x2 + 1 = 2x + 5x2 combined like terms
5x + 1 − 5x
2 2
= 2x + 5x − 5x
2 2
subtracted 5x 2 on both sides
1 = 2x combined like terms
104
CHAPTER 2. LINEAR EQUATIONS 2. SOLVING LINEAR EQUATIONS
2.2. Isolating The Variable. Here we present a procedure for solving linear equations as
well as many types of non-linear equations, as long as they turn out to be equivalent to linear
equations. It is called isolating the variable because it boils down to finding an equivalent
(Definition 1.1.2) equation of the form
x=c
where x is the unknown variable and c is a number. Here x is isolated on the left side in the
sense that it is there all alone (no other terms, no coefficient), and there is no x on the other
side. Even though it is traditional to isolate x on the left side, there is nothing wrong with
isolating it on the right: c = x.
THEOREM 2.2.1 (Isolating The Variable). The following procedure will isolate the variable
x in linear and some non-linear equations.
(1) If the equation contains fractions, multiply both sides by the LCD. Cancel com-
mon factors in each fraction to get rid of denominators.
(2) Use the distributive property to get rid of parentheses, and then combine the like
terms on both sides.
(3) Use the addition property to cancel the variable terms on one side and the con-
stant terms on the other side of the equation, and combine the like terms again.
By now the equation should have assumed one of the three forms:
• If the equation looks like ax = b then divide both sides of the equation by the
coefficient a. Now it looks like x = c and c is the unique solution. The set of
solutions can be written as {c}.
• If the equation looks like 0 = 4 or something equally wrong, then it is false for
all x. There are no solutions: the set of solutions is empty, or ∅.
• If the equation looks like 0 = 0, then it is true for all x. Every real number is a
solution: the set of solutions is the set of all reals, or R.
When a unique solution exists, it can be checked by substituting it for the variable in the original
equation and making sure the equation holds. When the solution set is R, it can be partially
checked by substituting a few random numbers for the variable and making sure the equation
holds for all of them.
EXAMPLE 2.2.1. Solve the equation
3(x − 3) + 1 = 4 + x
SOLUTION: No fractions here, so we start by removing parentheses and combining the
like terms:
3x − 9 + 1 = 4+ x distributivity
3x − 8 = 4+ x
105
2. SOLVING LINEAR EQUATIONS CHAPTER 2. LINEAR EQUATIONS
Then use the addition property of the equation to cancel the terms with x on one side,
and the constants on the other side, and combine the like terms again:
3x − 8 − x + 8 = 4+ x − x +8
2x = 12
Finally, we divide both sides by the coefficient of the variable term and obtain a solution:
2x 12
=
2 2
x = 6
To check the answer, we substitute 6 for x in the original equation and simplify:
3(x − 3) + 1 = 4+ x
3(6 − 3) + 1 = 4+6
3(3) + 1 = 10
10 = 10
The equation holds, so 6 is a solution.
ANSWER: {6}
EXAMPLE 2.2.2. Solve the equation
4x − (3 + x) = −3(1 − x)
SOLUTION: No fractions here, so we remove the parentheses and combine the like terms:
4x − 3 − x = −3 + 3x distributivity
3x − 3 = −3 + 3x
Now we cancel variable terms on one side and constants on the other side, and combine
the like terms again:
3x − 3 + 3 − 3x = −3 + 3x + 3 − 3x
0 = 0
This equation is true for all values of x. In other words, any real number makes the
equation true when substituted for x.
ANSWER: R
106
CHAPTER 2. LINEAR EQUATIONS 2. SOLVING LINEAR EQUATIONS
EXAMPLE 2.2.3. Solve the equation
2(7 + 5x) = 1 − (3 − 10x)
SOLUTION: No fractions here, so we start by removing parentheses and combining the
like terms:
14 + 10x = 1 − 3 + 10x
14 + 10x = −2 + 10x
After that we use the addition property to eliminate variable terms on one side and con-
stants on the other side, and combine the like terms again:
14 + 10x − 10x − 14 = −2 + 10x − 10x − 14
0 = −16
This equation is false for all values of x. In other words, no real number can make the
original equation true. There are no solutions.
ANSWER: ∅
EXAMPLE 2.2.4. Solve the equation
1 2 1
x +4= − x
3 3 2
SOLUTION: Here the LCD is 6, so we start by multiplying both sides by the LCD:
1 2 1
6 x +4 =6 − x
3 3 2
As we distribute, the fractions cancel:
1 2 1
6· x +6·4 = 6· −6· x
3 3 2
2x + 24 = 4 − 3x
Now we eliminate variable terms on one side and constants on the other side, combine
the like terms, and use the multiplication property to isolate the x variable:
2x + 24 + 3x − 24 = 4 − 3x + 3x − 24
5x = −20
5x −20
= divided both sides by 5
5 5
x = −4
107
2. SOLVING LINEAR EQUATIONS CHAPTER 2. LINEAR EQUATIONS
To check the answer, we substitute −4 for x in the original equation and simplify:
1 2 1
x +4 = − x
3 3 2
1 2 1
· (−4) + 4 = − · (−4)
3 3 2
−4 2
+4 = +2
3 3
−4 12 2 6
+ = + find LCD to add fractions
3 3 3 3
8 8
=
3 3
The equation holds, so −4 is a solution.
ANSWER: {−4}
EXAMPLE 2.2.5. Solve the equation
41 5 2 1
= x+ − x
9 2 3 3
SOLUTION: Here we have some fractions locked up inside the parentheses, so let us
distribute before we try to figure out the LCD:
41 5 2 1
= x+ − x
9 2 3 3
41 5 5 2 1
= x+ · − x
9 2 2 3 3
41 5 10 1
= x+ − x
9 2 6 3
The LCD is 18, so we multiply both sides of the equation by 18 and simplify until all the
denominators cancel:
41 5 10 1
18 = 18 x+ − x multiply both sides by LCD
9 2 6 3
18 · 41 18 · 5 18 · 10 18 · 1
= x+ − x distributed on the right side
9 2 6 3
82 = 45x + 30 − 6x simplified fractions
82 = 39x + 30 combined like terms
108
CHAPTER 2. LINEAR EQUATIONS 2. SOLVING LINEAR EQUATIONS
Now we finish solving by isolating the variable:
82 = 39x + 30
82 − 30 = 39x + 30 − 30
52 = 39x
52 39x
=
39 39
4
= x
3
We can check the solution by substituting it into the original equation and making sure it
holds:
41 5 2 1
= x+ − x
9 2 3 3
41 5 4 2 1 4
= + − ·
9 2 3 3 3 3
41 5 6 4
= −
9 2 3 9
41 30 4
= −
9 6 9
41 4
= 5−
9 9
41 45 4
= −
9 9 9
41 41
=
9 9
4
§ ª
ANSWER:
3
109
2. SOLVING LINEAR EQUATIONS CHAPTER 2. LINEAR EQUATIONS
2.3. Applications. We will solve applications by describing the variables used to represent
the quantities of interest, and stating appropriate equations. As you construct your own solu-
tions to applications, you may be using different variable names, and your equations may also
be somewhat different, depending on how you translate English sentences. The final numerical
answers, however, should always come out the same, regardless of translation.
EXAMPLE 2.3.1. Alice works two jobs. Her full-time job pays four times more than her
part-time job, and together they pay 3800 dollars per month. Find how much each job
pays.
SOLUTION: Let p be the amount the part-time job pays and let f be the amount the
full-time job pays in dollars per month. Let us translate the given statements:
full-time job pays 4 times more than part-time job
f = 4p
And the statement ”together both jobs pay $3800” can be written as
full-time pay + part-time pay = total pay
f + p = 3800
Since f = 4p, we can substitute (4p) for f into the second equation and obtain a linear
equation in one variable:
f +p = 3800
(4p) + p = 3800 substitute (4p) for f
5p = 3800 combine like terms
5p 3800
= multiplication property
5 5
p = 760
Finally, we substitute 760 for p in the first equation to find that
f = 4p = 4(760) = 3040
To check the answer, we can verify that the two paychecks add up to the stated total:
760 + 3040 = 3800
ANSWER:
p and f are part-time and full-time salaries respectively, in dollars
f = 4p
f + p = 3800
solution: p = 760, f = 3040
110
CHAPTER 2. LINEAR EQUATIONS 2. SOLVING LINEAR EQUATIONS
EXAMPLE 2.3.2. Rita bought rice and truffles a grocery store, and she paid $67.5 more for
truffles than she paid for the rice. Find the price of each item if the total bill was $72.
SOLUTION: Let r and t be the prices in dollars of the rice and the truffles respectively. It
may be helpful to rephrase given statements to make them easier to translate:
she paid $67.5 more for truffles than she paid for the rice
the price of truffles is 67.5 dollars greater than the price of rice
t = r + 67.5
And the other statement
the total bill was $72
can be translated as
r + t = 72
Since t = r + 67.5, we can substitute (r + 67.5) for t in the last equation, and then solve
for r:
r+t = 72
r + (r + 67.5) = 72
2r + 67.5 = 72 combined like terms
2r + 67.5 − 67.5 = 72 − 67.5 addition property
2r = 4.5 combined like terms
2r 4.5
= multiplication property
2 2
r = 2.25
Finally we use the first equation find that
t = r + 67.5
t = 2.25 + 67.5
t = 69.75
ANSWER:
r and t are the prices of rice and truffles respectively, in dollars
t = r + 67.5
r + t = 72
solution: r = 2.25, t = 69.75
111
2. SOLVING LINEAR EQUATIONS CHAPTER 2. LINEAR EQUATIONS
EXAMPLE 2.3.3. Alice, Bob, and Charlie stand to inherit 12000 dollars, and the will stipu-
lates that the inheritance is to be split in such a way that Bob gets two times more money
than Alice, while Charlie should get 1600 dollars less than Alice. Find the inheritance
amount for each person.
SOLUTION: Let a, b, and c be the amounts in dollars inherited by Alice, Bob, and Charlie
respectively. Bob gets twice as much as Alice, which we can write as
b = 2a
Charlie gets 1600 dollars less than Alice, which we can write as
c = a − 1600
Together, their amounts should add up to the total, so we can write this equation:
Alice’s share + Bob’s share + Charlie’s share = total inheritance
a + b + c = 12000
a + (2a) + (a − 1600) = 12000
This is a linear equation in one variable, so we can solve it the usual way. There are no
fractions here, so we start by removing parentheses and combining the like terms:
a + 2a + a − 1600 = 12000 remove parentheses
4a − 1600 = 12000 combined like terms
4a = 12000 + 1600 added 1600 on both sides
4a = 13600 combined like terms
4a 13600
= multiplication property
4 4
a = 3400
So Alice’s share is 3400 dollars, Bob’s share is 2a = 2 · 3400 = 6800 dollars, and Charlie’s
share is (a − 1600) = 3400 − 1600 = 1800 dollars. We can see that the original equation
holds, since the individual shares add up to the total:
3400 + 6800 + 1800 = 12000
ANSWER:
a, b, and c are inheritance amounts in dollars for Alice, Bob, and Charlie respectively,
b = 2a
c = a − 1600
a + b + c = 12000
solution: a = 3400, b = 6800, c = 1800
112
CHAPTER 2. LINEAR EQUATIONS 2. SOLVING LINEAR EQUATIONS
EXAMPLE 2.3.4. Katy invested some money into bonds, and the amount of interest earned
after one year was 25 times smaller than the original investment. Find the interest and
original investment if the total amount in her account is $546.
SOLUTION: Let P be the investment (also called principal) and let I be the interest, in
dollars. The interest is 25 times smaller:
I = P ÷ 25
The total amount, which is the principal plus the interest, is $546:
P + I = 546
Substituting P ÷ 25 for I in the second equation gives us an equation in one variable we
can solve:
P+I = 546
P
P+ = 546 multiply both sides by LCD
25
P
25 P + = 25(546) to get rid of fractions
25
25P
25P + = 25(546) distributed
25
25P + P = 25(546) simplified fraction
26P = 25(546) combined like terms
26P 25(546)
= multiplication property
26 26
P = 525
Finally, substituting 525 for P in the first equation gives us
I = P ÷ 25 = 21
ANSWER:
P and I are the principal and the interest respectively, in dollars
I = P ÷ 25
P + I = 546
solution: P = 525, I = 21
113
2. SOLVING LINEAR EQUATIONS CHAPTER 2. LINEAR EQUATIONS
Homework 2.2.
Solve each equation: 3 21
22. (1 + p) =
5 20
1. 4x = 6x − 28
23. 2( y − 3.5) = −2(14 − y)
2. −2 = −2m + 12
9 11
24. 2b + =−
3. −2z − 10 = 5z + 18 5 5
n 2x − 17 17 − 2x
4. 5 + =4 25. =
4 5 10
5. 0.2x − 0.1 = 0.6x − 2.1 5 5 3
26. − = r−
x 8 4 2
6. −11 = −8 +
2 7 3
27. z + = −z
x x x 5 5
7. − = +1
2 3 6
28. −2(1 − 7p) = 8(p − 7)
8. 2 − (−3a − 8) = 1
29. 7( y + 3) − 5 = 11 y − 4(1 + y)
9. −5(4x − 3) + 2 = −20x + 17
30. −8n − 19 = −2(8n − 3) + 3n
10. 66 = 6(6 + 5x)
3(z − 5) 2z + 10
1.37 − x 31. =
11. − 1.29 = −1 2 3
4
45 3 7 19
32. + n= n−
12. −16n + 12 = 39 − 7n 16 2 4 16
13. 3( y − 1) = y + 6 33. 0.25(6t + 2) − 10 = 3t − 7.1
14. −(3 − 5n) = 12 34. −76 = 5(1 + 3b) + 3(3b − 3)
15. −4x + 10 = −2(3x + 1) 0.6x 6 − 0.15x
35. +2=
−12 3
16. 2(4x − 4) = −20 − 4x
11 3 163
36. + r=
15 − 20x 4 4 32
17. −(4x − 3) =
5
1 29 4 2
37. n + =2 n+
18. −(n + 8) + n = −8n + 2(4n − 4) 3 6 3 3
19. 0.4x + 0.2(10 − x) = 3 1 2 3 7 83
38. − x− − x =−
2 3 4 2 24
20. −6(8k + 4) = −8(6k + 3) − 2
z+2
21. = 2z + 3
3
114
CHAPTER 2. LINEAR EQUATIONS 2. SOLVING LINEAR EQUATIONS
Solve applications by describing the vari- was invested and how much interest was
ables used to represent the quantities of in- earned if the total amount in the account
terest, and stating appropriate equations. is $1, 785
39. An éclair has five times as many calo- 45. A monthly pass for light rail is 1.2
ries as a latte, and together they have 264 times cheaper than the a bundle with 20
calories. Find how many calories each item day passes. Find the price of the monthly
has. pass and the price of the bundle, if the bun-
dle costs $10 more than the monthly pass.
40. Cassie has 74 more stickers than
Margo, and together they have 400 stick- 46. Annie made $270 more in tips in June
ers. Find how many stickers each person than she did in May. Find how much An-
has. nie made in tips in either month if the
amount in June was 1.5 times greater than
41. The width of a rectangular parking lot the amount in May.
is 40 feet shorter than its length. Find the
dimensions of the lot if its perimeter is 640 47. A bread recipe calls for a mix of
feet. bleached flour, whole wheat flour, and rye
flour. The volume of whole wheat should
42. The width of a rectangular computer be two times greater than the volume of
screen is 16/9 times greater than its height. bleached, and the volume of rye should be
Find the height of the screen if the width is 1 cup less than the volume of bleached.
1920 pixels. Find the volume in cups for each type of
flour if the recipe is asking for 15 cups of
43. Consider an astronomical triangle flour in total.
with sides a, b, and c, such that the length
of the side b is 3 light-years longer the 48. Three paper supply company employ-
length of a, and the length of c is 6 times ees: Jim, Pam, and Leslie, made a bet about
shorter than the length of a. Find the who can make the most sales in a month.
length of each side if the perimeter of the Jim won the bet, with Pam making 6 fewer
triangle is 55 light-years. sales than Jim, and Leslie making 7 fewer
sales than Jim. Find how many sales each
44. The interest earned by John’s savings employee has made, if the total amount of
account in 2017 is 50 times smaller than sales for the month was 113.
his original investment. Find how much
115
2. SOLVING LINEAR EQUATIONS CHAPTER 2. LINEAR EQUATIONS
Homework 2.2 Answers.
1. {14} 39.
L and E are calorie amounts in a latte and
3. {−4} an éclair respectively
E = 5L
5. {5} E + L = 264
7. ∅ solution: L = 44, E = 220
41.
9. R
w and l are width and length respectively,
in feet
11. {0.21}
w = l − 40
9 2w + 2l = 640
§ ª
13.
2 solution: w = 140, l = 180
15. {−6} 43.
a, b, and c are the side lengths in light-
17. R years,
b = a+3
19. {5}
c = a÷6
a + b + c = 55
7
§ ª
21. −
5 solution: a = 24, b = 27, c=4
23. ∅ 45.
m and b the prices of the monthly pass and
25. {17/2} the bundle respectively, in dollars
m = b ÷ 1.2
1
§ ª
27. − b = m + 10
4
solution: m = 50, b = 60
29. ∅
47.
31. {13} b, w, and r are the amounts of bleached,
whole wheat, and rye flour respectively, in
33. {−1.6} cups,
w = 2b
35. R r = b−1
b + w + r = 15
3
§ ª
37.
2 solution: b = 4, w = 8, r =3
116
CHAPTER 2. LINEAR EQUATIONS 3. LINEAR FORMULAS
3. Linear Formulas
3.1. Isolating a Variable.
DEFINITION 3.1.1 (Formula). While the word formula has no traditional formal definition,
this will be our word of choice for equations in more than one variable.
BASIC EXAMPLE 3.1.1. Some formulas useful in applications:
P = 2l + 2w
is a formula for the perimeter P of a rectangle with length l and width w, and
V = lwh
is a formula for the volume V of a box with length l, width w, and height h.
DEFINITION 3.1.2. We say that a formula involving a variable y is solved for y if it is
written in the form
y = expression
where one side of the equation is just y, and the expression on the other side does not
contain y at all.
BASIC EXAMPLE 3.1.2. Here are some formulas which are solved for a variable, and some
that are not.
E = mc 2 solved for E
x = 10 y 3 − a b solved for x
4x 2 − 3x
y = +1 solved for y
5a
y = 7(4x − yz) not solved for y because y appears on the right side
y2 = x2 + 3 not solved for y because neither side is y by itself
x + 40 = y +4 not solved for x because neither side is x by itself
EXAMPLE 3.1.1. Solve the formula for x:
6 y − 7x − 10 = 0
SOLUTION: We will isolate the variable x by following the same procedure (Theorem
2.2.1) we used to solve linear equations. We just have to pretend that x is the only variable.
There are no parentheses, so we use the addition property to isolate the terms with x on
117
3. LINEAR FORMULAS CHAPTER 2. LINEAR EQUATIONS
one side:
6 y − 7x − 10 = 0
6 y − 7x − 10 − 6 y + 10 = 0 − 6 y + 10 addition property
−7x = −6 y + 10 combined like terms
Finally, we divide by the coefficient of the term with x:
−7x −6 y + 10
=
−7 −7
−6 y + 10
x =
−7
−6 y + 10
ANSWER: x =
−7
Note that the answer to the previous example is an equation, not a number. Note also that the
answer is not unique. In a solved formula such as
−6 y + 10
x=
−7
the right side can be stated in a variety of ways, all equivalent, and all correct:
−6 y + 10 10 − 6 y 6 y 10
x= = = − = ...
−7 −7 7 7
EXAMPLE 3.1.2. Solve the formula for z:
5z − x = 3(4 − z)
SOLUTION: z is locked up in parentheses on the right side, so use distributivity to get rid
of these parentheses first:
5z − x = 3(4 − z)
5z − x = 12 − 3z
Use the addition property to isolate terms with z on one side, and then combine like terms:
5z − x + x + 3z = 12 − 3z + x + 3z addition property
8z = 12 + x combined like terms
118
CHAPTER 2. LINEAR EQUATIONS 3. LINEAR FORMULAS
Finally, divide by the coefficient of the term with z:
8z 12 + x
=
8 8
12 + x
z =
8
12 + x
ANSWER: z =
8
EXAMPLE 3.1.3. Solve the photon energy formula for h:
hc
E=
λ
SOLUTION: There is only one term with h, and it is already isolated on the right side, so
we just need to use the multiplication property to get rid of the extra factors. Multiply
both sides by the reciprocal of each factor on the right side besides h:
hc
E =
λ
λ hc λ
(E) =
c λ c
Eλ hcλ
= cancel common factors on the right
c λc
Eλ
= h
c
Eλ
ANSWER: h =
c
EXAMPLE 3.1.4. Solve the formula for X :
aX = b(X − c)
SOLUTION: The variable X is locked up inside the parentheses on the right side, so we
distribute first:
aX = bX − bc
119
3. LINEAR FORMULAS CHAPTER 2. LINEAR EQUATIONS
Now we use the addition property to isolate the terms with X on the same side:
aX − bX = bX − bc − bX addition property
aX − bX = −bc combined like terms
We would like to combine the terms with X , but they are not, strictly speaking, similar,
having different variable factors. We can, however, distribute and then multiply by the
reciprocal of the resulting factor anyway:
X (a − b) = −bc distributivity
X (a − b) −bc
= assuming (a − b) 6= 0
a−b a−b
−bc
X =
a−b
−bc
ANSWER: X =
a−b
120
CHAPTER 2. LINEAR EQUATIONS 3. LINEAR FORMULAS
Homework 2.3.
Solve the formula for the given variable 15. 4x − 5 y = 8 for y
assuming that all variable expressions in-
volved are non-zero. 5
16. C = (F − 32) for F
9
1. a b = c for b
17. 4x − 7b = 4 for b
h
2. g = for h 18. 4x − 5 y = 8 for x
i
3. a + c = b for c 1 c
19. +b= for b
a a
4. x − f = g for x
20. lwh = V for w
5. P = n(p − c) for n
πr 2 h
21. V = for h
6. S = L − 2B for L 3
22. r t = d for r
7. 2m + p = 4m + q for m
23. ax + b = c for a
8. q = 6(L − p) for L
f 24. R = aT + b for T
9. x=b for x
g ym c
25. = for y
b d
3y
10. p = for y
q 26. E = mc 2 for m
4y
11. h = v t − 16t 2 for v 27. c = for y
m+n
12. S = πrh + πr 2 for h rs
28. =k for r
a−3
13. 3x + 2 y = 7 for y
29. 4x − y = 5(x + y) − 1 for x
14. 5a − 7b = 4 for a
30. 10(a + b) = 6(a − 2b) for a
121
3. LINEAR FORMULAS CHAPTER 2. LINEAR EQUATIONS
Homework 2.3 Answers.
c 4 − 4x
1. b = 17. b =
a −7
3. c = b − a
c 1
P 19. b = −
5. n = a a
p−c
3V
p−q 21. h =
7. m = πr 2
2
bg c−b
9. x = 23. a =
f x
h + 16t 2 cb
11. v = 25. y =
t dm
7 − 3x
13. y = c(m + n)
2 27. y =
4
8 − 4x
15. y = 29. x = 1 − 6 y
−5
122
CHAPTER 2. LINEAR EQUATIONS 4. PERCENT
4. Percent
4.1. Percent Units and Notation.
DEFINITION 4.1.1. 1 percent, written also as 1%, is one hundredth of a unit, or 0.01 in
decimal representation. The sentence “a part is k% of the whole” or
p is k% of w
can be written as
k
p= ·w
100
where p is the part of the whole w, and k/100 is the decimal representation of k%.
BASIC EXAMPLE 4.1.1. The decimal representation of k percent, which is k/100, may be
useful in equations because it is easier to manipulate algebraically. To convert k% into a
decimal form, simply divide k by 100. To convert the decimal into percent, multiply it by
100. For example,
k percent decimal representation
3% 0.03
35% 0.35
150% 1.5
0.6% 0.006
There are three easy question types which can be translated into the percent equation in terms
of the unknown variable x.
Finding the part, knowing the whole and the percentage:
What is k% of w?
k
x = · w
100
Finding the percentage, knowing the whole and the part:
p is what % of w?
x
p = · w
100
Finding the whole, knowing the percentage and the part:
p is k% of what number?
k
p = · x
100
123
4. PERCENT CHAPTER 2. LINEAR EQUATIONS
EXAMPLE 4.1.1. What is 14% of 50?
SOLUTION: This translates as
14
x= · 50 = 7
100
Alternatively, we can convert the percent into a decimal form and then solve a similar
equation where the fraction is replaced by a decimal:
x = 0.14 · 50 = 7
ANSWER: 7
EXAMPLE 4.1.2. The number 21 is what percent of 70?
SOLUTION: This question translates as
x
21 = · 70
100
We solve this equation for x:
100 x 100
(21) · = · 70 ·
70 100 70
30 = x
Note that x is already in percent.
ANSWER: 30%
EXAMPLE 4.1.3. The number 330 is 120% of what number?
SOLUTION: This question translates as
120
330 = ·x solve this equation for x
100
100 120 100
(330) · = ·x ·
120 100 120
275 = x
So 330 is 120% of 275.
ANSWER: 275
124
CHAPTER 2. LINEAR EQUATIONS 4. PERCENT
4.2. Percent Increase and Decrease.
DEFINITION 4.2.1 (Percent Increase and Decrease). We say that a quantity A is obtained
by increasing the quantity P by k% if
k
A = P+ P
100
Similarly, we say that A is obtained by decreasing P by k% if
k
A = P− P
100
Note that in both cases the percentage is applied to the old quantity in order to obtain the
new quantity, not the other way around.
In many cases we need to solve these equations for the old quantity P, and then it is more
convenient to use equivalent formulas, obtained by applying the distributive property to the
right sides:
A = P(1 + k/100) percent increase
A = P(1 − k/100) percent decrease
Percent increase and decrease formulas have many applications. Examples of percent increase
include
• the sales tax, when the total price is the original price plus the tax, which is a percent-
age of the original price
• the price markup, like when the price of a popular item goes up by so many percent
• the bank account or the investment amount increase due to accruing the interest,
given the simple interest rate k
Examples of percent decrease include
• the income tax, when the total income is the original amount minus the tax, which is
a percentage of the original income
• the price discount, like when a store puts up items for sale at k% off
125
4. PERCENT CHAPTER 2. LINEAR EQUATIONS
EXAMPLE 4.2.1. A department store puts up shirts on sale at 35% discount. Find the
original price of a shirt before the discount if the new price is $26.
SOLUTION: We will need to solve the percent decrease equation for P, which is the original
price in dollars. The new price A = 26, and the percent decrease is k = 35:
A = P(1 − k/100)
26 = P(1 − 35/100)
26 = P(1 − 0.35)
26 = 0.65P
(26)/0.65 = (0.65P)/0.65
40 = P
ANSWER: $40
EXAMPLE 4.2.2. A fast food chain raised the price of the beef burger from $5 to $6, and
lowered the price of the chicken burger from $6 to $5. Find the percent increase for the
price of the beef burger and the percent decrease for the price of the chicken burger.
SOLUTION: For the beef burger, the old price is P = 5 and the new price is A = 6 dollars,
so the percent increase equation is
k
A = P+ ·P
100
k
6 = 5+ ·5
100
We solve it for k, which is exactly the percent increase:
5k
6−5 = 5+ −5 subtracted 5 on both sides
100
5k
1 =
100
k
1 = lowest terms
20
1
1 · 20 = k · 20
20
20 = k
So the price of the beef burger went up by 20%.
126
CHAPTER 2. LINEAR EQUATIONS 4. PERCENT
The reader may be tempted to think that the price of the chicken burger went down by
20%, but this is not the case. If the old price is P = 6 and the new price is A = 5 dollars,
then the percent decrease equation we have to solve is
k
A = P− ·P
100
k
5 = 6− ·6
100
6k
5−6 = 6− −6 subtracted 6 on both sides
100
6k
−1 = − combined like terms
100
3k
−1 = − lowest terms
50
50 3k 50 3
(−1) − = − − multiplied both sides by reciprocal of −
3 50 3 50
50
= k
3
16 23 = k mixed numbers are great for answers
So the price of the chicken burger went down by 16 32 %.
ANSWER: Up by 20% and down by 16 32 % respectively
127
4. PERCENT CHAPTER 2. LINEAR EQUATIONS
Homework 2.4.
1. What is 6% of 14? much did the dealer pay for the van if the
price after the markup is $1189?
2. What is 9% of 15?
17. Alice invested $1230 into a savings
3. 4 is 25% of what number? account with 1.4% monthly interest rate.
What is the balance in the account after one
4. 15 is 5% of what number? month?
5. 20 is what percent of 50? 18. Bob invested $6500 into a mutual fund
with 3.5% monthly interest rate. What
6. 16 is what percent of 64?
is the balance in the account after one
month?
7. 100 is what percent of 10?
19. Brent paid $5824 for a car, and this
8. What is 350% of 20?
price included the 4% sales tax. How much
9. 10.5 is 5% of what number? did the car cost before the sales tax was ap-
plied?
10. 1.8 is what percent of 30?
20. A tea importer’s total revenue is
11. What is 0.4% of 5000? $7689.60, and this figure includes the 8%
sales tax which needs to be sent to the
12. 2.2 is 5.5% of what number? state. Find the total revenue without the
sales tax applied.
21. Roger paid $238 for a webcam during
Solve the applications using the percent in- a 15%-off sale. Find the original price of
crease and decrease formulas. the webcam.
13. Suzy earned $208 in tips after 20% in- 22. Sasha paid $68 for a painting frame
come tax. What was her pre-tax income? during a 20% sale event. Find the original
price of the frame.
14. Tom’s pre-tax earnings were $1170.
How much money did Tom get after pay- 23. Colin took a $275 pay increase with
ing 30% income tax? his promotion. What percent increase does
that represent if his old salary was $2250
15. A department store puts dresses up for per month?
sale at 30% discount. If the discounted
price of a dress is $42, what was the origi- 24. Rosie took a $700 pay cut when she
nal price before the discount? took a job which allowed her to work from
home. What percent decrease does that
16. A used car dealer buys an old van and represent if her new job pays her $2800 per
puts it up for sale with 45% markup. How month?
128
CHAPTER 2. LINEAR EQUATIONS 4. PERCENT
Homework 2.4 Answers.
1. 0.84 13. $260
3. 16 15. $60
5. 40% 17. $1247.22
7. 1000% 19. $5600
9. 210 21. $280
11. 20 23. 12 29 %
129
5. APPLICATIONS CHAPTER 2. LINEAR EQUATIONS
5. Applications
5.1. Distance and Work. This equation of motion at a constant speed is well known:
d = rt
Here d is the distance traveled, r is the constant speed, and t is the duration of the motion.
For example, a car moving at r = 45 miles per hour for t = 2 hours will cover the distance of
d = r t = 45·2 = 90 miles. Note that one has to be careful to make a choice of units and stick to
it throughout the equation. Here we went with miles for the distance, hours for the duration,
and miles per hour for the speed. If the duration was given in, say, minutes, we would have to
convert it to hours before using that number in the equation.
We can operate on units of measurement, the way physicists do, to make sure that our equation
makes sense. Substituting units for the quantities should produce an equation with the same
units on both sides. Note that the equations below are not legitimate algebraic equations,
and the square brackets indicate that we are looking at the relationship between the units of
measurement, rather than numbers:
[d] = [r t]
miles
miles = · hours hours cancel
hours
miles = miles
This idea generalizes readily for any kind of work done at a constant rate:
w = rt
In the case with motion, the work is simply the distance traveled. But we can also write this type
of equation whenever a variable changes at a constant rate over time, even if that variable is not
traditionally thought of as work. One great way to understand these problems is by thinking
about the units of the rate, because the units of its numerator will be the units of work.
[w] = [r t]
units of work
units of work = · units of time units of time cancel
units of time
units of work = units of work
EXAMPLE 5.1.1. Ivan can grade 25 quizzes per hour. Describe the variables in the equation
w = r t by stating their units of measurement, and find how many quizzes Ivan can grade
in 4 hours.
SOLUTION: The rate of work is measured in quizzes per hour, so we let w be the number
of quizzes graded, let r be the rate of grading in quizzes per hour, and let t be the duration
130
CHAPTER 2. LINEAR EQUATIONS 5. APPLICATIONS
of grading in hours. Given t = 4 hours, the number of quizzes Ivan can grade is
w = 25 · 4 = 100
ANSWER:
w is the number of quizzes,
r is in quizzes per hour, t is in hours,
w = 100
EXAMPLE 5.1.2. Taylor makes 18 dollars per hour at her job. Describe the variables in the
equation w = r t by stating their units of measurement, and find how many dollars Taylor
earns in 3.5 hours.
SOLUTION: The earning rate is measured in dollars per hour, so we let w be the dollars
earned, let r = 18 be the rate in dollars per hour, and let t be the duration of work in
hours. Note that in this application we do not refer to Taylor’s actual job as “work”. Given
3.5 hours, Taylor will earn this many dollars:
w = 18 · 3.5 = 63
ANSWER:
w is the amount earned in dollars,
r is in dollars per hour, t is in hours,
w = 63
EXAMPLE 5.1.3. Uri can walk 7 km in 2 hours. Describe the variables in the equation
w = r t by stating their units of measurement, and find how fast Uri walks.
SOLUTION: We need to find Uri’s walking speed in km per hour, so we let w be the distance
in km, let r be the speed in km per hour, and let t be the duration of the walk in hours.
We can plug in what we know and solve the equation for r:
w = rt
7 = r ·2
(7)/2 = (r · 2)/2
3.5 = r
So Uri’s walking speed is 3.5 km per hour.
131
5. APPLICATIONS CHAPTER 2. LINEAR EQUATIONS
ANSWER:
w is the distance walked in km,
r is in km per hour, t is in hours,
r = 3.5
5.2. Applications of w = r t. Many problems related to distance and work can be solved
by writing an equation for the amount of work completed. For example, when two different
vehicles complete the same trip, one moving with speed r1 for t 1 units of time, the other moving
with speed r2 for t 2 units of time, we can write an equation
r1 t 1 = r2 t 2
to express the fact that the distance covered by each vehicle was the same.
EXAMPLE 5.2.1. Apu takes the express train from Springfield to Capital City, and a regular
train on his way back from Capital City to Springfield. The express completes the trip in
4 hours 30 minutes, while the regular train covers the same distance in 6 hours. Find the
speed of each train if the express runs 15 miles per hour faster.
SOLUTION: Let x be the speed of the express train and let r be the speed of the regular
train in miles per hour. The express is 15 miles per hour faster than the regular train:
x = r + 15
The distance they cover is the same, so we can also write
express distance x · 4.5 = r ·6 regular distance
We have to use 4.5 for the duration of the express trip because we must convert 4 hours
30 minutes into hours for the equation. Now we can substitute (r +15) for x in the second
equation and then solve for r:
x · 4.5 = r ·6
(r + 15) · 4.5 = r ·6
r · 4.5 + 15 · 4.5 = r ·6 distributivity
4.5r + 67.5 = 6r simplifying expressions
Now we use the addition property to isolate the term with r on the right side:
4.5r + 67.5 − 4.5r = 6r − 4.5r addition property
67.5 = 1.5r combined like terms
(67.5)/1.5 = (1.5r)/1.5 multiplication property
45 = r
132
CHAPTER 2. LINEAR EQUATIONS 5. APPLICATIONS
So the speed of the regular train is 45 miles per hour, and we can find the speed of the
express from the first equation:
x = r + 15
x = 45 + 15
x = 60
ANSWER:
x and r are the speeds of the express and the regular train respectively, in miles per hour
x = r + 15
4.5x = 6r
solution: x = 60, r = 45
Another common situation is when two workers each work towards a common goal, which is
to complete a known amount of work. Then we can write that the total work completed is
equal to the sum of individual work amounts:
w1 + w2 = w
r1 t 1 + r2 t 2 = w rewrite w1 and w2 in terms of rates
EXAMPLE 5.2.2. Kay and July are opening the incoming mail together, with the goal of
processing 400 envelopes. Find how much time they need to complete the task if Kay can
process 15 envelopes per hour, and July can process 35 envelopes per hour.
SOLUTION: We need to find t, which is the duration of time they both work together, in
hours. The total amount of work w is 400 envelopes, so we can write that
Kay’s work + July’s work = 400
15 · t + 35 · t = 400
This equation can be easily solved for t.
50t = 400
(50t)/50 = (400)/50
t = 8
So it will take 8 hours of them working together to process 400 envelopes.
ANSWER:
t is the duration of work, in hours,
equation: 15t + 35t = 400
solution: t = 8
133
5. APPLICATIONS CHAPTER 2. LINEAR EQUATIONS
EXAMPLE 5.2.3. One morning Aristotle starts walking from Athens to Megara, and at the
same time Socrates starts on a journey from Megara to Athens. Being younger, Aristotle
is walking 1.8 times faster than Socrates. After a 5 hour journey, they meet somewhere
in the middle of the way. Find the speed of each traveler if the total distance between
Athens and Megara is 42 km.
SOLUTION: Let a and s be the walking speeds of Aristotle and Socrates respectively, in
km per hour. What we call work here is measured in km, so we are just talking about the
distance traveled. Since Aristotle’s speed is 1.8 times greater than Socrates’ speed, we can
write
a = 1.8s
And since they both walk for 5 hours and then meet in the middle of a 42 km long road,
we can also write a distance equation:
A’s distance + S’s distance = total distance
A’s speed · duration + S’s speed · duration = total distance
a·5 + s·5 = 42
Now we can substitute 1.8s for a in this equation and solve for Socrates’ speed s:
(1.8s) · 5 + (s) · 5 = 42
9s + 5s = 42
14s = 42 combined like terms
(14s)/14 = (42)/14
s = 3
Now that we know that Socrates was walking at 3 km per hour, we can get Aristotle’s
speed from the first equation:
a = 1.8s
a = 1.8(3)
a = 5.4
So Aristotle was walking at 5.4 km per hour.
ANSWER:
a and s are the speeds of Aristotle and Socrates respectively, in km per hour
a = 1.8s
5a + 5s = 42
solution: a = 5.4, s=3
134
CHAPTER 2. LINEAR EQUATIONS 5. APPLICATIONS
Homework 2.5.
Solve applications of w = r t by describing 10. It takes Leela 56 days to inspect 14
the variables and their units of measure- buildings. Find the rate of the building in-
ment. spection in buildings per day.
1. The top speed of a bullet train is 210 km
per hour. Find how long it takes to travel
168 km at top speed.
Solve applications by describing the vari-
2. The average speed of a cheetah is 16 ables used to represent the quantities of in-
meters per second. Find how far the chee- terest, and stating appropriate equations.
tah can run in 24 seconds at that speed.
11. Alice paddled for 3 hours upstream
3. A standard computer hard drive rotates and then for 2 hours downstream. When
at 5400 revolutions per minute. Find the going downstream, her speed was 10 mph
number of revolutions over 10 seconds. greater than her speed when going up-
(Hint: Convert seconds into minutes.) stream. The total distance traveled was 30
miles. How fast did Alice travel upstream
4. Morgan’s chicken coop produces on av- and downstream?
erage 8 eggs per day. Find how many days
are needed to produce 52 eggs. 12. Bob rode for 3 hours on the train and
then for 2 hours on the bus. The bus is 50
5. Eve typed up a 4050 word document km per hour slower than the train. The to-
over two and a half hours. Find Eve’s typ- tal distance traveled by Bob was 225 km.
ing rate in words per minute. (Hint: Con- Find the the speed of the train and the
vert hours into minutes.) speed of the bus.
6. Jeff’s household used 630 gallons of wa- 13. Ridwan can plant 4 trees per hour, and
ter over the three winter months. Find the Sue can plant 5.5 trees per hour. Find how
rate of water consumption in gallons per much time they need to plant 285 trees, if
month. they are working together.
7. Jamie charges $80 per hour for her par- 14. Quinn runs at 14 km per hour and
alegal work. Find how long she has to work walks at 6 km per hour. During his week-
in order to earn $2000. end exercise, Quinn walks and runs for an
equal amount of time. One weekend he ran
8. Ivan drinks 3.5 cups of coffee per day. and walked for a total of 24 km. Find the
Find how long it takes for Ivan to drink 42 amount of time he walked.
cups of coffee.
15. A passenger and a freight train start
9. It takes Bongo two and a half hours to toward each other at the same time from
mow the front lawn. Find how many times two points 300 miles apart. If the rate of
he can mow the same lawn in 40 hours. the passenger train exceeds the rate of the
freight train by 15 miles per hour, and they
meet after 4 hours, what must the rate of
each be?
135
5. APPLICATIONS CHAPTER 2. LINEAR EQUATIONS
16. A car and a truck are 276 miles apart together, a 1200 gallon tank can be emp-
and start at the same time to travel toward tied in 12 minutes. Find the rates in gallons
each other. The car is 5 miles per hour per minute for both pumps.
faster than the truck. If they meet after 6
hours, find the rate of each. 19. Two automobiles started at the same
time from a point, but traveled in oppo-
17. Lab 1 can process blood samples 3 site directions. Their rates were 25 and
times faster than Lab 2. Working together, 35 miles per hour respectively. After how
the two labs can process 84 samples over 7 many hours were they 180 miles apart?
days. Find the rate of work for each Lab, in
samples per day. 20. Two trains travel toward each other
from points which are 195 miles apart.
18. Pump 1 pumps water at the rate which They travel at rate of 25 and 40 miles per
is 10 gallons per minute slower than the hour respectively. If they start at the same
rate of Pump 2. With both pumps working time, how soon will they meet?
136
CHAPTER 2. LINEAR EQUATIONS 5. APPLICATIONS
Homework 2.5 Answers.
1. solution: u = 2, d = 12
w is the distance traveled in km,
r is in km per hour, t is in hours,
t = 0.8 (48 minutes) 13.
t is the duration of work in hours,
3. equation: 4t + 5.5t = 285
w is the number of revolutions, solution: t = 30
r is in revolutions per minute, t is in min-
utes,
15.
w = 900
p and f are the speeds of the passenger
5. train and the freight train respectively, in
w is the number of words, mph
r is in words per minute, t is in minutes, p = f + 15
r = 27 4p + 4 f = 300
7. solution: p = 45, f = 30
w is wages earned in dollars,
r is in dollars per hour, t is in hours,
t = 25 17.
r1 and r2 are the rates for Lab 1 and Lab 2
9. respectively, in samples per day
w is lawns mowed,
r1 = 3r2
r is in lawns mowed per hour, t is in hours,
7r1 + 7r2 = 84
w = 16
solution: r1 = 9, r2 = 3
11.
u and d are the speeds of the upstream and
the downstream travel respectively, in mph 19.
t is the duration of travel in hours,
d = u + 10
equation: 25t + 35t = 180
3u + 2d = 30 solution: t = 3
137
6. LINEAR INEQUALITIES CHAPTER 2. LINEAR EQUATIONS
6. Linear Inequalities
6.1. Solution Sets. Recall that we call an equation linear if every term is either a constant
or a product of a constant and the first power of a single variable.
DEFINITION 6.1.1. A linear inequality is an inequality relation applied to two linear ex-
pressions. There are four types of linear inequalities corresponding to the four types of
inequality relations: x < y, x > y, x ≤ y, and x ≥ y.
We say that a number is a solution for the given inequality in variable x if substituting
that number for x makes the inequality true. The collection of all such numbers forms
the solution set for the inequality.
EXAMPLE 6.1.1. Determine whether 0, 10, and −5, are solutions for the inequality
x + 3 < 4 − 2x
SOLUTION: Let’s try x = 0:
(0) + 3 < 4 − 2(0)
3 < 4
This is true, so 0 is a solution. Let’s try x = 10 next:
(10) + 3 < 4 − 2(10)
33 < 4 − 20
33 < − 16
This is false, so 10 is not a solution. Finally we check x = −5:
(−5) + 3 < 4 − 2(−5)
−2 < 4 + 10
−2 < 14
This is true, so −5 is a solution.
ANSWER: 0 and −5 are solutions, and 10 is not
EXAMPLE 6.1.2. Describe the set of solutions of the inequality x ≥ 6.
SOLUTION: 6 is clearly a solution, and so is every number greater than 6. At the same
time, no number less than 6 is a solution. So the set of solutions is very large: it includes
6, 7, 7.5, and infinitely many other numbers. It would be unhelpful to state it as a roster:
{6, 7, 7.5, 6.12, 100, . . .}
138
CHAPTER 2. LINEAR EQUATIONS 6. LINEAR INEQUALITIES
But the set builder notation provides an unambiguous way of describing this set of num-
bers:
ANSWER: x x ≥6
6.2. Interval Notation and Graphs. Inequality solutions in set builder notation are cum-
bersome, so here we introduce a different notation, which is well-suited for describing large
portions of the real line. But before we introduce the notation, let us look at what solutions of
linear inequalities look like graphically. In the following table we list the four types of inequal-
ities in one variable x, the corresponding graphs of their solution sets, and the new notation
describing these sets.
inequality shaded points solve the inequality solution set in interval notation
x<a (−∞, a)
a
x≤a (−∞, a]
a
x>a (a, ∞)
a
x≥a [a, ∞)
a
Note that the graphs of solutions for x < a and x > a leave the point a empty, because a < a
and a > a are false, and so a does not belong in either solution set. For a similar reason the
graphs of x ≤ a and x ≥ a fill the point a and mark it as a solution.
The right column displays the description of the solution set in the interval notation.
DEFINITION 6.2.1 (Interval Notation). The interval notation (a, b) describes the set of all
real numbers between a and b. When we want to include the endpoints, we use square
brackets: [a, b]. When we want the interval to extend indefinitely, we use the infinity
sign: (a, ∞). The nuances are worked out in the following examples.
139
6. LINEAR INEQUALITIES CHAPTER 2. LINEAR EQUATIONS
BASIC EXAMPLE 6.2.1. Let us graph the solution set and describe it in the interval notation:
x ≤ 17
This inequality is true for 17 and all numbers below it:
17
The interval notation
(−∞, 17]
uses −∞ for the left endpoint meaning that the set extends indefinitely to the left on
the real number line, and 17 for the right endpoint. On the left, the round parenthesis (
indicates that −∞ does not solve the inequality, and so not in the solution set. On the
right, the square bracket ] indicates that 17 does solve the inequality, and so is included
in the solution set.
BASIC EXAMPLE 6.2.2. Let’s graph the solution set and describe it in the interval notation:
x > −4
This inequality is true for all numbers above −4, false for −4 and all numbers below it:
−4
The interval notation
(−4, ∞)
uses −4 for the left endpoint and ∞ for the right endpoint, meaning that the set extends
indefinitely to the right on the real number line. The round parentheses ( ) indicate that
neither endpoint solves the inequality.
BASIC EXAMPLE 6.2.3. Let’s graph the solution set and describe it in the interval notation:
x < 28
This inequality is true for all numbers below 28, false for 28 and all numbers above it:
28
The interval notation
(−∞, 28)
uses −∞ for the left endpoint, meaning that the set extends indefinitely to the right on
the real number line, and 28 for the right endpoint. The round parentheses ( ) indicate
that neither endpoint solves the inequality.
140
CHAPTER 2. LINEAR EQUATIONS 6. LINEAR INEQUALITIES
BASIC EXAMPLE 6.2.4. Let’s graph the solution set and describe it in the interval notation:
20 ≤ x
Note that we can also write this inequality in an equivalent form
x ≥ 20
Either way, this inequality is true for 20 and all the numbers above 20:
20
The interval notation
[20, ∞)
uses 20 for the left endpoint and ∞ for the right endpoint, meaning that the set extends
indefinitely to the right on the real number line. The square bracket [ on the left indicates
that 20 solves the inequality and is an element of the solution set. The round parenthesis
) on the right indicates that ∞ does not solve the inequality.
6.3. Addition and Multiplication Properties. Much like we did with the equations, we
will solve inequalities by constructing equivalent inequalities with well-known solution sets.
DEFINITION 6.3.1. Inequalities are equivalent if they have the same solution sets.
THEOREM 6.3.1 (Addition Property for Inequalities). Adding the same number or expres-
sion on both sides of an inequality produces an equivalent inequality.
In the following examples, we will describe each solution set by drawing a graph and stating
it in the interval notation.
EXAMPLE 6.3.1. Solve the inequality
x −4≤6
SOLUTION:
x −4 ≤ 6
x −4+4 ≤ 6+4 addition property
x ≤ 10 combined like terms
ANSWER: (−∞, 10],
10
141
6. LINEAR INEQUALITIES CHAPTER 2. LINEAR EQUATIONS
EXAMPLE 6.3.2. Solve the inequality: 4x + 3 > 3(x − 1)
SOLUTION: Just as we did with equations, we will need to isolate the term with the
variable on one side, so we start by getting rid of parentheses:
4x + 3 > 3(x − 1)
4x + 3 > 3x − 3 distributivity
4x + 3 − 3 > 3x − 3 − 3 isolating constants on the right side
4x > 3x − 6 combined like terms
4x − 3x > 3x − 6 − 3x isolating variable terms on the left side
x > −6 combined like terms
ANSWER: (−6, ∞),
−6
THEOREM 6.3.2 (Multiplication Property for Inequalities). Multiplying both sides of an
inequality by the same positive number produces an equivalent inequality. Multiplying
both sides of an inequality by the same negative number and then reversing the inequal-
ity sign produces an equivalent inequality.
Formally, the following three inequalities in x are all equivalent to each other for all real
numbers a and all positive numbers m:
x < a, mx < ma, −mx > −ma
Similarly, the following three inequalities in x are all equivalent to each other:
x ≤ a, mx ≤ ma, −mx ≥ −ma
EXAMPLE 6.3.3. Solve the inequality: 15x ≥ 3
SOLUTION: Dividing by 15 is the same as multiplying by its reciprocal 1/15, which is
positive, so the inequality sign does not change:
15x ≥ 3
(15x)/15 ≥ 3/15
x ≥ 1/5
ANSWER: [1/5, ∞), 1/5
142
CHAPTER 2. LINEAR EQUATIONS 6. LINEAR INEQUALITIES
EXAMPLE 6.3.4. Solve the inequality
2(1 − 6x) < −14x + 5
SOLUTION: First we get rid of parentheses and isolate the variable terms on one side, and
then we use the multiplication property to isolate the variable:
2(1 − 6x) < − 14x + 5
2 − 12x < − 14x + 5
2 − 12x + 12x < − 14x + 5 + 12x added 12x to both sides
2 < − 2x + 5 combined like terms
2−5 < − 2x + 5 − 5 subtracted 5 from both sides
−3 < − 2x combined like terms
−3 −2x
> dividing both sides by −2 reverses the sign
−2 −2
3/2 > x this is equivalent to x < 3/2
ANSWER: [−∞, 3/2), 3/2
143
6. LINEAR INEQUALITIES CHAPTER 2. LINEAR EQUATIONS
Homework 2.6.
Write an inequality corresponding to the 15. −47 ≥ 8 − 5x
shown solution set, and state it in the in-
6+ x
terval notation: 16. ≤ −1
12
1.
−2 17. −2(3 + k) < −44
2. 18. −7n − 10 ≥ 60
1
3. 19. 24 ≥ −6(m − 6)
5
20. −8(n − 5) ≥ 0
4.
−5
21. −r − 5(r − 6) < −18
5.
−3 22. −60 ≥ −4(−6x − 3)
6. 23. 4(2 y − 3) ≤ −44
4
x
24. 2 > 9 −
5
For each inequality, graph the solution set
and describe it using the interval notation. 4
25. (3x + 4) ≤ 20
5
7. 2 + x < 3
26. 3(2 y − 3) > 21
8. 4 y ≥ 12 x
27. +4≤1
x 3
9. ≥ 10
11 2 x 4
28. − <
n 3 5 15
10. −2 ≤
13
29. 24 + 4b < 4(1 + 6b)
n
11. 8 + ≥ 6
3 30. −8(2 − 2n) ≥ −16 + n
x 31. −5v − 5 < −5(4v + 1)
12. 11 > 8 +
2
32. −36 + 6x > −8(x + 2) + 4x
a−2
13. 2 >
5 33. 3(n + 3) + 7(8 − 8n) < 5n + 5 + 2
v−9
14. ≤2 34. −(4 − 5p) + 3 ≥ −2(8 − 5p)
−4
144
CHAPTER 2. LINEAR EQUATIONS 6. LINEAR INEQUALITIES
Homework 2.6 Answers.
1. x < −2, (−∞, −2) 21. (8, ∞)
3. x ≥ 5, [5, ∞)
8
5. x > −3, (−3, ∞)
7. (−∞, 1) 23. (−∞, −4]
−4
1
9. [110, ∞)
25. (−∞, 7]
110
7
11. [−6, ∞)
27. (−∞, −9]
−6
13. (−∞, 12) −9
12 29. (1, ∞)
15. [11, ∞)
1
11
31. (−∞, 0)
17. (19, ∞)
0
19
19. [2, ∞) 33. (1, ∞)
2 1
145
6. PRACTICE TEST 2 CHAPTER 2. LINEAR EQUATIONS
Practice Test 2
1. Solve the equation: 12. 384 is 96% of what number?
20x − 7 = −5x − 2
13. The monthly cost of living in a met-
2. Solve the equation: ropolitan area went down from 1400 to
1330. What percent decrease does that
7(x + 4) − 20 = 4(x + 2) + 3x represent?
3. Solve the equation: 14. A savings account has $571.20 at the
x x −1 end of the year, after the 2% simple interest
−1=
5 5 is added. How much money was originally
invested into the account at the beginning
4. Matt makes 35 dollars per week less of the year?
than Jane, and together they make 1245
dollars per week. How much money does 15. Train A, traveling 70 miles per hour,
each person make per week? leaves Westfall heading toward East River,
260 miles away. At the same time Train
5. The perimeter of a rectangular cloth is B, traveling 60 miles per hour, leaves East
84 inches, and one side of the rectangle is River heading toward Westfall. How soon
2.5 shorter than the long side. Find the di- do the two trains meet?
mensions of the cloth.
16. A bus leaves the station at noon. Two
6. Solve the equation: hours later a car leaves the same station,
x − 11 + 2(x − 2) = 6(5 − x) going 10 miles per hour faster than the bus,
and follows the same route. Find the speed
7. Solve the formula for D: of each vehicle if the car catches up to the
AD 7 bus 8 hours after the bus leaves the station.
=
3 2B
(Hint: Both vehicles end up covering the
8. Solve the formula for y: same distance, although the car travels
−5 y + 4x − 12 = 18 over a shorter period of time.)
9. Solve the formula for w: 17. Solve the inequality:
w + 6x = −4(2 y − w) 8(x + 1) ≥ 4x − 24
10. What is 14% of 280? 18. Solve the inequality:
1
11. 7 is what percent of 5? 7x − 3 > (15x + 4)
2
146
CHAPTER 2. LINEAR EQUATIONS 6. PRACTICE TEST 2
Practice Test 2 Answers.
1. {1/5} 10. 39.2
2. R 11. 140%
3. ∅ 12. 400
4. 13. 5% decrease
m and j are amounts for Matt and Jane re-
spectively, in dollars per week 14. $560
m = j − 35
15.
m + j = 1245 t is the duration of travel until trains meet,
solution: m = 605, j = 640 in hours,
equation: 70t + 60t = 260
5. solution: t = 2
l and w are length and width of the cloth,
in inches 16.
b and c are the speeds of the bus and the
2l + 2w = 84
car respectively, in mph
w = l ÷ 2.5
c = b + 10
solution: l = 30, w = 12 b·8= c·6
6. {5} solution: b = 30, c = 40
21 17. [−8, ∞)
7. D =
2AB
−8
30 − 4x
8. y =
−5 18. (−∞, −10)
6x + 8 y
9. w = −10
3
147
CHAPTER 3
Graphing
1. Reading and Constructing Graphs
1.1. Cartesian Plane.
DEFINITION 1.1.1 (Ordered Pair). An ordered pair is a pair of numbers with a definite order
to them. The notation (a, b) denotes an ordered pair with number a first and number b
second. Here a is the x coordinate, or the first coordinate, or the first component of the
pair, and b is the y coordinate, or the second coordinate, or the second component of the
pair.
Two ordered pairs are equal if their first components are equal and their second compo-
nents are equal.
BASIC EXAMPLE 1.1.1. The pairs (10, 25) and (5 + 5, 52 ) are equal.
The pairs A = (10, 25) and B = (10, 10) are distinct even though their first components are
equal, since the second component of A is 25, and it is not equal to the second component
of B, which is 10.
DEFINITION 1.1.2. A Cartesian plane is an infinite flat plane equipped with a Cartesian
coordinate system. In this system, an ordered pair of real numbers (a, b) corresponds to
the point of intersection of two lines: the vertical line through the point a on the x-axis,
and the horizontal line through the point b on the y-axis.
y
(a, b)
b
x
a
Note that the distance from the point (a, b) to the x-axis is |b|, while the distance to the
y-axis is |a|.
149
1. READING AND CONSTRUCTING GRAPHS CHAPTER 3. GRAPHING
DEFINITION 1.1.3. We can graph a set of points on a coordinate plane by shading all points
in that set with a specific color.
BASIC EXAMPLE 1.1.2. There are two major visual styles for the coordinate plane. The
illustration on the left shows the coordinate axes as thick black lines, and is typically used
when the origin (0, 0) is visible on the graph. The illustration on the right is styled like a
box and is often used when showing the axes is inconvenient.
80
y
70
40
60
50
30
y
40
30
20
20
10
x
75 100 125
50 100 150 200 x
BASIC EXAMPLE 1.1.3. Finite sets of points look like collections of dots. The graph on the
left shows the set containing points (−2, 0) and (1, 2):
3 3 3
y y y
2 2 2
1 1 1
x x x
−3 −2 −1 1 2 3 −3 −2 −1 2 3 −3 −2 −1 1 2 3
−1 −1 −1
−2 −2 −2
−3 −3 −3
Infinite sets of points often look like lines and curves. The middle graph shows the set
containing every point with x coordinate equal to 1.
Even larger sets of points, usually defined by inequalities, often appear as solid shaded
regions of the plane. The graph on the right shows the set of all points with x coordinate
between 0 and 1 and y coordinate between 1 and 2.
150
CHAPTER 3. GRAPHING 1. READING AND CONSTRUCTING GRAPHS
EXAMPLE 1.1.1. Consult the given graph and find
(1) all points with x coordinate 20
(2) all points with y coordinate 30
(3) the point with the lowest y coordinate
(4) the point with the highest x coordinate
100
y
90
80
70
60
50
40
30
20
10
x
10 20 30 40 50 60 70 80 90 100 110 120
SOLUTION: To find the all points with x coordinate 20, we can follow the vertical line
consisting of points with x coordinate equal to 20 and see where this line meets the graph.
In our case, the only such point is (20, 90):
100
y
90
80
70
60
50
40
30
20
10
x
10 20 30 40 50 60 70 80 90 100 110 120
151
1. READING AND CONSTRUCTING GRAPHS CHAPTER 3. GRAPHING
To find the all points with y coordinate 30, we can follow the horizontal line consisting
of points with y coordinate equal to 30 and see where this line meets the graph. In our
case, we can see two such points, (40, 30) and (80, 30):
100
y
90
80
70
60
50
40
30
20
10
x
10 20 30 40 50 60 70 80 90 100 110 120
To find the point with the lowest y coordinate, we look at how low the graph goes, since
points which appear to be down in the bottom have lower y coordinates. In our case, the
graph goes as low as y = 10, and the only point this far down is (60, 10):
100
y
90
80
70
60
50
40
30
20
10
x
10 20 30 40 50 60 70 80 90 100 110 120
To find the point with the highest x coordinate, we look at how far to the right the graph
goes, since points which appear to be on the right side have higher x coordinates. In our
case, the graph goes as far to the right as x = 100, and the only point this far on the right
is (100, 90):
152
CHAPTER 3. GRAPHING 1. READING AND CONSTRUCTING GRAPHS
100
y
90
80
70
60
50
40
30
20
10
x
10 20 30 40 50 60 70 80 90 100 110 120
ANSWER:
(1) (20, 90)
(2) (40, 30) and (80, 30)
(3) (60, 10)
(4) (100, 90)
EXAMPLE 1.1.2. Plot points A(1, 2), B(2, 1), C(−3, 1), and D(0, −4) on the same plane.
SOLUTION: Note that A and B are different points: the first coordinate of A(1, 2) is 1, so
A is located above 1 on the x-axis, whereas B(2, 1) is located above 2 on the x-axis.
5
y
4
3
A(1, 2)
2
C(−3, 1) B(2, 1)
1
ANSWER: x
−5 −4 −3 −2 −1 1 2 3 4 5
−1
−2
−3
D(0, −4)
−4
−5
153
1. READING AND CONSTRUCTING GRAPHS CHAPTER 3. GRAPHING
EXAMPLE 1.1.3. Plot points A(−40, 0.3), B(−20, 0.7), C(0, −0.6), and D(20, 0.2) on the
same plane.
SOLUTION: We may have to scale the coordinate axes in order to make the points spread
out in a visually appealing fashion. In this case, the x coordinate ranges from −40 to 20,
while the absolute value of the y coordinate never goes above 1.0.
0.8
y
0.7
B(−20, 0.7)
0.6
0.5
0.4
A(−40, 0.3)
0.3
0.2
0.1
ANSWER: x
−50 −40 −30 −20 −10 10 20 30 40 50
−0.1
−0.2
D(20, −0.2)
−0.3
−0.4
−0.5
−0.6
C(0, −0.6)
−0.7
−0.8
154
CHAPTER 3. GRAPHING 1. READING AND CONSTRUCTING GRAPHS
Homework 3.1.
Plot all given points on the same coordinate 7.
plane. A(10, 15)
B(−5, −5)
1.
C(15, −10)
A(1, 3) D(−5, 15)
B(−2, 4)
C(−1, −1) 8.
A(0, 0)
D(2, 0)
B(0, 400)
2. C(200, 300)
A(0, 1) D(400, 0)
B(2, −3)
C(1, −5)
D(−5, 0)
3.
A(−1, −60)
B(0, −20)
C(1, 80)
D(2, 90)
4.
A(−120, 0)
B(−20, −2)
C(40, 2)
D(100, −3)
5.
A(−0.4, 0.4)
B(0.1, −0.6)
C(0.5, 0.1)
D(0.7, 0)
6.
A(0, −0.8)
B(−0.5, −0.7)
C(0.4, 0)
D(−0.4, 0.5)
155
1. READING AND CONSTRUCTING GRAPHS CHAPTER 3. GRAPHING
For each graph shown, find coordinates of 11.
the points with required properties. 5
y
4
9.
5 3
y
4 2
3 1
x
2
−5 −4 −3 −2 −1 1 2 3 4 5
1 −1
x −2
−5 −4 −3 −2 −1 1 2 3 4 5
−1 −3
−2 −4
−3 −5
−4
(1) all points with x coordinate zero
−5 (2) all points with y coordinate zero
(3) all points with x coordinate −3
(1) all points with x coordinate zero (4) the point with the lowest y coor-
(2) all points with y coordinate zero dinate
(3) all points with x coordinate −4
(4) all points with y coordinate −1 12.
14
y
13
10.
7 12
y 11
6
10
5
9
4 8
3 7
2 6
1 5
x 4
−7 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6 7 3
−1
2
−2
1
−3 x
−4 −5−4−3−2−1
−1 1 2 3 4 5 6 7 8 9 10 11
−5 −2
−6
−7 (1) all points with x coordinate zero
(2) all points with y coordinate zero
(1) all points with x coordinate zero (3) all points with x coordinate 6
(2) all points with y coordinate zero (4) the point with the highest y coor-
(3) all points with x coordinate −2 dinate
(4) all points with y coordinate −6
156
CHAPTER 3. GRAPHING 1. READING AND CONSTRUCTING GRAPHS
13. 15.
y 100
4 y
90
3
80
2
70
1 60
x
50
−4 −3 −2 −1 1 2 3 4
−1 40
30
−2
20
−3
10
−4 x
10 20 30 40 50 60 70 80 90 100
(1) all points with x coordinate zero
(1) all points with x coordinate 40
(2) all points with y coordinate zero
(2) all points with y coordinate 50
(3) all points with x coordinate 2
(3) the point with the highest x coor-
(4) all points with the lowest y coor-
dinate
dinate
(4) the point with the lowest y coor-
14. dinate
5
y
4
16.
80
y
3
70
2
60
1
x 50
−5 −4 −3 −2 −1 1 2 3 4 5
−1 40
−2 30
−3 20
−4
10
−5 x
25 50 75 100 125 150 175 200
(1) all points with x coordinate zero
(2) all points with y coordinate zero (1) all points with x coordinate 125
(3) all points with x coordinate −2 (2) all points with y coordinate 40
(4) the point with the highest y coor- (3) the point with the lowest y coor-
dinate dinate
(4) the point with the highest y coor-
dinate
157
1. READING AND CONSTRUCTING GRAPHS CHAPTER 3. GRAPHING
Homework 3.1 Answers.
1. 7.
y y
B
4 20
A D A
2 10
D x x
−4 −2 C 2 4 −20 −10 B 10 20
C
−2 −10
−4 −20
9.
3.
100 y D
C
(1) (0, −2)
(2) (4, 0)
50 (3) (−4, −4)
(4) (2, −1)
x
11.
−2 −1 B 1 2
A −50 (1) (0, 0)
(2) (−4, 0) and (0, 0)
(3) (−3, −3)
−100 (4) (−2, −4)
13.
5.
1 y
(1) (0, 1)
(2) (−3, 0), (−1, 0), (1, 0), (3, 0)
A 0.5 (3) (2, −1)
(4) (−2, −1) and (2, −1)
C
D x
15.
−1 −0.5 0.5 1
−0.5 B (1) (40, 30), (40, 50), (40, 70)
(2) (20, 50), (40, 50), (60, 50)
(3) (60, 50)
−1 (4) (40, 30)
158
CHAPTER 3. GRAPHING 2. GRAPHING LINEAR EQUATIONS
2. Graphing Linear Equations
2.1. Solution Sets. Recall that equation is a particular kind of relation, which we infor-
mally defined as a true-or-false statement about numbers. Consider an equation in two vari-
ables x and y. It may be true for some ordered pairs (a, b) and false for others.
DEFINITION 2.1.1. A solution to an equation in two variables x and y is any ordered pair
(a, b) such that substituting a for x and b for y makes the equation true. The set of all
such pairs is the solution set for that particular equation.
EXAMPLE 2.1.1. Determine whether (−5, 4) is a solution for the equation 2x + 3 y = 3
SOLUTION: The left side of the equation is
2x + 3 y = 2(−5) + 3(4)
= −10 + 12
= 2
while the right side is 3. The sides are unequal, the equation is false, so this is not a
solution.
ANSWER: No
EXAMPLE 2.1.2. Determine whether (3, −2) a solution for the equation 2x 2 + y 3 = 4x + y
SOLUTION: The left side of the equation is
2x 2 + y 3 = 2(3)2 + (−2)3
= 2(9) + (−8)
= 10
while the right side is
4x + y = 4(3) + (−2)
= 10
The sides are equal, the equation is true, so this is a solution.
ANSWER: Yes
159
2. GRAPHING LINEAR EQUATIONS CHAPTER 3. GRAPHING
BASIC EXAMPLE 2.1.1. Some equations have empty solution sets, for example
x + y = x + y + 17
Some equations have solution sets which consist of all possible ordered pairs of numbers,
for example
x+y=x+y
But most practically useful equations have solution sets which include some pairs, and
exclude others.
EXAMPLE 2.1.3. Describe the solution set for the equation y = x.
SOLUTION: Any pair of the form (a, a), where a is a real number, will solve the equation.
At the same time, any pair of distinct numbers (a, b) such that a 6= b will make the equation
false, so the solution set consists of all pairs (a, a) where the components are equal to
each other. For example, (1, 1) is a solution, and so is (3, 3), (−10, −10), and so on. As is
typically the case with equations in two variables, the solution set is infinite.
ANSWER: The set of pairs (a, a), for each real number a.
2.2. Graphing Solution Sets.
DEFINITION 2.2.1. A graph of a relation, and in particular of an equation, consists of all
points on the plain for which that particular relation is true.
EXAMPLE 2.2.1. Graph the equation x = y.
SOLUTION: The reader can refer to the example 2.1.3 for the description of the solution
set, which consists of all the points (a, a), for each real number a. As we start plotting
some of the solutions on the Cartesian plane, we realize they are all located on the so-
called main diagonal.
3
y
2
1
ANSWER: x
−3 −2 −1 1 2 3
−1
−2
−3
160
CHAPTER 3. GRAPHING 2. GRAPHING LINEAR EQUATIONS
EXAMPLE 2.2.2. Graph the equation y = 5 − x 2 .
SOLUTION: Whenever an equation is solved for y, we can attempt to sketch it by evalu-
ating it for convenient values of x, and then connecting the dots, so to speak.
If x = 0, then y = 5 − (0)2 = 5 − 0 = 5
If x = 0.5 then y = 5 − (0.5)2 = 5 − 0.25 = 4.75
If x = 1, then y = 5 − (1)2 = 5 − 1 = 4
If x = 2, then y = 5 − (2)2 = 5 − 4 = 1
If x = 3, then y = 5 − (3)2 = 5 − 9 = −4
If x = −0.5, then y = 5 − (−0.5)2 = 5 − 0.25 = 4.75
In fact, negative x will result in the same y values as their positive inverses, so we also
get the points (−1, 4), (−2, 1) and (−3, −4). We can summarize the points we have found
in a table of the corresponding x and y coordinates:
x −3 −2 −1 −0.5 0 0.5 1 2 3
y −4 1 4 4.75 5 4.75 4 1 −4
We plot these points on the Cartesian plane:
y
6
4
2
x
−3 −2 −1 1 2 3
−2
−4
−6
At last, we draw a smooth curve through the dots.
161
2. GRAPHING LINEAR EQUATIONS CHAPTER 3. GRAPHING
ANSWER:
y
6
4
2
x
−3 −2 −1 1 2 3
−2
−4
−6
2.3. Graphing Lines.
THEOREM 2.3.1. A graph of a linear equation in variables x and y
Ax + B y = C
where not both A and B are zero, is always a straight line. Conversely, every straight line
on the Cartesian plane is the solution set of some linear equation in this form.
So the task of graphing a linear equation can be reduced to plotting just two solutions, and
then drawing a straight line through these points.
EXAMPLE 2.3.1. Graph the equation y = 2x − 3
SOLUTION: We can substitute any two numbers for x, but the best choices would make it
easy to find the corresponding y, so we will stick to small integers.
If x = 0, then y = 2(0) − 3 = −3
If x = 1, then y = 2(1) − 3 = −1
We plot the points (0, −3) and (1, −1) on the plane:
162
CHAPTER 3. GRAPHING 2. GRAPHING LINEAR EQUATIONS
3
y
2
1
x
−4 −3 −2 −1 1 2 3 4
−1
−2
−3
−4
−5
Finally, we draw a straight line through the points we found.
ANSWER:
3
y
2
1
x
−4 −3 −2 −1 1 2 3 4
−1
−2
−3
−4
−5
EXAMPLE 2.3.2. Graph the equation 3x − 5 y = 10
SOLUTION: Guessing the solutions for this equation is inconvenient, so we solve it for y.
The result is an equivalent equation which makes it really easy to find solution pairs.
3x − 5 y = 10
−5 y = 10 − 3x
−5 y 10 − 3x
=
−5 −5
3
y = −2 + x distributivity
5
It looks like we will have easier time if we substitute multiples of 5 for x.
3
If x = 0, then y = −2 + (0) = −2
5
163
2. GRAPHING LINEAR EQUATIONS CHAPTER 3. GRAPHING
3
If x = 5, then y = −2 + (5) = −2 + 3 = 1
5
We plot the points (0, −2) and (5, 1) on the plane:
3
y
2
1
x
−5 −4 −3 −2 −1 1 2 3 4 5 6 7 8 9
−1
−2
−3
−4
Finally, we draw a straight line through the points we found.
ANSWER:
3
y
2
1
x
−5 −4 −3 −2 −1 1 2 3 4 5 6 7 8 9
−1
−2
−3
−4
164
CHAPTER 3. GRAPHING 2. GRAPHING LINEAR EQUATIONS
Homework 3.2.
Graph the equation and highlight two dis- 10. y = 4
tinct points on the line.
11. x + y = −1
1. y = x + 2
12. x − y = −3
2. y = x − 2
13. x + 5 y = −15
3. y = 2x
14. 4x + y = 5
1
4. y = x
3 15. 8x − y = 5
5. y = −2x + 1 16. 3x + 4 y = 16
6. y = −x − 3 17. y + 3 = 0
4 18. x − 2 = 0
7. y = − x − 3
5
19. 2x + 3 y = 0
3
8. y = x − 5
2 20. −4x + 3 y = 0
9. x = −1
165
2. GRAPHING LINEAR EQUATIONS CHAPTER 3. GRAPHING
Homework 3.2 Answers.
1. 7.
6 y 6 y
5 5
4 4
3 3
2 2
1 1
x x
−6 −5 −4 −3 −2 −1 1 2 3 4 5 6 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6
−1 −1
−2 −2
−3 −3
−4 −4
−5 −5
−6 −6
3. 9.
5
y 3 y
4
2
3
2
1
1
x x
−5 −4 −3 −2 −1 1 2 3 4 5 −3 −2 −1 1 2 3
−1
−1
−2
−3
−2
−4
−5 −3
5. 11.
5 4
y y
4
3
3
2
2
1
1
x x
−5 −4 −3 −2 −1 1 2 3 4 5 −4 −3 −2 −1 1 2 3 4
−1
−1
−2
−2
−3
−3
−4
−5 −4
166
CHAPTER 3. GRAPHING 2. GRAPHING LINEAR EQUATIONS
13. 17.
6 5
y y
5 4
4
3
3
2
2
1 1
x x
−6 −5 −4 −3 −2 −1 1 2 3 4 5 6 −5 −4 −3 −2 −1 1 2 3 4 5
−1 −1
−2
−2
−3
−3
−4
−5 −4
−6 −5
15. 19.
7 5
y y
6
4
5
4 3
3 2
2
1
1
x x
−7 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6 7 −5 −4 −3 −2 −1 1 2 3 4 5
−1
−1
−2
−3 −2
−4 −3
−5
−4
−6
−7 −5
167
3. INTERCEPTS CHAPTER 3. GRAPHING
3. Intercepts
3.1. x and y-intercepts.
DEFINITION 3.1.1. For any equation in variables x and y we define x-intercepts as the
points where the graph of the equation meets the x-axis. In other words, x-intercepts are
all the points (a, 0) which solve the given equation.
Similarly, we define y-intercepts as the points where the graph of the equation meets
the y-axis. In other words, y-intercepts are all the points (0, b) which solve the given
equation.
EXAMPLE 3.1.1. Find the intercepts for the pictured graph:
y
5
x
−10 −5 5 10
−5
ANSWER: There are three x-intercepts: (−6, 0), (−2, 0), and (5, 0). There is one
y-intercept: (0, −3).
3.2. Finding Intercepts.
THEOREM 3.2.1. The most straightforward way to find x-intercepts from an equation is
to replace y by 0 and then solve for x. Similarly, the standard way to find y-intercepts is
to replace x by 0 in the equation and then solve for y.
EXAMPLE 3.2.1. Find the intercepts for the equation 4x − 5 y = 40, then use them to plot
the line.
SOLUTION: To find x-intercepts, substitute 0 for y and solve for x:
4x − 5(0) = 40
4x = 40
x = 10
168
CHAPTER 3. GRAPHING 3. INTERCEPTS
We found the x coordinate of the intercept, and the y coordinate is 0 by definition.
To find y-intercepts, substitute 0 for x and solve for y:
4(0) − 5 y = 40
−5 y = 40
y = −8
We found the y coordinate of the intercept, and the x coordinate is zero by definition.
Now we plot the x-intercept (10, 0) and the y-intercept (0, −8) on the plane, and draw a
line through them.
15 y
10
5
x
−5 5 10 15 20
−5
−10
−15
ANSWER: x-intercept (10, 0), y-intercept (0, −8)
EXAMPLE 3.2.2. Find the intercepts for the equation 2x = 17.
SOLUTION: There is no y in this equation, so it is a vertical line, and the only intercept is
an x-intercept. We can still follow the steps, though. To find the x-intercept, we solve for
x and get x = 8.5. To find the y-intercept, we need to plug in 0 for x and solve for y, but
the equation 0 = 17 has no solutions, so there are no y-intercepts.
ANSWER: x-intercept (8.5, 0), y-intercept: none
169
3. INTERCEPTS CHAPTER 3. GRAPHING
EXAMPLE 3.2.3. Find the intercepts for the equation −3 y + 6 = 0.
SOLUTION: There is no x in this equation, so it’s a vertical line, and the only intercept is a
y-intercept. To find the y-intercept, we solve for y and get y = 2. To find the x-intercept,
we need to plug in 0 for y and solve for x, but the equation 6 = 0 has no solutions, so
there are no x-intercepts.
ANSWER: x-intercept: none, y-intercept (0, 2)
170
CHAPTER 3. GRAPHING 3. INTERCEPTS
Homework 3.3.
Find the coordinates of all x-intercepts and 15. x − 2 y = 4
all y-intercepts without graphing the line.
16. 3x − y = 9
1. 2x − 7 y = 14
17. y = 2x + 6
2. 3x + 5 y = 15
18. y = −3x + 5
3. 9x + 2 y = 36
19. 3x − 9 = 3 y
4. 10x + 3 y = 60
20. 4x − 5 y = 20
5. x − 3 = 0
21. −5x + 3 y = 180
6. 0 = 4 + y
22. 10x + 7 y = 280
7. y = 2x + 6
23. y = 30 − 3x
8. y = 3x − 12
24. 40 + y = 5x
9. 5x = 4 y
25. x + 5 = 1
10. −5x − 6 y = 120
26. y = 3
11. 4x + y = 10
27. −4x = 20 y + 80
12. 3x = 20 + 6 y
28. 60 − 20x = 3x
29. 3 y = −x
Find the intercepts, and plot the line with
intercepts clearly shown. 30. 2 y − 4x = 0
13. 3x + 4 y = 12 31. y = −2
14. 2x − y = 6 32. x = 7
171
3. INTERCEPTS CHAPTER 3. GRAPHING
Homework 3.3 Answers.
1. x-intercept (7, 0), y-intercept (0, −2) 17.
y
6
3. x-intercept (4, 0), y-intercept (0, 18) 5
5. x-intercept (3, 0), y-intercept: none x
−4 −3 −2 2 4
7. x-intercept (−3, 0), y-intercept (0, 6)
−5
9. x-intercept (0, 0), y-intercept (0, 0)
11. x-intercept (2.5, 0), y-intercept (0, 10)
19.
y
13.
y 4
4
3 2
2
x
x −4 −2 2 3 4
−5 4 5
−2
−2 −3
−4
−4
15. 21.
y y
4
60
50
2
x x
−5 4 5 −50−36 50
−2
−50
−4
172
CHAPTER 3. GRAPHING 3. INTERCEPTS
23. 29.
y y
40
5
30
20
x x
−10 10 −5 5
−20
−5
−40
25. 31.
y
y
5 5
x
x
−5−4 5
−5 5
−2
−5
−5
27.
y
5
x
−20 20
−4
−5
173
4. SLOPE CHAPTER 3. GRAPHING
4. Slope
4.1. Slope of a Line. The slope of a line can be understood as a real number which quan-
tifies the how steep the line is. One of the basic applications of the slope is the grade of the
road.
DEFINITION 4.1.1. The grade of a straight segment of a road is the ratio of the vertical
displacement, known as rise, and the horizontal displacement, known as run. It is tra-
ditionally stated in percent, so 9% grade corresponds to the ratio 0.09. In the following
illustration, the road with upward slope looks like a hypotenuse of the right triangle with
the horizontal leg corresponding to the run, and the vertical leg corresponding to the rise.
rise road
grade = rise
run
run
EXAMPLE 4.1.1. Find the grade of the road if it rises 400 feet over the horizontal distance
of 1500 feet.
SOLUTION: The grade is 450/1500 = 0.3, or 30%
ANSWER: 30%
The slope of a line on a coordinate plane is defined similarly, except that the rise carries a sign,
and positive slopes correspond to lines going upward and to the right, while negative slopes
correspond to lines going downward and to the right.
DEFINITION 4.1.2. The slope of a segment with distinct endpoints (x 1 , y1 ) and (x 2 , y2 ) is
denoted as m and can be computed using the slope formula, displayed below. The slope
of a vertical segment is left undefined because in that case x 1 = x 2 , the denominator of
the slope formula is zero, and the result of division is undefined.
y2
y2 − y1 y2 − y1
y
m=
x2 − x1 y1
x2 − x1
x1 x2
x
174
CHAPTER 3. GRAPHING 4. SLOPE
DEFINITION 4.1.3. The slope of a line is the slope of any segment with two distinct end-
points on that line. The slope of a vertical line is left undefined.
The slope of a segment with endpoints (x 1 , y1 ) and (x 2 , y2 ) can be understood as the ratio of
rise and run, with rise represented by y2 − y1 and run represented by x 2 − x 1 . The slope can
also be understood as the change of y coordinate per unit change of x coordinate, or as the
rate of change of y with respect to x.
4.2. Using Slope and Intercept to Plot.
EXAMPLE 4.2.1. Plot the line with slope 1/2 and y-intercept (0, 1)
SOLUTION: The slope 1/2 means that
1 rise
=
2 run
so the line rises 1 unit every time the x coordinate increases by 2 units. We can plot the
y-intercept (0, 1), which is a point on the line, then count 2 units to the right, 1 unit up,
and plot the point (2, 2). Having found two points, we can complete the graph by drawing
a line through them.
ANSWER:
5
y
4
3
2
1
x
−5 −4 −3 −2 −1 1 2 3 4 5
−1
−2
−3
−4
−5
175
4. SLOPE CHAPTER 3. GRAPHING
EXAMPLE 4.2.2. Plot the line with slope −3 and y-intercept (0, 4)
SOLUTION: The slope −3 can be thought of as a fraction
−3 rise
=
1 run
so the line falls 3 units every time the x coordinate increases by 1 unit. We can plot the
y-intercept (0, 4), which is a point on the line, then count 1 unit to the right, 3 units down,
and plot the point (1, 1). Having found two points, we can complete the graph by drawing
a line through them.
ANSWER:
6 y
5
4
3
2
1
x
−6 −5 −4 −3 −2 −1 1 2 3 4 5 6
−1
−2
−3
−4
−5
−6
176
CHAPTER 3. GRAPHING 4. SLOPE
Homework 3.4.
Plot the line using the slope and the inter- 21. (10, 18), (−11, −10)
cept.
22. (−3, 6), (−20, 13)
1. Slope 2/3, y-intercept (0, −1)
23. (−16, −14), (11, −14)
2. Slope 3/5, y-intercept (0, 1)
24. (13, 15), (2, 10)
3. Slope 0, y-intercept (0, 1)
25. (7, −14), (−8, −9)
4. Slope 0, y-intercept (0, −5)
26. (−18, −5), (14, −3)
5. Slope 3, y-intercept (0, 4)
27. (−5, 7), (−18, 14)
6. Slope −3, y-intercept (0, 2)
28. (19, 15), (5, 11)
7. Slope −2, y-intercept (0, −3)
8. Slope 2, y-intercept (0, 0)
9. Slope undefined, x-intercept (−3, 0)
10. Slope undefined, x-intercept (4, 0)
11. Slope −4/5, y-intercept (0, 6)
12. Slope −1/3, y-intercept (0, 5)
13. Slope 1/2, y-intercept (0, 0)
14. Slope 5/3, y-intercept (0, −2)
Find the slope of a segment with given end-
points.
15. (2, 10), (3, 15)
16. (3, 4), (7, 12)
17. (−2, 10), (−2, −15)
18. (1, 2), (−6, −14)
19. (−15, 10), (16, −7)
20. (13, −2), (7, 7)
177
4. SLOPE CHAPTER 3. GRAPHING
Find the slope of the pictured line. 32.
6 y
5
29.
7 4
y
6
3
5
2
4
1
3
x
2
−6 −5 −4 −3 −2 −1 1 2 3 4 5 6
1 −1
x
−2
−7 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6 7
−1
−3
−2
−4
−3
−4 −5
−5 −6
−6
−7 33.
7
y
6
30.
5 5
y
4
4
3
3 2
1
2 x
1 −7 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6 7
−1
x −2
−5 −4 −3 −2 −1 1 2 3 4 5 −3
−1
−4
−2 −5
−6
−3
−7
−4
−5 34.
5
y
31. 4
4
y 3
3 2
2 1
x
1 −5 −4 −3 −2 −1 1 2 3 4 5
−1
x
−4 −3 −2 −1 1 2 3 4 −2
−1 −3
−2 −4
−5
−3
−4
178
CHAPTER 3. GRAPHING 4. SLOPE
35. 36.
5 6
y y
4 5
4
3
3
2
2
1 1
x x
−5 −4 −3 −2 −1 1 2 3 4 5 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6
−1 −1
−2
−2
−3
−3
−4
−4 −5
−5 −6
179
4. SLOPE CHAPTER 3. GRAPHING
Homework 3.4 Answers.
1. 7.
6 5
y y
5 4
4
3
3
2
2
1 1
x x
−6 −5 −4 −3 −2 −1 1 2 3 4 5 6 −5 −4 −3 −2 −1 1 2 3 4 5
−1 −1
−2
−2
−3
−3
−4
−5 −4
−6 −5
3. 9.
4
y 5
y
3
4
3
2
2
1
1
x x
−4 −3 −2 −1 1 2 3 4 −5 −4 −3 −2 −1 1 2 3 4 5
−1 −1
−2
−2
−3
−3
−4
−4 −5
5. 11.
6 7
y y
5 6
4 5
4
3
3
2
2
1 1
x x
−6 −5 −4 −3 −2 −1 1 2 3 4 5 6 −7 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6 7
−1 −1
−2
−2
−3
−3
−4
−4
−5
−5 −6
−6 −7
180
CHAPTER 3. GRAPHING 4. SLOPE
13. 4
4 21.
y 3
3
23. 0
2
1
1 25. −
3
x
−4 −3 −2 −1 1 2 3 4
7
−1 27. −
13
−2
3
−3 29.
2
−4
31. undefined
15. 5
5
17. undefined 33.
6
17
19. − 35. −1
31
181
5. SLOPE-INTERCEPT FORM CHAPTER 3. GRAPHING
5. Slope-Intercept Form
5.1. Slope-Intercept Form.
DEFINITION 5.1.1. Every non-vertical line with slope m and y-intercept b is a solution set
for an equation in the slope-intercept form
y = mx + b
If either m or b is zero, the corresponding term is not shown.
This definition can be thought of as a theorem with an easy proof.
THEOREM 5.1.1. Given any two reals m and b, the slope of the line
y = mx + b
is m, and the y-intercept is b.
PROOF. We can find the y-intercept of y = mx + b by making x zero and solving for y:
y = mx + b
y = m·0+ b
y = b
To show that the slope is m, take any two points on the line, like (0, b) and (1, m + b), and use
the slope formula
y2 − y1 (m + b) − b m
slope = = = =m
x2 − x1 1−0 1
BASIC EXAMPLE 5.1.1. Some lines in slope-intercept form:
y = 3x − 2 has slope 3 and y-intercept (0, −2)
y = x +6 has slope 1 and y-intercept (0, 6)
y = 17 has slope 0 and y-intercept (0, 17)
y = −x has slope −1 and y-intercept (0, 0)
BASIC EXAMPLE 5.1.2. Some lines not in slope-intercept form:
2 y = 3x + 4 is not solved for y
y = −2(x + 5) has a product instead of a sum on the right
182
CHAPTER 3. GRAPHING 5. SLOPE-INTERCEPT FORM
4x − 5
y= has a fraction instead of a sum on the right
2
EXAMPLE 5.1.1. Find the slope and the y-intercept of the line
y = −6x − 14
SOLUTION: This equation is already in the slope-intercept form, so −6 is the slope and
(0, −14) is the y-intercept.
ANSWER: slope −6, y-intercept (0, −14)
To put a linear equation in the slope-intercept form, one can solve for y, remove parentheses,
and state the right side in a simplified form.
EXAMPLE 5.1.2. Find the slope, the y-intercept, and the equation in the slope-intercept
form for the line
x − 2y − 4 = 8
SOLUTION: We will actually find the slope-intercept form first, which will provide us with
everything else. We start by solving the equation of the line for the y variable.
x − 2y − 4 = 8 isolate the term with y on the left
x − 2y − 4 + 4 − x = 8+4− x by adding 4 and −x on both sides
−2 y = −x + 12 combined like terms
−x + 12
y = divided both sides by −2
−2
−x 12
y = + rewrote as a sum of two terms
−2 −2
1
y = x −6
2
This is the slope-intercept form, so 1/2 is the slope and (0, −6) is the y-intercept.
ANSWER: slope 1/2, y-intercept (0, −6)
183
5. SLOPE-INTERCEPT FORM CHAPTER 3. GRAPHING
EXAMPLE 5.1.3. If possible, find the slope, the y-intercept, and the equation in the slope-
intercept form for the line
−2 y + 3 = −5
SOLUTION: Solve the equation for the y variable:
−2 y + 3 = −5
−2 y + 3 − 3 = −5 − 3
−2 y = −8
y = 4
There is no x, but we can pretend that its coefficient is zero:
y = 0x + 4
This is the slope-intercept form, so 0 is the slope and (0, 4) is the y-intercept.
ANSWER: slope 0, y-intercept (0, 4)
EXAMPLE 5.1.4. If possible, find the slope, the y-intercept, and the equation in the slope-
intercept form for the line
2x = 3x + 5
SOLUTION: There is no y, so it is not possible to solve this equation for y. If there are any
solutions at all, this is a vertical line, and the slope is undefined. We solve this equation
for x to find the x-intercept:
2x = 3x + 5
2x − 3x = 3x + 5 − 3x
−x = 5 not yet solved for x
x = −5 multiplied both sides by −1
This is a vertical with the x-intercept (−5, 0), so the slope is undefined and there are no
y-intercepts.
ANSWER: slope undefined, no y intercepts
184
CHAPTER 3. GRAPHING 5. SLOPE-INTERCEPT FORM
EXAMPLE 5.1.5. Determine the equation of the pictured line and state the answer in the
slope-intercept form.
4
y
3
2
1
x
−4 −3 −2 −1 1 2 3 4
−1
−2
−3
−4
SOLUTION: The y-intercept is (0, −2), so b = −2. To find the slope, take any two points
on the line, like (0, −2) and (2, 1), and use the slope formula with (0, −2) for (x 1 , y1 ) and
(2, 1) for (x 2 , y2 ):
y2 − y1 1 − (−2) 3
m= = =
x2 − x2 2−0 2
3
ANSWER: y = x −2
2
EXAMPLE 5.1.6. Determine the equation of the pictured line and state the answer in the
slope-intercept form.
y
60
40
20
x
−10 −5 5 8 10
185
5. SLOPE-INTERCEPT FORM CHAPTER 3. GRAPHING
SOLUTION: The y-intercept is (0, 40), so b = 40. To find the slope, take any two points
(x 1 , y1 ) and (x 2 , y2 ) on the line, like (0, 40) and (8, 0), and use the slope formula:
y2 − y1 0 − 40
m= = = −5
x2 − x2 8−0
ANSWER: y = −5x + 40
5.2. Parallel Lines. More than 2000 years ago, Euclid provided an axiomatic definition
of parallel lines. The classical axioms of geometry imply that given a line and a point not on
the line, there exists a unique and distinct parallel line passing through that point, and that
parallel lines have no points in common. On a coordinate plane, we can provide a consistent
definition involving the slope.
DEFINITION 5.2.1. Two distinct lines are parallel if their slopes are equal. Just as in clas-
sical geometry, a line is not parallel to itself, even though it has the same slope as itself.
This tradition, however, has very little bearing on anything we consider in this text.
y
6
5
m1 = m2
4
Parallel lines never intersect:
3
2
1
x
1 2 3 4 5 6 7 8
EXAMPLE 5.2.1. Determine whether the two given lines are parallel.
y = 4x − 5 line 1
8x − 2 y = 0 line 2
SOLUTION: We will find the slopes and compare them. The first equation is in slope-
intercept form, so the slope of the first line m1 = 4, and its y-intercept is (0, −5). To find
186
CHAPTER 3. GRAPHING 5. SLOPE-INTERCEPT FORM
the slope of the second line, we solve its equation for the y variable:
8x − 2 y = 0
8x − 2 y − 8x = 0 − 8x
−2 y = −8x
−2 y −8x
=
−2 −2
y = 4x
This is now also in the slope-intercept form, so the slope of the second line m2 = 4 and its
y-intercept is (0, 0). Since the slopes are equal and y-intercepts are different, we conclude
that the lines are parallel. (If y-intercepts were the same, we would be looking at two
equivalent equations for a single line.)
ANSWER: parallel
EXAMPLE 5.2.2. Determine whether the two given lines are parallel.
y = −2x + 4 line 1
−3 y = 6x − 12 line 2
SOLUTION: Divide the equation of the second line by −3 on both sides to solve it for y:
−3 y 6x − 12
=
−3 −3
6x 12
y = −
−3 −3
y = −2x + 4
So the two equations are equivalent, and the solution sets are the same.
ANSWER: a single line
187
5. SLOPE-INTERCEPT FORM CHAPTER 3. GRAPHING
5.3. Perpendicular Lines. Another useful relationship between two lines is perpendicu-
larity. A geometrical definition requires a right angle between the lines (angle measure of 90◦
or π/2 radians), but on a coordinate plane we can provide a consistent definition involving the
slope.
DEFINITION 5.3.1. Two lines are perpendicular (or orthogonal) if their slopes m1 and m2
are negative reciprocals of each other:
m1 m2 = −1
Moreover, every vertical line is perpendicular to every horizontal line, even though the
slope of a vertical line is undefined.
THEOREM 5.3.1. If neither line is vertical, the definition can be restated in a form that
shows the negative reciprocals explicitly. Two lines with non-zero slopes m1 and m2 are
perpendicular if either
1
m1 = −
m2
or
1
m2 = −
m1
Perpendicular lines intersect at the right angle:
y
6
5
m1 m2 = −1
4
3
90◦
2
1
x
1 2 3 4 5 6 7 8
188
CHAPTER 3. GRAPHING 5. SLOPE-INTERCEPT FORM
EXAMPLE 5.3.1. Determine whether the two given lines are parallel, perpendicular, or
neither.
y = −3x + 8 line 1
3y − x = 4 line 2
SOLUTION: We will find the slopes and compare them. The first equation is in slope-
intercept form, so m1 = −3. To find the slope of the second line, we solve its equation for
the y variable:
3y − x = 4
3y − x + x = x +4
3y = x +4
3y x +4
= need to rewrite the right side as a sum
3 3
x 4
y = +
3 3
This is now in the slope-intercept form, so the slope m2 = 1/3. The slopes are not equal,
so we check whether they are negative reciprocals of each other:
1
m1 m2 = −3 · = −1
3
They are, so the lines are perpendicular.
ANSWER: perpendicular
EXAMPLE 5.3.2. Determine whether the two given lines are parallel, perpendicular, or
neither.
y = 6 line 1
x = −6 line 2
SOLUTION: The first line is horizontal and the second one is vertical, so they are perpen-
dicular.
ANSWER: perpendicular
189
5. SLOPE-INTERCEPT FORM CHAPTER 3. GRAPHING
EXAMPLE 5.3.3. Determine whether the two given lines are parallel, perpendicular, or
neither.
2x + y = 4 line 1
x + 2y = 2 line 2
SOLUTION: We will find the slopes and compare them. The first equation is equivalent to
y = −2x + 4
so m1 = −2. To find the slope of the second line, we solve its equation for the y variable
and put it in the slope-intercept form:
x + 2y = 2
x + 2y − x = 2− x
2y = −x + 2
2y −x + 2
=
2 2
−x 2
y = + rewrote right side as a sum
2 2
1
y = − x +1 slope-intercept form
2
so m2 = −1/2. The slopes are not equal, so the lines are not parallel. Next we check
whether they are negative reciprocals of each other:
1
m1 m2 = (−2) − =1
2
The result differs from −1, so the slopes are not negative reciprocals of each other, and
the lines are not perpendicular.
ANSWER: neither
190
CHAPTER 3. GRAPHING 5. SLOPE-INTERCEPT FORM
Homework 3.5.
Find the slope and the y intercept (if any) Determine the equation of the pictured line
from the equation by solving for y and find- and state the answer in the slope-intercept
ing an equivalent equation in the slope- form.
intercept form. Do not graph the line.
17.
8 4
y
1. y = x − 4
5 3
5 2
2. y = − x + 13
7 1
3. y = −60x x
−4 −3 −2 −1 1 2 3 4
4. y = −30x − 14 −1
−2
5. 5x = 7 y
−3
6. 8x − 9 y = 0
−4
7. 12x − 6 y = 9
18.
6 y
8. 2x − 5 y = 8 5
4
9. y − 3 = 5
3
2
10. y = 3
1
x
11. x = 4 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6
−1
12. 1 = 7 − x −2
−3
13. 3x + 4 y = 12 −4
−5
14. 4x + 2 y = 8 −6
15. −4x + y = 7
16. 3x − y = 6
191
5. SLOPE-INTERCEPT FORM CHAPTER 3. GRAPHING
19. 22.
7 6
y y
6 5
5
4
4
3
3
2
2
1 1
x x
−7 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6 7 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6
−1 −1
−2
−2
−3
−3
−4
−4
−5
−6 −5
−7 −6
20. 23.
6 y 6 y
5 5
4 4
3 3
2 2
1 1
x x
−6 −5 −4 −3 −2 −1 1 2 3 4 5 6 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6
−1 −1
−2 −2
−3 −3
−4 −4
−5 −5
−6 −6
21. 24.
6 y 6 y
5 5
4 4
3 3
2 2
1 1
x x
−6 −5 −4 −3 −2 −1 1 2 3 4 5 6 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6
−1 −1
−2 −2
−3 −3
−4 −4
−5 −5
−6 −6
192
CHAPTER 3. GRAPHING 5. SLOPE-INTERCEPT FORM
For each pair of lines, determine whether 30.
they are parallel, perpendicular, or neither.
6x − 3 y = 9
25.
3 2x − y = 3
y = x −3
4
31.
3x − 4 y = −2
26. x = 7
8x − 2 y = 7 x −8 = 0
3x + 12 y = 9 32.
27.
x = 4
6x − 4 y = 5
y = 4
8x + 12 y = 3
33.
28.
3x − 4 y = 12 x − 4y = 8
3 4x + y = 2
y = x +3
4
34.
29.
2x − 4 y = 6 2x + 5 y = 3
x − 2y = 3 5x − 2 y = 6
193
5. SLOPE-INTERCEPT FORM CHAPTER 3. GRAPHING
Homework 3.5 Answers.
1. Slope 8/5, y-intercept (0, −4) 19. y = −5
3. Slope −60, y-intercept (0, 0) 1
21. y = − x + 2
2
5. Slope 5/7, y-intercept (0, 0)
2
23. y = x −3
7. Slope 2, y-intercept (0, −3/2) 5
9. Slope 0, y-intercept (0, 8) 25. parallel
11. Slope undefined, no y-intercept 27. perpendicular
13. Slope −3/4, y-intercept (0, 3) 29. neither (same line)
15. Slope 4, y-intercept (0, 7) 31. parallel
17. y = 2x − 1 33. perpendicular
194
CHAPTER 3. GRAPHING 6. POINT-SLOPE FORM
6. Point-Slope Form
6.1. A Point and a Slope Determine a Line.
DEFINITION 6.1.1. Every non-vertical line with slope m and passing through the point
(x 1 , y1 ) is a solution set for an equation in the point-slope form
y − y1 = m(x − x 1 )
This definition can be thought of as a theorem with an easy proof.
THEOREM 6.1.1. Given any real m and any point (x 1 , y1 ), the line
y − y1 = m(x − x 1 )
has slope m and passes through the point (x 1 , y1 ).
PROOF. Substitute x 1 for x and y1 for y into the equation of the line to check that (x 1 , y1 )
is a solution:
y1 − y1 = m(x 1 − x 1 )
0 = m·0
0 = 0
The equation holds, so the line passes through the point (x 1 , y1 ). To find the slope, find the
slope-intercept form by solving for the y variable:
y − y1 = m(x − x 1 )
y − y1 = mx − mx 1
y − y1 + y1 = mx − mx 1 + y1
y = mx − mx 1 + y1
This does not look like the slope-intercept form, but rewriting the right side as a sum of two
terms helps:
y = mx + (−mx 1 + y1 )
So −mx 1 + y1 is the y-intercept, while m is the slope.
195
6. POINT-SLOPE FORM CHAPTER 3. GRAPHING
EXAMPLE 6.1.1. Find an equation in point-slope form for a line with slope 7, passing
through the point (3, 4).
SOLUTION: Substitute 7 for m and (3, 4) for (x 1 , y1 ) in the point-slope formula
y − y1 = m(x − x 1 )
to obtain the answer.
ANSWER: y − 4 = 7(x − 3)
EXAMPLE 6.1.2. Find an equation in point-slope form for a line with slope −3/2, passing
through the point (−2, 5).
SOLUTION: After we substitute −2 for x 1 and 5 for y1 in the point-slope formula we get
3
y − 5 = − (x − (−2))
2
Replacing −(−2) by +2 makes the answer look nicer, and is considered point-slope form.
3
ANSWER: y − 5 = − (x + 2)
2
EXAMPLE 6.1.3. Determine the slope and the coordinates of at least one point on the line
y + 1 = −5(x − 8)
SOLUTION: This equation is in the point-slope form, so m = −5 must be the slope, and
(x 1 , y1 ) = (8, −1) must be a point on the line.
ANSWER: Slope −5, point (8, −1)
6.2. Two Points Determine a Line. The very first Euclidean postulate states that two dis-
tinct points determine a line, so it is natural to ask for an equation of a line, given two points
on the coordinate plane.
EXAMPLE 6.2.1. Find the slope and an equation in the point-slope form for the line passing
through the points (1, −7) and (−4, 3).
196
CHAPTER 3. GRAPHING 6. POINT-SLOPE FORM
SOLUTION: Let (x 1 , y1 ) = (1, −7), (x 2 , y2 ) = (−4, 3), and find the slope using the slope
formula:
y2 − y1 3 − (−7) 10
m= = = = −2
x2 − x1 (−4) − 1 −5
Substitute −2 for m and (1, −7) for (x 1 , y1 ) in the point-slope formula to get the answer:
y − (−7) = −2(x − 1)
ANSWER: y + 7 = −2(x − 1)
EXAMPLE 6.2.2. Find the slope and an equation in the slope-intercept form for the line
passing through the points (−12, 3) and (−3, 6).
SOLUTION: We will find an equation of this line in the point-slope form first, and then
solve it for y to get the slope-intercept form. Let (x 1 , y1 ) = (−12, 3), (x 2 , y2 ) = (−3, 6),
and find the slope using the slope formula:
y2 − y1 6−3 3 1
m= = = =
x2 − x1 (−3) − (−12) 9 3
Substitute 1/3 for m and (−12, 3) for (x 1 , y1 ) in the point-slope formula to get the equation
in the point-slope form:
1
y − 3 = (x − (−12))
3
Solve the equation for y and put the right side in the slope-intercept form:
1
y −3 = (x + 12) point-slope form
3
1 1
y −3 = x + · 12 distributivity
3 3
1
y −3 = x +4
3
1
y −3+3 = x +4+3 add 3 to both sides
3
1
y = x +7 slope-intercept form
3
1
ANSWER: y = x +7
3
197
6. POINT-SLOPE FORM CHAPTER 3. GRAPHING
6.3. Using Point and Slope to Plot.
EXAMPLE 6.3.1. Plot the line passing through the point (−4, −5) with slope 2.
SOLUTION: We can think of the slope 2 as
rise 2
=
run 1
The point (−4, −5) is on the line, and going 2 units up and 1 unit to the right gives us
another point on the line (−3, −3).
ANSWER:
7
y
6
5
4
3
2
1
x
−7 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6 7
−1
−2
−3
−4
−5
−6
−7
198
CHAPTER 3. GRAPHING 6. POINT-SLOPE FORM
Homework 3.6.
Given a slope and a point on the line, state 16. (2, 3) and (4, 1)
the equation of the line in point-slope form.
17. (−3, 1) and (3, 5)
1. Slope −1, point (3, 6)
18. (−3, 4) and (3, 1)
2. Slope 1, point (2, 8)
19. (−3, 5) and (−1, −3)
3. Slope −3, point (−2, −5)
20. (−4, −1) and (1, 9)
4. Slope −2, point (−3, −1)
21. (5, 0) and (0, −2)
5. Slope 7/2, point (2, −4)
22. (−2, 0) and (0, 3)
6. Slope −4/5, point (−2, 6)
7. Slope −3/5, point (−4, −8) Graph the line with the given slope and
point.
8. Slope 1/3, point (4, 1)
23. Slope 3/4, point (1, 2)
24. Slope 2/5, point (−3, −4)
Determine the slope and the coordinates of
at least one point on the line. 25. Slope −4/3, point (−2, 5)
9. y − 17 = −6(x + 1) 26. Slope −3/2, point (1, 0)
10. y + 4 = 3(x − 2)
1 Graph the equation.
11. y − 7 = (x + 4)
2
2
27. y + 2 = (x − 1)
2 3
12. y − 1 = − (x − 13)
3
3
28. y − 1 = (x + 5)
13. y + 12 = −0.4(x + 0.1) 4
1 1 1
14. y − = 9 x − 29. y − 1 = − (x − 3)
3 4 4
1
30. y − 1 = − (x − 3)
2
Given two points on the line, find the slope 1
and the equation of the line and state the 31. y + 4 = (x − 1)
2
answer in the slope-intercept form.
1
15. (6, 8) and (3, 5) 32. y + 2 = (x + 1)
3
199
6. POINT-SLOPE FORM CHAPTER 3. GRAPHING
Homework 3.6 Answers.
1. y − 6 = −1(x − 3) 25.
6 y
5
3. y + 5 = −3(x + 2) 4
3
7 2
5. y + 4 = (x − 2) 1
2 x
−6 −5 −4 −3 −2 −1 1 2 3 4 5 6
−1
3
7. y + 8 = − (x + 4) −2
5
−3
−4
9. Slope −6, point (−1, 17)
−5
−6
1
11. Slope , point (−4, 7)
2
27.
6 y
13. Slope −0.4, point (−0.1, −12) 5
4
3
15. Slope 1, y = x + 2
2
1
2 x
17. Slope 2/3, y = x + 3 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6
3 −1
−2
19. Slope −4, y = −4x − 7 −3
−4
2 −5
21. Slope 2/5, y = x −2 −6
5
23. 29.
6 y 6 y
5 5
4 4
3 3
2 2
1 1
x x
−6 −5 −4 −3 −2 −1 1 2 3 4 5 6 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6
−1 −1
−2 −2
−3 −3
−4 −4
−5 −5
−6 −6
200
CHAPTER 3. GRAPHING 6. POINT-SLOPE FORM
31.
6 y
5
4
3
2
1
x
−6 −5 −4 −3 −2 −1 1 2 3 4 5 6
−1
−2
−3
−4
−5
−6
201
7. FUNCTION NOTATION CHAPTER 3. GRAPHING
7. Function Notation
7.1. Notation. In modern mathematics, a function is a certain kind of collection of ordered
pairs. In this text, however, we have no real need for this kind of formalism, so we will only
define a specific kind of a function, which is determined by an algebraic expression.
DEFINITION 7.1.1 (Functions defined by expressions). If a formula with variables x and
y is solved for y, then we can say that it defines a function. We can give this function any
name we want, like f , and then f will denote a rule for taking a number x, called input
or argument of the function f , and computing the number y, called output or value of the
function f .
BASIC EXAMPLE 7.1.1. Let the function F be defined by the equation
y = x 2 − 2x
When the argument x = 4, the value of F is
(4)2 − 2(4) = 8
When the argument x = −6, the value of F is
(−6)2 − 2(−6) = 36 + 12 = 48
In other words, finding the value of F amounts to substituting the argument into the
formula which defines the function, and computing the result.
DEFINITION 7.1.2. Given a function f , the notation f (x) denotes the value of f corre-
sponding to the argument x. While this looks exactly like a product of variables f and
x, the awkward placement of parentheses is usually enough to help the reader realize
that f is not a variable, but a name of a function. The parentheses here are not the usual
algebraic parentheses, and so they do not obey the distributive property. The only way to
get rid of f (x) in an expression is to replace it by the corresponding value of f .
f (x) reads “ f of x” or “the value of f at x”.
It is very traditional to create functions which share the name with the variable used to define
them. For example, a function defined by
y = x +1
may also be defined by
y(x) = x + 1
and then it is also called y. Even so, it is usually easy to determine the meaning of an expres-
sion, since function names do not appear in arithmetic expressions without the parentheses
and the argument.
202
CHAPTER 3. GRAPHING 7. FUNCTION NOTATION
EXAMPLE 7.1.1. Let y be a function defined by
y(x) = x − 4
Find y(15).
SOLUTION: Substitute 15 for x in the expression x − 4 to find the value of y(15):
y(15) = (15) − 4 = 11
ANSWER: 11
EXAMPLE 7.1.2. Let g be a function defined by
|x + 6|
g(x) =
x2
Find g(−10).
SOLUTION: We have to evaluate the expression inside the absolute value bars before we
can remove them:
|(−10) + 6|
g(−10) =
(−10)2
| − 4|
=
100
4
=
100
= 0.04
ANSWER: 0.04
The absolute value is actually a traditional notation for a function with the following rule:
if the argument (or input) is non-negative, then the value of the function is the same as the
argument; if the argument is negative, then the value is the opposite of the argument.
203
7. FUNCTION NOTATION CHAPTER 3. GRAPHING
EXAMPLE 7.1.3. Let h be a function defined by
h(x) = x 3 + 2
Find h(4) + h(5).
SOLUTION: We can find values h(4) and h(5) separately and then add them:
h(4) = (4)3 + 2 = 64 + 2 = 66
h(5) = (5)3 + 2 = 125 + 2 = 127
h(4) + h(5) = 66 + 127 = 193
ANSWER: 193
EXAMPLE 7.1.4. Let f be a function defined by f (x) = 12 − 5x. Find f (0) − f (10).
SOLUTION: Another way to simplify this kind of expression is to substitute f (0) and f (10)
with expressions right away, but then we have to introduce the substitution parentheses:
f (0) − f (10) = (12 − 5(0)) − (12 − 5(10))
= 12 − (−38)
= 50
ANSWER: 50
7.2. Applications of Linear Functions.
DEFINITION 7.2.1. A function f is linear if it can be defined by a linear expression
f (x) = mx + b
A graph of this linear function is the graph of the corresponding linear equation
y = mx + b
DEFINITION 7.2.2. A linear model is a linear function which is designed to model a rela-
tionship between two quantities in an application.
204
CHAPTER 3. GRAPHING 7. FUNCTION NOTATION
EXAMPLE 7.2.1. A water pump fills a water tank with water at a constant rate. There
are 14 gallons of water in the tank to begin with, and 30 seconds after the pump starts
working the amount of water in the tank goes up to 24 gallons. Let x represent the
number of seconds the pump was working, and let y(x) represent the amount of water
in the tank at the corresponding time.
(1) Find a linear model y(x) for the amount of water in the tank x seconds after the
pump starts working.
(2) Use the model to predict the amount of water in the tank 2 minutes after the
pump starts working.
(3) Use the model to predict the number of seconds needed to fill the tank with 100
gallons of water.
SOLUTION: When the pump turns on, the timer is at x = 0 and the number of gallons
in the tank is y(x) = 14. After 30 seconds, x = 30 and y(x) = 24. It may be helpful to
represent the two given data points as a table:
x y(x)
0 14
30 24
(1) We need an equation of the line which goes through the points (0, 14) and (30, 24).
The first point happens to be a y-intercept, but we still need to find the slope:
24 − 14 10 1
m= = =
30 − 0 30 3
Using the slope-intercept form y = mx + b we can write the linear model:
1
y(x) = x + 14
3
(2) The variable x measures time in seconds, so we need to convert 2 minutes into
120 seconds before using our linear model:
1
y(120) = · (120) + 14 = 40 + 14 = 54
3
So 2 minutes after the pump starts, there are 54 gallons of water in the tank.
(3) Substitute 100 for y(x) in the equation of the model and solve for x:
1
100 = x + 14
3
100 − 14 = 13x + 14 − 14 addition property
1
86 = x combined like terms
3
205
7. FUNCTION NOTATION CHAPTER 3. GRAPHING
To finish solving for x we multiply both sides by 3, which is the reciprocal of 1/3:
1
(86) · 3 = x ·3 multiplication property
3
258 = x
So the pump has to work for 258 seconds before the tank has 100 gallons of
water.
ANSWER:
1
(1) y(x) = x + 14
3
(2) 54 gallons
(3) 258 seconds
EXAMPLE 7.2.2. In year 2000, Brent’s private library contained 240 historical manuscripts,
and by 2017 the collection grew to contain 461 historical manuscripts. Let x represent
the year since 2000 and let y(x) represent the number of manuscripts in the library in
that year.
(1) Find a linear model y(x) for the number of manuscripts in a given year.
(2) Use the model to predict the number of manuscripts in year 2024.
(3) Use the model to predict the year when the library obtains the 1000th manu-
script.
SOLUTION: Since x represents the year since 2000, x and the year are related by
year = 2000 + x
It may be helpful to represent the two given data points as a table:
Year x y(x)
2000 0 240
2017 17 461
(1) We need an equation of the line which goes through the points (0, 240) and
(17, 461). The first point happens to be a y-intercept, but we still need to find
the slope:
461 − 240 221
m= = = 13
17 − 0 17
Using the slope-intercept form y = mx + b we can write the linear model:
y(x) = 13x + 240
206
CHAPTER 3. GRAPHING 7. FUNCTION NOTATION
(2) Year 2024 corresponds to x = 24:
y(24) = 13 · 24 + 240 = 552
So the library will have 552 manuscripts in year 2024.
(3) Substitute 1000 for y(x) in the equation of the model and solve for x:
1000 = 13x + 240
760 = 13x
6
58 13 = x
x is somewhat over 58, so 1000th manuscript will be added sometime during the
year 2058.
ANSWER:
(1) y(x) = 13x + 240
(2) 552 manuscripts
(3) year 2058
207
7. FUNCTION NOTATION CHAPTER 3. GRAPHING
Homework 3.7.
1. Find f (4) if f (x) = 3x + 1 (1) Find a linear model T (x) for the
coffee temperature x minutes af-
2. Find g(2) if g(t) = t 3 + 1 ter it was served.
(2) Use the model to predict the tem-
3. Find h(−6) if h(x) = 2 − 3x perature 8 minutes after the cof-
fee was served.
4. Find p(−3) if p(t) = t 2 + t
(3) Use the model to determine
the time when the temperature
5. Find F (−4) if F (s) = |1 + s|
reaches 74 ◦ F.
6. Find G(0) if G(x) = 2(3x − 7)
13. The population of a small town was
t −4 32.6 thousand people in 1990 and it grew
7. Find H(−2) if H(t) = to 33.2 thousand people by year 2000. Let
t
x be the number of years since 1990, and
x2 − 1 let y(x) be the population in thousands of
8. Find K(−5) if K(x) =
6 people.
9. Find P(−1) if P(x) = x 4 − 4x 3
(1) Find a linear model P(x) for
10. Find Q(−3) if Q(s) = 2s3 − s2
the town population in the corre-
11. The temperature of the air was 41 ◦ F at sponding year.
sunrise, and 56 ◦ F 30 minutes after the sun- (2) Use the model to predict the pop-
rise. Let x represent the number of min- ulation in year 2030.
utes since sunrise and let T (x) represent (3) Use the model to determine the
the temperature at that time. time when the population reaches
34.4 thousand people.
(1) Find a linear model T (x) for the
air temperature x minutes after 14. A racing car accelerates from zero to
the sunrise. 60 km/s over 2.5 seconds. Let t be the
(2) Use the model to predict the tem- elapsed time in seconds and let v(t) be the
perature 40 minutes after the sun- speed of the car in km/s at that time.
rise.
(3) Use the model to determine
the time when the temperature (1) Find a linear model v(t) for the
reaches 46 ◦ F. speed of the car t seconds after
start.
12. A cup of hot coffee is served at 97 ◦ F, (2) Use the model to predict the
and 5 minutes later it cools down to 89 ◦ F. speed after 3 seconds.
Let x be the number of minutes since the (3) Use the model to determine the
coffee was served, and let T (x) be the tem- time when the speed reaches 108
perature at that time. km/s.
208
CHAPTER 3. GRAPHING 7. FUNCTION NOTATION
Homework 3.7 Answers.
1. 13 11.
1
(1) T (x) = x + 41
3. 20 2
(2) 61 ◦ F
(3) 10 minutes
5. 3
13.
7. 3 (1) P(x) = 0.06x + 32.6
(2) 35 thousand people
9. 5 (3) year 2020
209
7. PRACTICE TEST 3 CHAPTER 3. GRAPHING
Practice Test 3
1. Graph the equation and highlight two 10. Determine the equation of the pictured
distinct points on the line. line and state it in the slope-intercept form:
1
y =− x +3 4
y
2
3
2. Find the coordinates of all x-intercepts
2
and all y-intercepts:
1
3x − y = 12
x
3. Find the coordinates of all x-intercepts −4 −3 −2 −1 1 2 3 4
−1
and all y-intercepts:
7 3 −2
y= x−
2 2 −3
4. Plot the line with slope 2 and y- −4
intercept (0, −4).
11. Find an equation of the line with slope
5. Find the slope of the segment with end- 0.5 passing through the point (4, −6) and
points (−6, −3) and (−3, 2). state it in slope-intercept form.
6. Find an equivalent equation in slope- 12. Find an equation of the line passing
intercept form: through the points (−1, 4) and (4, 14), and
10x − 4 y = 8 state it in slope-intercept form.
7. Find the slope of the line passing 13. Evaluate the function F (−4) if
through the points (3, 0) and (3, −4). F (x) = 1 − x − x 2
8. Determine whether the lines are paral- 14. Find an equation of the line with un-
lel, perpendicular, or neither: defined slope passing through the point
(−7, 7).
y+x = 4
2y − 7 = 2x 15. Find an equation of the line passing
through the point (5, 0) and parallel to the
9. Find an equation of the line with slope line
−5 passing through the point (−2, 3) and x+ y =3
state it in point-slope form. State the answer in slope-intercept form.
210
CHAPTER 3. GRAPHING 7. PRACTICE TEST 3
Practice Test 3 Answers.
1. 5. 5/3
7
y
6
5
5
4 6. y = x −2
3 2
2
1
x
7. undefined
−7 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6 7
−1
−2
−3
8. perpendicular
−4
−5
−6
−7 9. y − 3 = −5(x + 2)
2. x-intercept (4, 0), y-intercept (0, −12)
1 4
3 3 10. y = − x +
3. x-intercept , 0 , y-intercept 0, − 3 3
7 2
4.
6 y 11. y = 0.5x − 8
5
4
3 12. y = 2x + 6
2
1
x
13. −11
−6 −5 −4 −3 −2 −1 1 2 3 4 5 6
−1
−2
−3 14. x = −7
−4
−5
−6 15. y = −x + 5
211
CHAPTER 4
Linear Systems
1. Graphing Systems of Equations
1.1. Solution Sets. Recall that solutions to a linear equation in two variables x and y are
ordered pairs, which we can plot as points on the Cartesian plane. Recall also that the solution
set for any linear equation is a straight line.
DEFINITION 1.1.1. A system of linear equations in two variables is two or more linear equa-
tions in the same variables x and y, and it is written with a curly brace like this:
x+y = 5
x−y = 4
An ordered pair is a solution for a system if it is a solution for each and every equation in
the system.
x+y = 5
EXAMPLE 1.1.1. Determine whether (2, 3) is a solution for the system
x−y = 4
SOLUTION: Substitute 2 for x and 3 for y in the first equation:
x+y = 5
(2) + (3) = 5
5 = 5
So it is a solution for the first equation. Now we substitute 2 for x and 3 for y in the
second equation:
x−y = 4
(2) − (3) = 4
−1 = 4
So it is not a solution for the second equation, and therefore not a solution for the system.
ANSWER: No
213
1. GRAPHING SYSTEMS OF EQUATIONS CHAPTER 4. LINEAR SYSTEMS
EXAMPLE 1.1.2. Determine whether (4.5, 0.5) is a solution for the system
x+y = 5
x−y = 4
SOLUTION: Substitute 4.5 for x and 0.5 for y in the first equation:
x+y = 5
(4.5) + (0.5) = 5
5 = 5
So it is a solution for the first equation. Now we substitute 4.5 for x and 0.5 for y in the
second equation:
x−y = 4
(4.5) − (0.5) = 4
4 = 4
So it is a solution for the second equation, and therefore a solution for the system.
ANSWER: Yes
Knowing that solution sets of individual linear equations look like lines, we can try to figure
out the possibilities for the solution sets for systems of two equations. Ordered pairs which
satisfy both equations will look like points which belong to both graphs. So there are only
three major possibilities.
Two lines intersect at a Two lines are parallel, Two lines coincide, yield-
point, yielding a single so- yielding an empty solu- ing infinitely many solu-
lution. tion set. tions
4 4 4
y y y
2 2 2
x x x
−4 −2 2 4 −4 −2 2 4 −4 −2 2 4
−2 −2 −2
−4 −4 −4
214
CHAPTER 4. LINEAR SYSTEMS 1. GRAPHING SYSTEMS OF EQUATIONS
1.2. Solving Systems by Graphing.
EXAMPLE 1.2.1. Solve the system by graphing:
x−y = 2
x+y = 6
SOLUTION: Let us plot both lines on the same grid. Here it will be convenient to use
intercepts for graphing. The first line x − y = 2 has intercepts (2, 0) and (0, −2). The
second line x + y = 6 has intercepts (6, 0) and (0, 6).
8
y
6
4
2
x
−4 −2 2 4 6 8
−2
−4
The lines seem to intersect precisely at (4, 2). We can check the solution by making sure
x = 4 and y = 2 satisfy both equations:
(4) − (2) = 2
(4) + (2) = 6
ANSWER: (4, 2)
y = 10x − 70
EXAMPLE 1.2.2. Solve the system by graphing:
y = −20x + 80
SOLUTION: Let us plot both lines on the same grid. Here it will be convenient to use the
slope-intercept form for graphing. The first line y = 10x − 70 has y-intercept (0, −70)
and slope 10, which we can plot as going up 10 units and 1 unit to the right. The second
line y = −20x + 80 has y-intercept (0, 80) and slope −20, which we can plot by finding
another point on the line 20 units down and 1 unit to the right.
215
1. GRAPHING SYSTEMS OF EQUATIONS CHAPTER 4. LINEAR SYSTEMS
100 y
80
60
40
20
x
−2 −1 1 2 3 4 5 6 7 8
−20
−40
−60
−80
The lines seem to intersect precisely at (5, −20). We can check the solution by making
sure x = 4 and y = 2 satisfy both equations:
(−20) = 10(5) − 70
(−20) = −20(5) + 80
ANSWER: (5, −20)
2 y = 8x
EXAMPLE 1.2.3. Solve the system by graphing:
y = 4x + 3
SOLUTION: The first line in slope-intercept form is y = 4x, which is a line with slope 4
passing through the origin. The second line has slope 4 and y-intercept 3.
5
y
4
3
2
1
x
−3 −2 −1 1 2 3
−1
−2
−3
−4
−5
216
CHAPTER 4. LINEAR SYSTEMS 1. GRAPHING SYSTEMS OF EQUATIONS
These lines have the same slope, but different y-intercepts, so they are parallel, there are
no intersections, and the solution set is empty.
ANSWER: no solutions
y+x = 4
EXAMPLE 1.2.4. Solve the system by graphing:
y = 4− x
SOLUTION: Solving the first equation for y shows that it is equivalent to the other one:
y+x = 4
y+x−x = 4− x
y = 4− x
Graphing either equation yields the same line:
6 y
4
2
x
−4 −2 2 4 6
−2
−4
Every point satisfying the first equation also satisfies the second equation, so there are
infinitely many solutions. Formally, we can write down the solution set in the set-builder
notation, indicating that any ordered pair which solves one of the equations will also solve
the entire system:
(x, y) y = 4 − x
Without going into specifics, we can also say that there are infinitely many solutions (as
many as there are points on a line).
ANSWER: infinitely many solutions
217
1. GRAPHING SYSTEMS OF EQUATIONS CHAPTER 4. LINEAR SYSTEMS
Homework 4.1.
6x + y = −3
Determine whether the order pair is a so-
lution for the given system of equations. 12.
x+y = 2
5x − 2 y = −3
3
1. (1, 4), y = − x +1
7x − 3 y = −5
4
13.
y = −3 x + 2
3x + 2 y = −5
2. (−15, 20), 4
4 y + 5x = 5
y = 2x − 4
3x − 2 y = 0
3. (3, 2), 14.
1
x + 2 y = 15 y =
x +2
2
3x − y = 5
5
4. (2, 1), y = x +4
x+y = 3
3
15.
y = −2 x − 3
x − 3y = 1
5. (−2, −1), 3
x + 2y = 0
1
y =
x +4
1 2
y = − x +3
16.
6. (4, 1),
2 y = 1x +1
y = 3x −2
2
4
y = −x − 2
17.
2
Solve the system by graphing. y =
x +3
3
y = −x + 1
x + 4 y = −12
7. 18.
y = −5x − 3 2x + y = 4
y = −3
2x + 3 y = −6
8. 19.
y = −x − 4 2x + y = 2
x + 3 y = −9
3x + 2 y = 2
9. 20.
5x + 3 y = 3 3x + 2 y = −6
y = 2x + 2
2x + y = −2
10. 21.
y = −x − 4 x + 3y = 9
x−y = 4
x −3 = 3
11. 22.
2x + y = 2 5x + 2 y = 8
218
CHAPTER 4. LINEAR SYSTEMS 1. GRAPHING SYSTEMS OF EQUATIONS
Homework 4.1 Answers.
1. Yes 11. (2, −2)
5
y
4
3
3. No
2
1
x
−5 −4 −3 −2 −1 1 2 3 4 5
−1
5. No
−2
−3
−4
−5
7. (−1, 2)
4
y
13. no solutions
3
5
2 y
4
1 3
x
2
−4 −3 −2 −1 1 2 3 4 1
−1 x
−2 −5 −4 −3 −2 −1 1 2 3 4 5
−1
−3 −2
−3
−4
−4
−5
9. (3, −4) 15. (−3, −1)
5 5
y y
4 4
3 3
2 2
1 1
x x
−5 −4 −3 −2 −1 1 2 3 4 5 −5 −4 −3 −2 −1 1 2 3 4 5
−1 −1
−2 −2
−3 −3
−4 −4
−5 −5
219
1. GRAPHING SYSTEMS OF EQUATIONS CHAPTER 4. LINEAR SYSTEMS
17. (−3, 1) 21. (−3, 4)
5 5
y y
4 4
3 3
2 2
1 1
x x
−5 −4 −3 −2 −1 1 2 3 4 5 −5 −4 −3 −2 −1 1 2 3 4 5
−1 −1
−2 −2
−3 −3
−4 −4
−5 −5
19. (3, −4)
5
y
4
3
2
1
x
−5 −4 −3 −2 −1 1 2 3 4 5
−1
−2
−3
−4
−5
220
CHAPTER 4. LINEAR SYSTEMS 2. SUBSTITUTION
2. Substitution
2.1. Solving Systems using Substitution. We can solve systems of linear equations al-
gebraically by eliminating one of the variables using substitution, and then solving a linear
equation for the remaining variable.
7x + y = 14
EXAMPLE 2.1.1. Solve the system
y − 3x = −6
SOLUTION: Choose a variable to eliminate. Any variable will do, but the ones with low
coefficients tend to produce the least tedium. We choose y in the first equation, since it
has coefficient 1. Then we find an equivalent equation which is solved for y:
(1) 7x + y = 14 first equation
(2) y = 14 − 7x use this equation later, when x is found
Now we can use algebraic substitution to eliminate y from the other equation, by replacing
it with (14 − 7x). This yields a linear equation in one variable x, which we solve by usual
means.
y − 3x = −6 second equation
(14 − 7x) − 3x = −6 substitute (14 − 7x) for y
14 − 10x = −6 combined like terms
14 − 10x − 14 = −6 − 14
−10x = −20
−10x −20
=
−10 −10
x = 2
Finally, we substitute 2 for x in the equation (2), which we solved for y.
y = 14 − 7x
y = 14 − 7(2)
y = 14 − 14
y = 0
So the only solution is the ordered pair (2, 0).
ANSWER: (2, 0)
221
2. SUBSTITUTION CHAPTER 4. LINEAR SYSTEMS
This solution technique can be summarized as follows.
THEOREM 2.1.1. To solve a linear system of two equations in two variables,
(1) Solve either equation for either variable to get
y = expression with x or x = expression with y
(2) Substitute the expression into the other equation, producing a linear equation in
one variable.
(3) Solve the resulting equation to get
x = number or y = number
(4) substitute the number you found into the equation from the first step to find the
other coordinate.
When we are dealing with systems which have empty or infinite solution sets, the third step
of the procedure will result in an equation which is always false or always true respectively, as
the following examples illustrate.
y = x +1
EXAMPLE 2.1.2. Solve the system
y = x −1
SOLUTION: The first equation is already solved for y, so we substitute (x + 1) for y in the
second equation:
(x + 1) = x −1
x +1− x = x −1− x subtract x on both sides
1 = −1
This equation is always false, so no ordered pair (x, y) can ever satisfy the system.
ANSWER: no solutions
x + 3 y = −10
EXAMPLE 2.1.3. Solve the system
6 y = −2x − 20
SOLUTION: We choose to eliminate x from the first equation because it has the most
convenient coefficient.
x + 3y = −10
x = −3 y − 10
222
CHAPTER 4. LINEAR SYSTEMS 2. SUBSTITUTION
Substitute (−3 y − 10) for x into the other equation:
6y = −2x − 20 the other equation
6y = −2(−3 y − 10) − 20 now we will solve this for y
6y = 6 y + 20 − 20
6y = 6y
We can keep going and subtract 6 y on both sides to get an equivalent equation 0 = 0, but
any time we get the same expression on both sides, we can instantly conclude that the
linear equation is true for all numbers, which implies that the system has infinitely many
solutions.
ANSWER: infinitely many solutions
223
2. SUBSTITUTION CHAPTER 4. LINEAR SYSTEMS
Homework 4.2.
y = −6
Solve using substitution.
11.
3x − 6 y = 30
y = −3x
1.
y = 6x − 9
6x − 4 y = −8
12.
y = −6x + 2
y = x +5
2.
y = −2x − 4
−2x − y = −5
13.
x − 8 y = −23
x = 2y + 1
3.
3x − 6 y = 2
6x + 4 y = 16
14.
−2x + y = −3
x = −3 y
4.
x + 4 y = 10
−6x + y = 20
15.
−3x − 3 y = −18
y = −2x − 9
5.
y = 2x − 1
7x + 5 y = −13
16.
x − 4 y = −16
y = −6x + 3
6.
y = 6x + 3
y − 2x = −6
17.
2y − x = 5
y = 6x − 6
7.
−3x − 3 y = −24
2x − y = 0
18.
y − 2x = −2
−x + 3 y = 12
8.
y = 6x + 21
2x + 3 y = −2
19.
2x − y = 9
y = 3x − 1
9. 1
x − y = 13
3
y = 2x + 5
20.
−2 y = −4x − 10
x − 3y = 7
10.
−4x + 12 y = 28
224
CHAPTER 4. LINEAR SYSTEMS 2. SUBSTITUTION
Homework 4.2 Answers.
1. (1, −3) 13. (1, 3)
3. ∅
15. (−2, 8)
5. (−2, −5)
17 16
7. (2, 6) 17. ,
3 3
9. infinitely many solutions
25 11
19. ,−
11. (−2, −6) 8 4
225
3. ELIMINATION CHAPTER 4. LINEAR SYSTEMS
3. Elimination
3.1. Equivalent Systems. Another algebraic technique for solving systems is known as
Gaussian elimination or simply elimination, or sometimes addition. It relies on two ways of
obtaining an equivalent system of equations. We do not need the full power of the Gaussian
elimination to solve a system of two equations in two variables, so we present a much simplified
version of the technique here.
DEFINITION 3.1.1. Systems of linear equations are equivalent if they have the same solu-
tion sets.
THEOREM 3.1.1. Given a system of equations, we can replace either of the equations
by an equivalent one to obtain an equivalent system. In particular, we can apply the
multiplication property to either of the equations. Formally, for any real m 6= 0 these two
systems are equivalent:
a = b =
a b
c = d m(c) = m(d)
THEOREM 3.1.2. Given a system of equations, we can apply the addition property to either
of the equations. In particular, we can use the fact that the two sides of the first equation
are equal, and add them to the two sides of the other equation. Formally, these two
systems are equivalent:
a = b =
a b
c = d c+a = d+b
3.2. Eliminating Variable with Addition. In the following examples we will use proper-
ties of systems to eliminate one of the variables in one of the equations, solve for that variable,
then substitute the value we found into the other equation, and find the other coordinate of
the solution.
EXAMPLE 3.2.1. Solve the system using elimination:
2x + y = 7
3x − y = 3
SOLUTION: Notice that coefficients of the y variable are opposites of each other. If we
add left sides of these equations, then y will cancel (will get eliminated) and we will be
able to solve for x. So we are going to replace the second equation in the system by a sum
of the given equations, by writing that the sum of left sides is equal to the sum of right
sides:
226
CHAPTER 4. LINEAR SYSTEMS 3. ELIMINATION
2x + y = 7
(3x − y) + (2x + y) = 3 + 7
Simplifying the second equation yields
(3x − y) + (2x + y) = 3+7
3x + 2x − y + y = 10
5x = 10 combined like terms
x = 2
Now we can substitute 2 for x into the first equation and solve for y:
2x + y = 7
2(2) + y = 7
4+ y = 7
y = 3
So the only solution is the point (2, 3).
ANSWER: (2, 3)
x − 4y = 6
EXAMPLE 3.2.2. Solve the system using elimination:
2x + y = 12
SOLUTION: If we add equations now, neither variable will get eliminated. But we can
start by applying the multiplication property so that coefficients for one of the variables
become opposites of each other. As before, it is probably easier to eliminate the variable
with smaller coefficients, so we will try to get rid of x first. Here we only need to multiply
the first equation by −2 on both sides:
−2(x − 4 y) = −2(6)
2x + y = 12
Simplifying the first equation yields an equivalent system:
−2x + 8 y = −12
2x + y = 12
Now that the coefficients for x are opposites of each other, we can replace the second
equation using the addition property for systems:
−2x + 8 y = −12
(2x + y) + (−2x + 8 y) = (12) + (−12)
227
3. ELIMINATION CHAPTER 4. LINEAR SYSTEMS
Simplifying the second equation yields
(2x + y) + (−2x + 8 y) = (12) + (−12)
2x − 2x + y + 8 y = 0
9y = 0
y = 0
Now we can substitute y = 0 into the other original equation x − 4 y = 6 and solve for x:
x − 4y = 6
x − 4(0) = 6
x = 6
So the only solution is the point (6, 0).
ANSWER: (6, 0)
2x − 5 y = 12
EXAMPLE 3.2.3. Solve the system using elimination:
3x + 2 y = −1
SOLUTION: This is, in a way, a worst case scenario, as far as coefficients are concerned.
Before we can eliminate a variable using addition, we need to create a pair of opposite
coefficients, and here we will have to multiply both equations. We will opt to eliminate
x first because the coefficients seem a bit simpler. In order to do so, we will multiply the
first equation by 3 on both sides, and the second equation by −2 on both sides:
3(2x − 5 y) = 3(12)
−2(3x + 2 y) = −2(−1)
Simplifying both equations yields
6x − 15 y = 36
−6x − 4 y = 2
Notice that now the coefficients for x are opposites of each other, so we can eliminate x
using the addition property:
6x − 15 y = 36
(−6x − 4 y) + (6x − 15 y) = 2 + 36
Simplifying the second equation yields
−19 y = 38
And solving it for y gives us
y = −2
228
CHAPTER 4. LINEAR SYSTEMS 3. ELIMINATION
Now we can substitute −2 for y into the original equation 2x − 5 y = 12 and solve for x:
2x − 5 y = 12
2x − 5(−2) = 12
2x + 10 = 12
2x = 2
x = 1
So the only solution is the point (1, −2).
ANSWER: (1, −2)
229
3. ELIMINATION CHAPTER 4. LINEAR SYSTEMS
Homework 4.3.
2x − y = 5
Solve using elimination.
11.
5x + 2 y = −28
4x + 2 y = 0
1.
−4x − 9 y = −28
−5x + 6 y = −17
12.
x − 2y = 5
−7x + y = −10
2.
−9x − y = −22
10x + 6 y = 24
13.
−6x + y = 4
−x − 2 y = −7
3.
x + 2y = 7
x + 3 y = −1
14.
10x + 6 y = −10
−9x + 5 y = −22
4.
9x − 5 y = 13
2x + 4 y = 24
15.
4x − 12 y = 8
−6x + 9 y = 3
5.
6x − 9 y = −9
−6x + 4 y = 12
16.
12x + 6 y = 18
5x − 5 y = −15
6.
5x − 5 y = −15
8x − 5 y = −9
17.
3x + 5 y = −2
4x − 6 y = −10
7.
4x − 6 y = −14
4x + 6 y = −1
18.
x − 3y = 2
−3x + 3 y = −12
8.
−3x + 9 y = −24
x + 9y = 1
19.
2x − 6 y = 10
−x − 5 y = 28
9.
−x + 4 y = −17
2x − 15 − 10 y = 40
20.
28 = x − 4 y
−10x − 5 y = 0
10.
−10x − 10 y = −30
230
CHAPTER 4. LINEAR SYSTEMS 3. ELIMINATION
Homework 4.3 Answers.
1. (−2, 4) 11. (−2, −9)
3. Infinite number of solutions 13. (0, 4)
15. (8, 2)
5. ∅
17. (−1, 0.2)
7. ∅
1
19. 4, −
9. (−3, −5) 3
231
4. APPLICATIONS OF SYSTEMS CHAPTER 4. LINEAR SYSTEMS
4. Applications of Systems
4.1. Constant Rate Problems.
EXAMPLE 4.1.1. Suzy pays $111 for an order of 4 black printer cartridges and 7 color
cartridges. When supplies start running low, she goes back to the same store and pays
$126 for an order of 3 black and 10 color cartridges. Find the price of one black cartridge
and the price of one color cartridge.
SOLUTION: Let b and c be the prices of one black and one color cartridge respectively, in
dollars. We can write an equation for the total price of each order. Since 4 black cartridges
cost 4b dollars and 7 color cartridges cost 7c dollars,
4b + 7c = 111
Similarly, the second order can be expressed as
3b + 10c = 126
To solve this system by elimination, we can multiply both sides of the first equation by
3, and both sides of the second equation by −4, so that we can add and eliminate the b
variable:
3(4b + 7c) = 3(111)
−4(3b + 10c) = −4(126)
12b + 21c = 333
simplified both equations
−12b − 40c = −504
Now the coefficients for b are opposites of each other, and after adding equations we get
(12b + 21c) + (−12b − 40c) = 333 + (−504)
−19c = −171 canceled like terms
c = 9 divided both sides by −19
Now we substitute 9 for c in the original equation 3b + 10c = 126 and solve for b:
3b + 10(9) = 126
3b + 90 = 126
3b = 36 subtracted 90 on both sides
b = 12 divided both sides by 3
ANSWER:
b and c are prices of black and color cartridges respectively, in dollars
4b + 7c = 111
3b + 10c = 126
solution: b = 12, c=9
232
CHAPTER 4. LINEAR SYSTEMS 4. APPLICATIONS OF SYSTEMS
4.2. Mixing Problems. Mixing problems may assume many forms. Traditional applica-
tions revolve around mixing appropriate volumes of various chemical solutions in order to get
a solution with a specific concentration. In other applications, we could be “mixing” currency
bills of two different denominations or different kinds of coffee or grain.
EXAMPLE 4.2.1. A chemist receives an order for 10 liters of 48% isopropyl solution, that
is, a solution consisting of 48 parts of isopropyl alcohol and 52 parts of water by volume.
The stockroom only has some 15% solution and some 70% solution. How many liters of
each must be mixed in order to fulfill the order?
SOLUTION: As usual, we start by assigning variables to the quantities we need to find.
Let x and y be the needed volumes of 15% and 70% solutions respectively, measured in
liters.
We will solve this problem by writing down and solving a system of two equations. The
first one will say that the total volumes of the two solutions add up to 10 liters. The
second one will say that the volumes of just the isopropyl in the two solutions add up
to the volume of isopropyl in the final product. Before stating the equations, it may be
helpful to construct the following table:
15% solution 70% solution 48% solution
total volume in liters x y 10
volume of isopropyl in liters 0.15x 0.70 y 0.48 · 10
Of course, mixing the solutions means that the corresponding volumes add up to the cor-
rect totals:
x + y = 10
0.15x + 0.7 y = 0.48 · 10
We can solve this system using substitution. Solving the first equation for y yields
(3) y = 10 − x
We can substitute (10 − x) for y in the second equation to find x:
0.15x + 0.7(10 − x) = 0.48 · 10
0.15x + 7 − 0.7x = 4.8 distributivity
−0.55x + 7 = 4.8 combined like terms
−0.55x = 4.8 − 7 subtracted 7 on both sides
−0.55x = −2.2 combined like terms
x = 4 divided both sides by −0.55
233
4. APPLICATIONS OF SYSTEMS CHAPTER 4. LINEAR SYSTEMS
Now we can substitute 4 for x in the equation (3) and find y:
y = 10 − x
y = 10 − (4)
y = 6
ANSWER:
x and y are volumes of 15% and 70% solutions respectively, in liters
x + y = 10
0.15x + 0.7 y = 0.48 · 10
solution: x = 4, y =6
EXAMPLE 4.2.2. Charlie has in his coin jar a certain number of quarters and nickels. There
are 77 coins in the jar, and together they are worth $6.05. Find how many quarters and
how many nickels are in the jar.
SOLUTION: Let q and n be quantities of quarters and nickels in the jar respectively. The
total number of coins is 77, which we can translate as
q+n = 77 first equation
Recall that a quarter is worth $0.25 and a nickel is worth $0.05. The total worth of coins
is $6.05, which we can write as
0.25q + 0.05n = 6.05 second equation
We can solve this system of two equations by substitution. First we solve the first equation
for n, because it has a convenient coefficient 1:
(4) q+n = 77
(5) n = 77 − q use this later to find n
Now we can substitute (77 − q) for n in the second equation and solve for q:
0.25q + 0.05(77 − q) = 6.05
0.25q + 3.85 − 0.05q = 6.05 distributivity
0.2q + 3.85 = 6.05 combined like terms
0.2q = 6.05 − 3.85 subtracted 3.85 on both sides
0.2q = 2.2 combined like terms
q = 11 divided both sides by 0.2
234
CHAPTER 4. LINEAR SYSTEMS 4. APPLICATIONS OF SYSTEMS
Finally we substitute 11 for q in the equation (5) and solve for n:
n = 77 − q
n = 77 − 11
n = 66
ANSWER:
q and n are the quantities of quarters and nickels respectively
q + n = 77
0.25q + 0.05n = 6.05
solution: q = 11, n = 66
235
4. APPLICATIONS OF SYSTEMS CHAPTER 4. LINEAR SYSTEMS
Homework 4.4.
1. A collection of dimes and quarters is the rest are five dollar bills. The total
worth $15.25. There are 103 coins in all. amount of cash in the box is $50. Find the
How many of each is there? number of each type of bill in the cash box.
2. A purse contains $3.90 made up of 9. A total of $9000 is invested, part of it at
dimes and quarters. If there are 21 coins in 10% and the rest at 12%. The total inter-
all, how many dimes and how many quar- est after one year is $1030. How much was
ters are there? invested at each rate?
3. There were 429 people at a play. Ad- 10. A total of $18000 is invested, part of
mission was $1 each for adults and 75 it at 6% and the rest at 9%. The total in-
cents each for children. The receipts were terest after one year is $1248. How much
$372.50. How many children and how was invested at each rate?
many adults attended?
11. At a coin-operated laundromat, using
4. There were 200 tickets sold for a the washer 6 times and then the dryer 6
women’s basketball game. Tickets for stu- times costs $13.50, while using the washer
dents were 50 cents each and for adults 75 4 times and then the dryer 8 times costs
cents each. The total amount of money col- $12. Find the cost of one wash and the cost
lected was $132.50. How many of each of one drying cycle.
type of ticket was sold?
12. In June, Luke worked 150 hours at the
5. A total of $27000 is invested, part of it regular rate and 10 hours at the overtime
at 12% and the rest at 13%. The total in- rate, for the total paycheck of $4150. In
terest after one year is $3385. How much July, Luke worked 145 hours at the regu-
was invested at each rate? lar rate and 16 hours at the overtime rate,
for the total paycheck of $4265. Find the
6. A total of $50000 is invested, part of it regular and the overtime hourly rates.
at 5% and the rest at 7.5%. The total in-
terest after one year is $3250. How much 13. A tea that is 20% jasmine is blended
was invested at each rate? with a tea that is 15% jasmine. How many
pounds of each tea are used to make 5 lb
7. A coin bank contains nickels and dimes. of tea that is 18% jasmine?
The number of dimes is 10 less than twice
the number of nickels. The total value of 14. How many pounds of coffee that is
all the coins is $2.75. Find the number of 40% java beans must be mixed with 80 lb
each type of coin in the bank. of coffee that is 30% java beans to make a
coffee blend that is 32% java beans?
8. A total of 26 bills are in a cash box.
Some of the bills are one dollar bills, and
236
CHAPTER 4. LINEAR SYSTEMS 4. APPLICATIONS OF SYSTEMS
Homework 4.4 Answers.
d = 2n − 10
1.
d and q are quantities of dimes and quar- 0.1d + 0.05n = 2.75
ters respectively
solution: n = 15, d = 20
0.1d + 0.25q = 15.25
d + q = 103 9.
solution: d = 70, q = 33 x and y are dollar amounts invested at
10% and 12% respectively
x + y = 9000
3.
a and c are quantities of adults and chil- 0.1x + 0.12 y = 1030
dren respectively
solution: x = 2500, y = 6500
1a + 0.75c = 372.5
a + c = 429 11.
w and d are the costs of one washer and
solution: a = 203, c = 226 one dryer cycle respectively, in dollars
6w + 6d = 13.5
5.
x and y are dollar amounts invested at 4w + 8d = 12
12% and 13% respectively solution: w = 1.50, d = 0.75
x + y = 27000
13.
0.12x + 0.13 y = 3385
x and y are weights of 20% and 15% teas
solution: x = 12500, y = 14500 in pounds
x+ y =5
7.
0.2x + 0.15 y = 0.18 · 5
n and d are quantities of nickels and dimes
respectively solution: x = 3, y =2
237
5. MULTIVARIATE LINEAR INEQUALITIES CHAPTER 4. LINEAR SYSTEMS
5. Multivariate Linear Inequalities
5.1. Solution Sets.
DEFINITION 5.1.1. An ordered pair (a, b) is a solution for an inequality in two variables x
and y if substituting a for x and b for y makes the inequality true. The solution set for an
inequality in two variables is a collection of all ordered pairs which solve the inequality.
EXAMPLE 5.1.1. Determine whether (4, −5) a solution for the inequality
x + 1 < 2y
SOLUTION: Substituting 4 for x and (−5) for y yields
4+1 < 2(−5)
5 < − 10
This is false, so (4, −5) is not a solution.
ANSWER: no
EXAMPLE 5.1.2. Determine whether (0, 12) a solution for the inequality
−4x + 2 y ≥ 1
SOLUTION: Substituting 0 for x and 12 for y yields
−4 · 0 + 2 · 12 ≥ 1
24 ≥ 1
This is true, so (0, 12) is a solution.
ANSWER: yes
5.2. Solving Inequalities.
THEOREM 5.2.1. The solution set for a linear inequality in two variables x and y is a half
of the coordinate plane. The equation of the line separating the solution set from the rest
of the plane can be obtained by replacing the inequality sign by =.
A strict inequality (< and >) solution set does not contain the points of the line, while a
non-strict (≤ and ≥) inequality solution set does contain the points on the line.
238
CHAPTER 4. LINEAR SYSTEMS 5. MULTIVARIATE LINEAR INEQUALITIES
EXAMPLE 5.2.1. Graph the solution set for the inequality
x+ y ≤2
SOLUTION: This is a non-strict inequality, so the points on the line
x+ y =2
belong to the solution set. We plot the solid line first:
4
y
3
2
1
x
−4 −3 −2 −1 1 2 3 4
−1
−2
−3
−4
The solution set also includes all points on one side of the line, but which side? One way
to find out is by testing any one point not on the line. The point (0, 0), for example, is
not on the line. If (0, 0) solves the inequality, then all points on the same side as (0, 0)
(which is below the line in this case) belong to the solution set. And if (0, 0) does not
solve the inequality, then the points on the other side of the line belong to the solution set.
Substituting 0 for both x and y in the inequality yields
0+0≤2
This is true, so (0, 0) is a solution, and the entire side of the coordinate plane with (0, 0)
consists of solutions, so we graph them all by shading that portion of the plane.
4
y
3
2
1
ANSWER: x
−4 −3 −2 −1 1 2 3 4
−1
−2
−3
−4
239
5. MULTIVARIATE LINEAR INEQUALITIES CHAPTER 4. LINEAR SYSTEMS
EXAMPLE 5.2.2. Graph the solution set for the inequality −2x + y > −1
SOLUTION: This is a strict inequality, so the points on the line
−2x + y = −1
do not belong to the solution set. We are still going to plot the line first, but it will be
dashed rather than solid, to indicate that the line itself is not a part of the solution set. The
equation of this line in slope-intercept form is
y = 2x − 1
So the slope is 2 and the y-intercept is (0, −1).
4
y
3
2
1
x
−4 −3 −2 −1 1 2 3 4
−1
−2
−3
−4
The origin (0, 0) is not on the line, so we can use it for locating the solution set. We
substitute 0 for both x and y in the inequality to see whether it holds.
−2 · 0 + 0 > −1
0 > −1
This is true, so (0, 0) is a solution, and the portion of the plane containing (0, 0) must be
the solution set, so we shade accordingly.
4
y
3
2
1
ANSWER: x
−4 −3 −2 −1 1 2 3 4
−1
−2
−3
−4
240
CHAPTER 4. LINEAR SYSTEMS 5. MULTIVARIATE LINEAR INEQUALITIES
EXAMPLE 5.2.3. Graph the solution set for the inequality
−2x ≥ 6
SOLUTION: This is a non-strict inequality, so the points on the line
−2x = 6
belong to the solution set. This line equation is equivalent to
x = −3
which is a vertical line with x-intercept (−3, 0). We plot the solid line first:
5
y
4
3
2
1
x
−5 −4 −3 −2 −1 1 2 3 4 5
−1
−2
−3
−4
−5
The origin (0, 0) is not on the line, so we can use it for locating the solution set. With 0
for both x and y in the inequality, we get
−2 · 0 ≥ 6
0 ≥ 6
This is false, so the side of the coordinate plane to the right of the line is not a part of the
solution set. Hence the points to the left of the line belong to the solution set, and we
shade accordingly:
5
y
4
3
2
1
ANSWER: x
−5 −4 −3 −2 −1 1 2 3 4 5
−1
−2
−3
−4
−5
241
5. MULTIVARIATE LINEAR INEQUALITIES CHAPTER 4. LINEAR SYSTEMS
EXAMPLE 5.2.4. Graph the solution set for the inequality
3x < 5 y
SOLUTION: This is a strict inequality, so we will start by graphing a dashed line
3x = 5 y
The slope-intercept form of this equation is
3
y= x
5
so the slope is 3/5 and the y-intercept is (0, 0). We can plot this line by starting at the
y-intercept (0, 0), and then using the slope to locate another point on the line by going
3 units up and 5 units to the right. This takes us to (5, 3), which is another point on this
line.
6 y
5
4
3
2
1
x
−6 −5 −4 −3 −2 −1 1 2 3 4 5 6
−1
−2
−3
−4
−5
−6
Unlike in previous examples, the origin (0, 0) is on the line, so it is useless for locating the
solution set. But we can pick any other point for that, as long as it is not on the line. Let
us try (1, 0), which is just below the line. Substituting 1 for x and 0 for y in the inequality
yields
3·1 < 5·0
3 < 0
This is false, so (1, 0) is not a solution, and the portion of the plane below the line is not a
part of the solution set. Hence the portion above the line must be the solution set, so we
shade accordingly:
242
CHAPTER 4. LINEAR SYSTEMS 5. MULTIVARIATE LINEAR INEQUALITIES
ANSWER:
6 y
5
4
3
2
1
x
−6 −5 −4 −3 −2 −1 1 2 3 4 5 6
−1
−2
−3
−4
−5
−6
243
5. MULTIVARIATE LINEAR INEQUALITIES CHAPTER 4. LINEAR SYSTEMS
Homework 4.5.
Determine whether the given point is a so- 10. −3 y + 9x > 0
lution for the given inequality.
11. y ≥ x − 1
1. (2, 6)
2x + y > 10 12. y < x + 4
2. (8, −1) 13. y > x
2x − y ≤ −11 14. 2x ≤ y
3. (−3, 2) 15. x + y ≥ 3
x ≥ −1
16. x + y < −1
4. (7, −4)
y <6 17. 2x − 6 y ≥ 6
5. (0, 5) 18. −4x − 8 y > 8
3x + 3 > y − 3
19. y < −2
6. (0, 0)
20. x < 1
14x < 17 y
21. 2x − 3 y < 12
22. 3x + 2 y ≥ −6
Graph the solution set for the given in-
equality. 23. −2x + 5 y < −10
7. y < x − 2 24. 5x + 4 y ≤ 20
8. y ≤ x + 5 25. 6 y ≥ 9x + 18
9. 4 y ≤ 6x 26. 5 y + 25x ≤ −25
244
CHAPTER 4. LINEAR SYSTEMS 5. MULTIVARIATE LINEAR INEQUALITIES
Homework 4.5 Answers.
1. no 11.
4
y
3
2
3. no
1
x
−4 −3 −2 −1 1 2 3 4
−1
5. yes
−2
−3
−4
7.
4
y
3 13.
3
y
2
2
1
x
1
−4 −3 −2 −1 1 2 3 4
−1 x
−3 −2 −1 1 2 3
−2
−1
−3
−4 −2
−3
9. 15.
6 y 5
y
5 4
4
3
3
2
2
1 1
x x
−6 −5 −4 −3 −2 −1 1 2 3 4 5 6 −5 −4 −3 −2 −1 1 2 3 4 5
−1 −1
−2
−2
−3
−3
−4
−5 −4
−6 −5
245
5. MULTIVARIATE LINEAR INEQUALITIES CHAPTER 4. LINEAR SYSTEMS
17. 23.
7 7
y y
6 6
5 5
4 4
3 3
2 2
1 1
x x
−7 −6 −5 −4 −3 −2 −1
−1 1 2 3 4 5 6 7 −7 −6 −5 −4 −3 −2 −1
−1 1 2 3 4 5 6 7
−2 −2
−3 −3
−4 −4
−5 −5
−6 −6
−7 −7
19. 25.
4
y 6 y
3 5
4
2
3
1 2
x 1
x
−4 −3 −2 −1 1 2 3 4
−6 −5 −4 −3 −2 −1 1 2 3 4 5 6
−1 −1
−2
−2
−3
−3 −4
−5
−4
−6
21.
7
y
6
5
4
3
2
1
x
−7 −6 −5 −4 −3 −2 −1
−1 1 2 3 4 5 6 7
−2
−3
−4
−5
−6
−7
246
CHAPTER 4. LINEAR SYSTEMS 5. PRACTICE TEST 4
Practice Test 4
1. Solve the system by graphing: 6. A total of $2400 is invested into simple
interest accounts, part of it at 5% and the
3y − x = 3
rest at 10%. The total interest after one
2x − 4 = y year is $175. How much was invested at
each rate?
2. Solve the system:
2x + y = 12
7. A pastry delivery is a mix of donuts and
y − 3x = 2 scones, 40 items in total. Each donut costs
$1.50, while each scone costs $2.50. Find
3. Solve the system: how many of each type of pastry there are
if the total price of the order is $76.
−2x + 3 y = −1
2x + 5 y = 25 8. How many liters of 20% saline solution
should be added to 40 liters of a 50% saline
4. Solve the system: solution to make a 30% solution? How
y = −3x − 4 many liters of 30% solution will result?
0 = 6x + 2 y 9. Graph the solution set for inequality:
5. A party with 3 adults and 2 children
pays 45 dollars for their tickets at Acme y < 5x − 2
Theater. Another party, with 4 adults and
4 children, pays 68 dollars. Find the price 10. Graph the solution set for inequality:
of one adult ticket and one the price of one
child ticket. 2 y + 6 ≥ −2(x + y − 3)
247
5. PRACTICE TEST 4 CHAPTER 4. LINEAR SYSTEMS
Practice Test 4 Answers.
1. (3, 2) d + s = 40
5 1.5d + 2.5s = 76
y
4 solution: d = 24, s = 16
3
2 8.
1 x and y are volumes of 20% and 30% so-
x lutions respectively, in liters
−5 −4 −3 −2 −1 1 2 3 4 5
x + 40 = y
−1
−2 0.2x + 0.5 · 40 = 0.3 · y
−3
solution: x = 80, y = 120
−4
−5 9.
5
y
2. (2, 8) 4
3
3. (5, 3) 2
1
4. no solutions x
−5 −4 −3 −2 −1 1 2 3 4 5
−1
5.
−2
a and c are prices of one adult and one
child ticket respectively, in dollars −3
−4
3x + 2 y = 45
−5
4x + 4 y = 68
solution: a = 11, c=6 10.
4
y
6. 3
x and y are amounts invested at 5% and 2
10% respectively, in dollars
1
x + y = 2400
x
0.05x + 0.1 y = 175 −4 −3 −2 −1 1 2 3 4
solution: x = 1300, y = 1100 −1
−2
7.
−3
d and s are quantities of donuts and scones
respectively −4
248
CHAPTER 5
Polynomial Expressions
1. Integer Exponent
2 y
1
x
−2 −1 1 2
y=x
y = x2
−1
y = x3
−2
1.1. Definition.
DEFINITION 1.1.1 (Integer Exponent). Recall that given a real number a and a positive
b · . . . · b}. If n = 0 and b is any real number, then
integer n, notation b n means |b · b · {z
n times
b n = b0 = 1
If the exponent is negative and the base b 6= 0, then the reciprocal of the base is raised to
the corresponding power: n
1 1
b =
−n
= n
b b
In particular, b−1 becomes another way to express the reciprocal of b.
249
1. INTEGER EXPONENT CHAPTER 5. POLYNOMIAL EXPRESSIONS
BASIC EXAMPLE 1.1.1. Some trivial uses of the definition:
x1 = x this is true for any real number x
3
1 1
2 −3
= = reciprocal of 2 cubed
2 8
−1
3 7
− = − exponent −1 yields the reciprocal of the base
7 3
(x 4 + 5x 2 + 13)0 = 1 exponent zero always results in 1 for any base
−2 2
7 5 25
= = take the reciprocal of the base, then square it
5 7 49
THEOREM 1.1.1 (Exponent with Base ±1). The exponent with base 1 is always 1. If n is
an integer, then
1n = 1
It is also true that (−1)n = 1 for all even n and (−1)n = −1 for all odd n, even if the
exponent n is negative.
When we simplify a product or a quotient of exponential expressions, we will attempt to bring
it into the form with a single numerical coefficient, with each base appearing only once, and
with all the exponents positive. We will also defer our exploration of the negative case until
the end of this chapter, since polynomial expressions only make use of the non-negative integer
exponent. It is of note, though, that the theorems which follow work for the negative exponent
just as well.
1.2. Products and Quotients of Exponents.
THEOREM 1.2.1 (Product Rule for Exponents). A product of exponential expressions with
the same base can be written as a single base with the sum of exponents. If m, n are
integers and b is a real number, then
b m b n = b m+n
BASIC EXAMPLE 1.2.1. It is easy enough to see why the product rule works if we follow the
definition to simplify a product with the same base raised to various powers.
b3 b2 = (bbb)(bb)
= bbbbb
= b5
250
CHAPTER 5. POLYNOMIAL EXPRESSIONS 1. INTEGER EXPONENT
EXAMPLE 1.2.1. Simplify the expression:
54 · 53 · 5
SOLUTION:
54 · 53 · 5 = 54 · 53 · 51
= 54+3+1
= 58
= 390625
ANSWER: 390625
EXAMPLE 1.2.2. Simplify the expression:
(1 + x)5 (1 + x)7
SOLUTION: Here the common base is (1 + x).
(1 + x)5 (1 + x)7 = (1 + x)5+7
= (1 + x)12
Recall that exponent does not distribute over addition, so we cannot simplify this answer
any more.
ANSWER: (1 + x)12
EXAMPLE 1.2.3. Simplify the expression:
(a2 b6 )(ab3 )
SOLUTION: Factors of a product can be multiplied in any order, so we bring similar bases
next to each other, then apply the product rule:
(a2 b6 )(ab3 ) = a2 a1 b6 b3
= a2+1 b6+3
= a3 b9
ANSWER: a3 b9
251
1. INTEGER EXPONENT CHAPTER 5. POLYNOMIAL EXPRESSIONS
EXAMPLE 1.2.4. Simplify the expression:
(−4x 3 )(−3x 17 )
SOLUTION:
(−4x 3 )(−3x 17 ) = (−4)(−3)x 3 x 17
= 12x 3+17
= 12x 20
ANSWER: 12x 20
THEOREM 1.2.2 (Quotient Rule for Exponents). A quotient of exponential expressions
with the same base can be written as a single base with the difference of exponents. If m
and n are integers and b 6= 0 is a real number, then
bm
= b m−n
bn
BASIC EXAMPLE 1.2.2. It is easy to see why the quotient rule works if we follow the defi-
nition.
b6 b bb bb b
= four pairs of b factors cancel
b4 b bb b
= bb
= b2
BASIC EXAMPLE 1.2.3. The quotient rule is actually a direct consequence of definition of
the negative exponent and the product rule. Recall that the fraction is in fact a product of
the numerator and the reciprocal of the denominator:
bm m 1
= (b ) n division is multiplication by reciprocal
bn b
= b m b−n by definition of negative exponent
= b m+(−n) product rule
= b m−n
252
CHAPTER 5. POLYNOMIAL EXPRESSIONS 1. INTEGER EXPONENT
EXAMPLE 1.2.5. Simplify the expression:
513
510
SOLUTION:
513
= 513−10
510
= 53
= 125
ANSWER: 125
EXAMPLE 1.2.6. Simplify the expression:
5x 6 y 8
15x 2 y
SOLUTION:
5x 6 y 8 1 x6 y8
= · lowest terms
15x 2 y 3 x2 y1
1 6−2 8−1
= x y
3
1 4 7
= x y
3
1 4 7
ANSWER: x y
3
253
1. INTEGER EXPONENT CHAPTER 5. POLYNOMIAL EXPRESSIONS
1.3. Exponents of Sums. It is a common mistake is to assume that exponent distributes
over addition as well as over multiplication, but this is not the case. Students faced with an
expression such as
(x + 2)2
are often tempted to rewrite it as x 2 + 22 or x 2 + 4, but this is not true in general, as anyone
can check by comparing the values of these expressions for a specific x. For example, if x = 1
then
(x + 2)2 = (1 + 2)2 = 32 = 9
x2 + 4 = 12 + 4 = 5
So these expressions are not equivalent. Instead, when faced with exponents of sums, we will
replace exponents by multiplication and then apply the distributivity of multiplication over
addition:
(x + 2)2 = (x + 2)(x + 2)
= x · (x + 2) + 2 · (x + 2)
= x · x + x ·2+2· x +2·2
= x 2 + 2x + 2x + 4
= x 2 + 4x + 4
For now, though, we will leave answers like (x + 2)2 in exponential form.
254
CHAPTER 5. POLYNOMIAL EXPRESSIONS 1. INTEGER EXPONENT
Homework 5.1.
Simplify the given expression. 13. (x 2 y 7 )(x y 2 )
1. (−1)1982 14. (n3 k10 )(n5 k4 )
2. (−1)1001 15. (−x 4 y)(−6x 2 y 11 )
3. 00 16. (−5a3 b0 )(−20a5 b7 )
4. (1 + x 2 )0 1146
17.
1144
5. 23 · 24
7108
3
6. 3 · 3 3 18.
7105
7. (−4)(−4)2 (−4)3 (−6)16 (−6)4
19.
(−6)15
8. (−5)3 (−5)2
(−0.9)13
9. X 1 X 20 X 300 20.
(−0.9)2 (−0.9)8
10. Y 20 Y 0 Y 22
60x 8 y 12
21.
11. (x − y)6 (x − y)7 −15x 5 y 11
12. (a + 2b)19 (a + 2b) 8a6 y 5
22.
24x 5 y 3
255
1. INTEGER EXPONENT CHAPTER 5. POLYNOMIAL EXPRESSIONS
Homework 5.1 Answers.
1. 1 13. x 3 y 9
3. 1
15. 6x 6 y 12
5. 128
17. 121
7. 4096
19. −7776
9. X 321
11. (x − y)13 21. −4x 3 y
256
CHAPTER 5. POLYNOMIAL EXPRESSIONS 2. PROPERTIES OF EXPONENT
2. Properties of Exponent
2.1. Exponent of Exponent.
THEOREM 2.1.1 (Power Rule for Exponents). When working with an exponent of an expo-
nential expression, we can rewrite it as the same base with the product of the exponents.
If m and n are integers and b is a real number, then
(b m )n = b m·n
BASIC EXAMPLE 2.1.1. It is easy to see why the power rule works if we follow the definition.
(b2 )3 = (bb)3
= (bb)(bb)(bb)
= bbbbbb
= b6
EXAMPLE 2.1.1. Simplify the expression: (23 )4
SOLUTION:
(23 )4 = 23·4
= 212
= 4096
ANSWER: 4096
EXAMPLE 2.1.2. Simplify the expression:
((x 5 )3 )7
SOLUTION:
((x 5 )3 )7 = (x 5·3 )7
= (x 15 )7
= x 15·7
= x 105
ANSWER: x 105
257
2. PROPERTIES OF EXPONENT CHAPTER 5. POLYNOMIAL EXPRESSIONS
2.2. Distributivity.
THEOREM 2.2.1 (Distributivity of Exponent over Multiplication). Integer exponent dis-
tributes over multiplication. If n is an integer and a, b are real numbers, then
(ab)n = a n b n
This rule extends naturally to products with more than 2 factors:
(x yz)n = x n y n z n
BASIC EXAMPLE 2.2.1. It is easy to see why distributivity works if we follow the definition.
(a b)4 = (a b)(a b)(a b)(ab)
= (aaaa)(b bb b) changed the order of multiplication
= a4 b4
EXAMPLE 2.2.1. Simplify the expression:
(−3x)4
SOLUTION: The base of this exponential expression is a product with two factors: −3 and
x, so we apply the distributive property:
(−3x)4 = (−3)4 x 4
= 81x 4
ANSWER: 81x 4
EXAMPLE 2.2.2. Simplify the expression:
(−10abc)3
SOLUTION:
(−10a bc)3 = (−10)3 a3 b3 c 3
= −1000a3 b3 c 3
ANSWER: −1000a3 b3 c 3
258
CHAPTER 5. POLYNOMIAL EXPRESSIONS 2. PROPERTIES OF EXPONENT
EXAMPLE 2.2.3. Simplify the expression:
(x 3 y 4 )5
SOLUTION: The base of the exponent has two factors: x 3 and y 4 , so we distribute the
exponent first, and then apply the rule for exponentiating the exponent:
(x 3 y 4 )5 = (x 3 )5 ( y 4 )5
= x 3·5 y 4·5
= x 15 y 20
ANSWER: x 15 y 20
EXAMPLE 2.2.4. Simplify the expression:
(−2ab2 )3 (a3 b)5
SOLUTION: Distribute exponents over the products first, and simplify exponents of expo-
nents:
(−2a b2 )3 (a3 b)5 = (−2)3 a3 (b2 )3 · (a3 )5 b5
= −8a3 b2·3 · a3·5 b5
= −8a3 b6 · a15 b5
The factors of this product can be multiplied in any order, so we bring similar bases next
to each other, and then apply the product rule:
−8a3 b6 · a15 b5 = −8 · a3 a15 · b6 b5
= −8 · a3+15 · b6+5
= −8a18 b11
ANSWER: −8a18 b11
259
2. PROPERTIES OF EXPONENT CHAPTER 5. POLYNOMIAL EXPRESSIONS
EXAMPLE 2.2.5. Simplify the expression:
(4x y)17
(4x y)14
SOLUTION: Here the common base is (4x y), so we can apply the quotient rule, and
then distribute the exponent over the product. Alternatively, we could distribute first, and
cancel common factors later.
(4x y)17
= (4x y)17−14
(4x y)14
= (4x y)3
= 43 x 3 y 3 distribute exponent over the product
= 64x y3 3
ANSWER: 64x 3 y 3
THEOREM 2.2.2 (Distributivity of Exponent in Fractions). Integer exponent distributes
over division as well as over multiplication. If n is an integer, a is real, and b is a non-zero
real, then a n a n
= n
b b
This rule extends naturally to fractions with more than 2 factors:
2x n 2n x n
= n n n
abc a b c
BASIC EXAMPLE 2.2.2. It is easy to see why distributivity works if we follow the definition.
a 3 aaa
=
b b b b
aaa
=
bbb
a3
=
b3
260
CHAPTER 5. POLYNOMIAL EXPRESSIONS 2. PROPERTIES OF EXPONENT
EXAMPLE 2.2.6. Simplify the expression:
2
7
c
SOLUTION:
2
7 72
=
c c2
49
=
c2
49
ANSWER:
c2
EXAMPLE 2.2.7. Simplify the expression:
3
−3d
x4
SOLUTION: The base of this exponential expression is a product of three factors: −3, d,
and the reciprocal of x 4 , so we can apply the distributive property:
−3d 3 (−3)3 d 3
=
x4 (x 4 )3
−27d 3
= power rule
x 4·3
−27d 3
=
x 12
−27d 3
ANSWER:
x 12
261
2. PROPERTIES OF EXPONENT CHAPTER 5. POLYNOMIAL EXPRESSIONS
EXAMPLE 2.2.8. Simplify the expression:
(−2bc 5 )3
6b2 c 2
SOLUTION: The base of the exponential expression in the numerator is a product of 3
factors: −2, b, and c 5 . We have to distribute the exponent over this product before we
can cancel common factors in the fraction:
(−2bc 5 )3 (−2)3 b3 (c 5 )3
= distributed exponent over the product
6b2 c 2 6b2 c 2
−8b3 c 5·3
= power rule
6b2 c 2
4 b3 c 15
= − lowest terms
3 b2 c 2
4
= − b3−2 c 15−2 quotient rule
3
4
= − bc 13
3
4
ANSWER: − bc 13
3
262
CHAPTER 5. POLYNOMIAL EXPRESSIONS 2. PROPERTIES OF EXPONENT
Homework 5.2.
Simplify the given expression. (m3 n2 )6
12.
(mn3 )4
1. (g 5 )8
13. (32 )3
2. (h )9 10
14. ((22 )2 )2
3. (−2k x)5
3(2 )
3
4. (cat) 7 15.
x(3 )
4
5. 30 + 5x 0 16.
3
6. (2c 0 )4 17. (x 5 y)(x y)2
7. (x y)7 ( yz)3 18. (x y)5 (x 3 y)3
2
8. (a2 b)4 (a5 b6 ) 4
x3
19.
9. (3m3 c 3 )(−3m10 c 6 )2 −2x y 5
3
10. (a7 b4 )2 (ab2 c)9 a5
20.
−3a3 b
(g h )
7 2 2
11.
(g 2 h)3
263
2. PROPERTIES OF EXPONENT CHAPTER 5. POLYNOMIAL EXPRESSIONS
Homework 5.2 Answers.
1. g 40 11. g 8 h
3. −32k5 x 5 13. 729
15. 6561
5. 6
17. x 21 y 9
7 10 3
7. x y z
x8
23 15 19.
9. 27m c 16 y 20
264
CHAPTER 5. POLYNOMIAL EXPRESSIONS 3. INTRODUCTION TO POLYNOMIALS
3. Introduction to Polynomials
6 y
4
2
x
−3 −2 −1 1 2 3
−2
y = x(x + 1)(x − 2)
−4
−6
3.1. Definitions.
DEFINITION 3.1.1. A monomial expression, or just monomial, is an algebraic expression
which is a product of numerical constants and/or variables, with each variable optionally
carrying a positive integer exponent.
A simplified form of a monomial expression has a single numerical factor in the very front,
called coefficient, and lists each variable base at most once.
BASIC EXAMPLE 3.1.1. This is a monomial with variables x and y and constants −3 and 2:
−3 · x 2 · y 5 · x · 2
To simplify, rearrange the factors and use the properties of exponent to combine the bases:
−3 · x 2 · y 5 · x · 2 = (−3)(2) · x 2 x 1 · y 5
= −6 · x 2+1 · y 5
= −6x 3 y 5
In simplified form, this monomial has 3 factors and coefficient −6.
These are not monomials:
x + y, seen as a product, has a single factor (x + y) which is neither a numerical constant
nor a variable. In general, no sum with more than one term is a monomial.
2/x has the reciprocal of x as a factor, which is the same as x −1 , but reciprocals of variables
and negative exponents are not allowed.
(3x)2 is not a monomial expression, since its only factor is neither a variable nor a numer-
ical constant. It is though equivalent to 9x 2 , which is a monomial with two factors and
coefficient 9.
265
3. INTRODUCTION TO POLYNOMIALS CHAPTER 5. POLYNOMIAL EXPRESSIONS
DEFINITION 3.1.2. The degree of a monomial is the sum of all the exponents of the vari-
ables. Variables without visible exponents are considered to have exponent 1. Non-zero
monomials with no variables have degree 0. The degree of the zero monomial is unde-
fined. Descriptive names exist for monomials of low degrees, and learning the names of
degrees 0 through 2 is particularly crucial for working with this text.
degree type of a monomial
0 constant
1 linear
2 quadratic
3 cubic
4 quartic
5 quintic
BASIC EXAMPLE 3.1.2. Here are some monomials and their degrees.
7x y, which is equivalent to 7x 1 y 1 , has degree 1 + 1 = 2, so it is a quadratic monomial
with coefficient 7.
−17 has degree 0, so it is a constant monomial with coefficient −17.
x y 2 has degree 1 + 2 = 3, so it is a cubic monomial with coefficient 1.
−x y 2 z 5 has degree 1 + 2 + 5 = 8 and coefficient −1.
0 is the zero monomial. Its degree is traditionally left undefined in order to satisfy the
following rule: the degree of a product of monomials is the sum of individual degrees. This
rule works for any two non-zero monomials, but if we let the degree of the zero monomial
to be 0, then any product involving it would also have degree 0, which is against the rule.
DEFINITION 3.1.3. A polynomial expression, or just polynomial, is a sum of monomial ex-
pressions. A polynomial in one variable, or univariate polynomial, uses at most one variable
throughout the expression. A polynomial in two variables, or bivariate polynomial, uses at
most two.
DEFINITION 3.1.4 (Standard Form for Polynomials). When writing down a polynomial in
one variable x, it is traditional to order monomials in the descending order of degree,
starting with the highest one, so the standard way of writing a polynomial like
x − 3 + 2x 2 − 5x 7 is − 5x 7 + 2x 2 + x − 3
The coefficient of the highest degree term is called the leading coefficient of the polynomial.
266
CHAPTER 5. POLYNOMIAL EXPRESSIONS 3. INTRODUCTION TO POLYNOMIALS
There are no firm conventions for writing down a polynomial in two or more variables, or for
defining the leading coefficient in that case.
DEFINITION 3.1.5 (Polynomial Names). The following names exist for polynomials with
a low number of terms:
number of terms type of a polynomial
1 monomial
2 binomial
3 trinomial
DEFINITION 3.1.6. The degree of a polynomial is the highest of all the monomial degrees.
The degree of the zero polynomial is undefined. The descriptive names for low degrees
are the same as for monomials.
BASIC EXAMPLE 3.1.3. Here are some polynomials with their properties described:
polynomial number of terms degree description
20 1 0 constant monomial
x −3 2 1 linear binomial
x + x yz
3
2 3 cubic binomial
2
x − x −1 3 2 quadratic trinomial
−x y + x y + x
2 2 6 5
3 7 7th degree trinomial
BASIC EXAMPLE 3.1.4. These are not polynomials:
1
x− has a term which is not a monomial.
y
4(x + 1) is not strictly speaking a polynomial expression, since it consists of a single term
which is not a monomial. It is equivalent though to 4x + 4, which is a linear binomial, so
later in the text we will refer to it as a polynomial in a fully factored form.
3.2. Combining Similar Monomial Terms.
DEFINITION 3.2.1. Monomial terms are like (used an an adjective) if their simplified forms
consist of exactly the same variables raised to exactly the same powers. Sometimes like
terms are also called similar terms.
267
3. INTRODUCTION TO POLYNOMIALS CHAPTER 5. POLYNOMIAL EXPRESSIONS
EXAMPLE 3.2.1. Simplify the polynomial by combining like terms.
5x − 17x
SOLUTION: We combine like terms by applying the distributive property to their common
factors, which is all the variables.
5x − 17x = (5 − 17)x
= −12x
ANSWER: −12x
EXAMPLE 3.2.2. Simplify the polynomial by combining like terms.
x y 2 + 13 y 2 x
SOLUTION: These terms don’t look the same, but they are in fact similar, since we can
rearrange the order of multiplication at will:
x y 2 + 13 y 2 x = x y 2 + 13x y 2
= (1 + 13)x y 2
= 14x y 2
ANSWER: 14x y 2
EXAMPLE 3.2.3. Simplify the polynomial by combining like terms.
3ab3 − 5 + a3 b + 4 − 6ab3
SOLUTION: Note that the terms 3a b3 and a3 b are not similar. Even though they have the
same variables, the corresponding exponents are different. We can rearrange the terms of
the sum so as to put like terms next to each other, and then apply the distributive property:
3a b3 − 5 + a3 b + 4 − 6ab3 = 3ab3 − 6ab3 − 5 + 4 + a3 b
= (3 − 6)ab3 − 1 + a3 b −5 + 4 = −1
= −3ab3 − 1 + a3 b
ANSWER: −3ab3 − 1 + a3 b
268
CHAPTER 5. POLYNOMIAL EXPRESSIONS 3. INTRODUCTION TO POLYNOMIALS
EXAMPLE 3.2.4. Simplify the polynomial by combining like terms.
0.3x − 0.6 + 2.3x 2 − 1.7x − x 2 + 0.6
SOLUTION: Put like terms next to each other and distribute:
0.3x − 0.6 + 2.3x 2 − 1.7x − x 2 + 0.6 = 2.3x 2 − x 2 + 0.3x − 1.7x − 0.6 + 0.6
= (2.3 − 1)x 2 + (0.3 − 1.7)x + 0
= 1.3x 2 − 1.4x
In answers, we will list the terms of polynomials in one variable in the descending order
of degree.
ANSWER: 1.3x 2 − 1.4x
3.3. Evaluating Polynomials in Applications.
EXAMPLE 3.3.1. Evaluate the expression x 2 − 2x − 6 for x = 5.
SOLUTION:
x 2 − 2x − 6 = (5)2 − 2 · (5) − 6
= 25 − 10 − 6
= 9
ANSWER: 9
EXAMPLE 3.3.2. Evaluate the expression −3x 2 + 2x y − 4 y 2 for x = 3 and y = −2.
SOLUTION:
−3x 2 + 2x y − 4 y 2 = −3 · (3)2 + 2(3)(−2) − 4(−2)2
= (−3)(9) + (−12) − 4(4)
= −27 − 12 − 16
= −55
ANSWER: −55
269
3. INTRODUCTION TO POLYNOMIALS CHAPTER 5. POLYNOMIAL EXPRESSIONS
EXAMPLE 3.3.3. Professor calculates the class grade (out of 100 points) by computing the
value of the following expression:
0.1H + 0.1Q + 0.6T + 0.2F
where H, Q, T , and F are the points accumulated for homework, quizzes, tests, and the
final respectively. Find the class grade for a student with H = 85, Q = 90, T = 76, and
F = 68.
SOLUTION:
0.1H + 0.1Q + 0.6T + 0.2F = 0.1(85) + 0.1(90) + 0.6(76) + 0.2(68)
= 8.5 + 9 + 45.6 + 13.6
= 76.7
ANSWER: 76.7 points
EXAMPLE 3.3.4. The volume of a pyramid with a square base is given by the expression
1 2
a h
3
where a is the side of the square and h is the height of the pyramid. Find the volume
of the Great Pyramid of Giza, with dimensions a = 230.4 meters and h = 146.5 meters.
(Actually, the pyramid’s top deteriorated over time, so the actual height is a bit lower, but
the historical height we are using here gives a better volume estimate.)
SOLUTION:
1 2 1
a h = (230.4)2 · 146.5
3 3
= 2592276
ANSWER: 2592276 cubic meters
270
CHAPTER 5. POLYNOMIAL EXPRESSIONS 3. INTRODUCTION TO POLYNOMIALS
Homework 5.3.
Determine whether the given expression is 16. − y 14
a polynomial expression.
17. 6x 3 − 3x 2 + 2x − 1
1. 7x − 3
18. 3x 8 + 12x 3 − 8
2. 2x + 9 − 7x
5 2
19. −x y 2 z 3
x +1
2
3.
x 2 − 5x + 1 20. m7 + x 4 y 4
4. −11 21. a + b + c + d + 3x
5. x 5 + x −5 22. ab − cd − x y
1 2
6. − 2+x
x x
Combine the like terms. For polynomials in
7. (x + 1)(x 2 − 10) one variable, state answers in the standard
form, with terms listed in the descending
1 2
8. x − x 10 order of degree.
3 7
23. 6n2 − 5 + 5n2
2 1
9. − x 3 + x 2 +
7 7 24. 4x + 7x 2 + 3x
10. x 3 − x(x + 1)
25. 3x 4 − 2x + 2x + x 4
26. 9a5 + 3a2 − 2a5 − 3a2
For each given polynomial, describe its de-
gree and the number of terms in words. For 27. 10 y 2 + 2 y 3 − 3 y 3 − 4 y 2 − 6 y 2 − y 4
example, x + 1 is a linear binomial, while
x 2 + x y + y 2 is a quadratic trinomial. Use 28. 12b6 − b3 + 8b6 + 4b3 − b7 − 3b3
numbers for describing polynomials of high
degree and/or with too many terms. For 29. 14x 2 − 5x y + x y 2 − x y − 7x 2 + y 2 x
example, x 99 + x 3 + x 2 + x is a polynomial
of degree 99 with 4 terms. 30. −a5 b2 + 4a2 b5 − 2a5 b2 − 6b5 a2
1 2 2 5 1
11. −5 31. 9x 3 + x − x + 7x 3 + x + x 2
2 3 3 4
12. 6x 1 4 12 3 1
32. y − 2 y3 + y − y − y4 + 6 y3
9 5 5 3
13. 5x y 2 − 3
14. x y + 1 − ab
1
15. a2 − bc
2
271
3. INTRODUCTION TO POLYNOMIALS CHAPTER 5. POLYNOMIAL EXPRESSIONS
Evaluate each polynomial if x = 3, y = −2, 42. The distance in meters traveled by a
and z = 0.1 projectile t seconds after the start of the
timer can be found as the value of the fol-
33. −3x + 9 lowing expression:
34. 12 − 10x 5 2
3.5 + 1.5t + t
2
35. 2x 2 − 3x + 6 Find the distance traveled by the projectile
10 seconds after the start of the timer.
36. −3x 2 − 4x + 5
1 2 43. The amount of drug in the blood-
37. 2 y 4 − y stream, measured in mcg/mL, can be es-
2
timated by the value of the expression
2
38. − y 3 + y −0.004t 4 + 0.004t 3 + 0.35t 2 + 0.6t
5
where t is the time in hours since the drug
39. x 2 − y 2 z was administered. Find the amount of drug
in the bloodstream 5 hours after the injec-
40. x 4 y 3 z 2
tion.
41. A chemist finds the amount (measured
44. When the technology arm of Acme Co
in liters) of acid in a mixture of x liters
produces x computer monitors, the pro-
of solution A with y liters of solution B by
jected profit in dollars can be found using
computing the value of the expression
the expression
0.3x + 0.7 y
250x − 1.1x 2 − 4000
Find the amount of acid in a mixture of
x = 1.2 liters of solution A with y = 0.85 Find the profit corresponding to the pro-
liters of solution B. duction of 190 monitors.
272
CHAPTER 5. POLYNOMIAL EXPRESSIONS 3. INTRODUCTION TO POLYNOMIALS
Homework 5.3 Answers.
1. yes 23. 11n2 − 5
3. no, because it is a fraction with a variable 25. 4x 4
denominator
27. − y 4 − y 3
5. no, because it has negative exponents
7. no, even though it can be shown that this 29. 7x 2 − 6x y + 2x y 2
product of binomials is equivalent to a cu-
bic polynomial 3 2
31. 16x 3 + x +x
4
9. yes, because rational coefficients are OK
33. 0
11. constant monomial
35. 15
13. cubic binomial
37. 30
15. quadratic binomial
39. 8.6
17. cubic polynomial with 4 terms
19. 6th degree monomial 41. 0.995 liters
21. linear polynomial with 5 terms 43. 9.75 mcg/mL
273
4. SUMS OF POLYNOMIALS CHAPTER 5. POLYNOMIAL EXPRESSIONS
4. Sums of Polynomials
We add polynomials by combining like terms.
EXAMPLE 4.0.1. Add polynomials by combining like terms:
(4x 2 + 6x − 5) + (x 2 − 4x + 5)
SOLUTION: Adding a sum amounts to adding each of its terms. To make like terms easier
to see, we change the order of summation and put them next to each other:
(4x 2 + 6x − 5) + (x 2 − 4x + 5) = 4x 2 + 6x − 5 + x 2 − 4x + 5
= 4x 2 + x 2 + 6x − 4x − 5 + 5
= (4 + 1)x 2 + (6 − 4)x + 0
= 5x 2 + 2x
ANSWER: 5x 2 + 2x
BASIC EXAMPLE 4.0.1. Subtracting a polynomial amounts to adding its opposite. To rewrite
the opposite of a polynomial as a polynomial expression, we remove the parentheses and
invert the sign of each term:
−(x 7 − x 4 a3 − x a6 + 7) = −x 7 + x 4 a3 + x a6 − 7
EXAMPLE 4.0.2. Subtract polynomials by combining like terms:
(5x 2 − 7x − 6) − (2x 2 + 3x − 4)
SOLUTION: The opposite of a sum is equal to the sum of opposites. This allows us to
remove the parentheses on the second polynomial by taking the opposite of each term:
(5x 2 − 7x − 6) − (2x 2 + 3x − 4) = 5x 2 − 7x − 6 − 2x 2 − 3x + 4
= 5x 2 − 2x 2 − 7x − 3x − 6 + 4
= (5 − 2)x 2 + (−7 − 3)x − 2
= 3x 2 − 10x − 2
ANSWER: 3x 2 − 10x − 2
274
CHAPTER 5. POLYNOMIAL EXPRESSIONS 4. SUMS OF POLYNOMIALS
EXAMPLE 4.0.3. Find the perimeter of the shape by adding lengths of all sides:
4x + 1
2x − 5 2x − 5
4x + 1
SOLUTION: The terms of a sum can be added in any order, so we can put like terms next
to each other and apply the distributive property.
(4x + 1) + (2x − 5) + (4x + 1) + (2x − 5) = 4x + 2x + 4x + 2x + 1 − 5 + 1 − 5
= (4 + 2 + 4 + 2)x − 8
= 12x − 8
ANSWER: 12x − 8
275
4. SUMS OF POLYNOMIALS CHAPTER 5. POLYNOMIAL EXPRESSIONS
Homework 5.4.
Simplify the given expression and state the Simplify the given expression and state the
answer as a polynomial expression. If a answer as a polynomial expression. If a
polynomial is in one variable, list the terms polynomial is in one variable, list the terms
of the answer in the decreasing order of de- of the answer in the decreasing order of de-
gree. gree.
1. (3x + 7) + (x + 2) 17. (3 y + 1) − (5 y + 8)
2. (x + 10) + (12x + 1) 18. (7 y − 3) − (11 y + 17)
3. (2t + 1) + (−8t + 7) 19. (4x 2 + x − 7) − (3 − 8x 2 − 4x 3 )
4. (4t + 2) + (−11t − 3) 20. (3a3 − 2a + 7) − (5a2 − 2a3 + 2a)
5. (7t 2 − 3t + 9) + (2t 2 + 4t − 6) 21.
1 3 3 2 7
6. (8a2 + 4a − 7) + (6a2 − 4a − 1) y + 2 y2 − − − y3 + 2 y2 +
5 10 5 1000
7. (3t 3 + 4t 2 − 1) + (−2t 2 − 4t + 5)
5 3 1 1 1 3 1 1
22. y − y− − − y + y−
8. (4.9x 3 +3.2x 2 −5.1x)+(2.1x 2 −3.7x +4.5) 8 4 3 2 4 3
4 2 1 1 2 1 1 23. (0.9x 3 +0.2x −5)−(0.7x 4 −0.3x −0.1)
9. x + x −3 + x − x−
3 2 3 8 2
24.
10.
0.07x 3 − 0.03x 2 + 0.01x − 0.02x 3 − 0.04x 2 − 1
3 4 1 3 2 2 4 1 3 3 2
y + y − y +3 + y − y − y
5 2 3 5 4 4
Find the perimeter of the shape by adding
11. (3x 2 +6x y − x y 2 −4)+(−x y − x y 2 +4)
lengths of all sides.
12. (7a2 b2 − 6ab) + (a4 + a2 b2 + 6ab)
25.
7x
14 − 2x 14 − 2x
Rewrite the opposites of polynomials as
polynomial expressions. 7x
13. −(−5x + 7) 26.
6 − 2x
14. −(x 2 + x)
3 + 3x 3 + 3x
15. −(3a3 − a2 + 8b − 7ab)
16. −(−12z 6 − d 3 − zd) 6 − 2x
276
CHAPTER 5. POLYNOMIAL EXPRESSIONS 4. SUMS OF POLYNOMIALS
27. 28.
5x 2 − 1 16x − 3
3x − x 2 3x 2 − 1
17 − x x 2 + 2x
277
4. SUMS OF POLYNOMIALS CHAPTER 5. POLYNOMIAL EXPRESSIONS
Homework 5.4 Answers.
1. 4x + 9 15. −3a3 + a2 − 8b + 7ab
3. −6t + 8 17. −2 y − 7
5. 9t 2 + t + 3 19. 4x 3 + 12x 2 + x − 10
7. 3t 3 + 2t 2 − 4t + 4 3 3 307
21. y −
5 1000
5 2 3 7
9. x + x−
3 8 2 23. −0.7x 4 + 0.9x 3 + 0.5x − 4.9
11. 3x 2 + 5x y − 2x y 2 25. 10x + 28
13. 5x − 7 27. 4x 2 + 2x + 16
278
CHAPTER 5. POLYNOMIAL EXPRESSIONS 5. PRODUCTS OF POLYNOMIALS
5. Products of Polynomials
5.1. Monomial Times Monomial. Since the order of multiplication does not affect the
value of a product, we can multiply monomials by multiplying their coefficients and adding
the variable powers using the product rule for exponent.
EXAMPLE 5.1.1. Simplify the expression:
(−5x 2 y 6 )(−2x y 3 z 4 )
SOLUTION:
(−5x 2 y 6 )(−2x y 3 z 4 ) = (−5)(−2) · x 2 x · y 6 y 3 · z 4
= 10 · x 2+1 · y 6+3 · z 4 product rule
= 10x 3 y 9 z 4
ANSWER: 10x 3 y 9 z 4
5.2. Monomial Times Polynomial.
EXAMPLE 5.2.1. Simplify the expression:
3x 2 (5x 2 − 4x − 1)
SOLUTION: This can be done by distributing, and then simplifying the resulting monomial
products:
3x 2 (5x 2 − 4x − 1) = (3x 2 )(5x 2 ) − (3x 2 )(4x) − (3x 2 )(1)
= 3 · 5 · x2 · x2 − 3 · 4 · x2 · x − 3 · 1 · x2
= 15x 2+2 − 12x 2+1 − 3x 2 product rule
= 15x 4 − 12x 3 − 3x 2
This is a univariate polynomial, so we state the answer in the standard form, by listing the
terms in the descending order of their monomial degree:
ANSWER: 15x 4 − 12x 3 − 3x 2
279
5. PRODUCTS OF POLYNOMIALS CHAPTER 5. POLYNOMIAL EXPRESSIONS
EXAMPLE 5.2.2. Simplify the expression:
−10x y 2 (x 2 − 3x y + 2 y 2 )
SOLUTION: Notice that when we apply the distributive property, we have to compute the
correct sign for each term:
−10x y 2 (x 2 − 3x y + 2 y 2 ) = −(10x y 2 )(x 2 ) + (10x y 2 )(3x y) − (10x y 2 )(2 y 2 )
= −10 · x 1 x 2 · y 2 + 30 · x 1 x 1 · y 2 y 1 − 20 · x · y 2 y 2
= −10x 1+2 y 2 + 30x 1+1 y 2+1 − 20x y 2+2
= −10x 3 y 2 + 30x 2 y 3 − 20x y 4
This polynomial uses more than one variable, so there is no standard form, and the terms
of the answer can be listed in any order:
ANSWER: −10x 3 y 2 + 30x 2 y 3 − 20x y 4
5.3. Polynomial Times Polynomial.
THEOREM 5.3.1. Multiplying two polynomials amounts to taking a sum of all monomials
obtained by multiplying each term of the first polynomial by each term of the second poly-
nomial. Since the product of any two monomials is a monomial, and a sum of monomials
is a polynomial, it follows that the product of two polynomials is a polynomial.
EXAMPLE 5.3.1. Simplify the expression:
(3x − 2)(x + 5)
SOLUTION: We start by distributing the multiplication over the first sum:
(3x − 2)(x + 5) = 3x(x + 5) − 2(x + 5)
Then we distribute again as we multiply binomials by monomials:
3x(x + 5) − 2(x + 5) = (3x)(x) + (3x)(5) − (2)(x) − (2)(5)
= 3x 2 + 15x − 2x − 10 simplified monomials
= 3x + (15 − 2)x − 10
2
combined like terms
= 3x + 13x − 10
2
ANSWER: 3x 2 + 13x − 10
280
CHAPTER 5. POLYNOMIAL EXPRESSIONS 5. PRODUCTS OF POLYNOMIALS
EXAMPLE 5.3.2. Simplify the expression:
(x + y − 1)(2x + 3 y + 5)
SOLUTION: This is a product of two trinomials, and we need to multiply each term of the
first one by each term of the second one. This results in a sum of 3 · 3 = 9 monomials,
which we simplify by combining the like terms.
(x + y − 1)(2x + 3 y + 5) = x(2x + 3 y + 5) + y(2x + 3 y + 5) − 1(2x + 3 y + 5)
= (2x 2 + 3x y + 5x) + (2x y + 3 y 2 + 5 y) + (−2x − 3 y − 5)
= 2x 2 + 3 y 2 + (3x y + 2x y) + (5x − 2x) + (5 y − 3 y) − 5
= 2x 2 + 3 y 2 + (3 + 2)x y + (5 − 2)x + (5 − 3) y − 5
= 2x 2 + 3 y 2 + 5x y + 3x + 2 y − 5
This is a polynomial in 2 variables, so there is no standard form, and we can list the terms
of the answer in any order.
ANSWER: 2x 2 + 3 y 2 + 5x y + 3x + 2 y − 5
There are many alternative ways to annotate the process of multiplying polynomials, like the
table method, but of course they all produce the same result.
EXAMPLE 5.3.3. Simplify the expression by using a table:
(x + y − 1)(2x + 3 y + 5)
SOLUTION: We will construct a table where the top row will list the terms of the first
polynomial, and the left column will list the terms of the second polynomial. We will then
fill out the middle of the table by multiplying the monomials from the corresponding row
and column. We will complete the task by adding all of the monomials from the middle
of the table and combining like terms.
+x +y −1
+2x +2x 2
+2x y −2x
+3 y +3x y +3 y 2 −3 y
+5 +5x +5 y −5
2x 2 + 2x y − 2x + 3x y + 3 y 2 − 3 y + 5x + 5 y − 5 = 2x 2 + 3 y 2 + 5x y + 3x + 2 y − 5
ANSWER: 2x 2 + 3 y 2 + 5x y + 3x + 2 y − 5
281
5. PRODUCTS OF POLYNOMIALS CHAPTER 5. POLYNOMIAL EXPRESSIONS
Homework 5.5.
Simplify the given expression and state the 17. (x + 3)(2x + 3)
answer as a polynomial expression. If a
polynomial is in one variable, list the terms 18. (3x + 1)(x + 5)
of the answer in the decreasing order of de-
gree. 19. (x − 0.6)(x − 0.7)
1. (4x 4 )(8) 20. (1.1x + 10)(x − 1.2)
2. (−7)(6 y 7 ) 21. (5x + 6 y 3 )(x + y)
3. (−x 5 )(x 8 ) 22. (x 5 − 2)( y 2 − x y)
1 1
4. (− y 3 )(−2 y 5 ) 23. −x x −4
2 3
1 2 2 3
5. t − t 2 3
5 3 24. 2x − x +6
3 2
3 9 3
6. − x − x 25. (x + 1)(x 2 − x + 1)
7 2
7. (−2x)(3x 2 y) 26. (x − 2)(x 2 + 2x + 4)
8. (−1.5a2 b2 )(−2b) 27. (x 2 − x + 3)(x + 1)
9. (3x)(5x)(−2x 7 ) 28. (a2 + a − 7)(a + 2)
2 2 29. (5x 2 − 3x − 1)(x 2 + 4)
10. x (−6x)(−5x 4 )
3
30. (1 + 2a2 )(4a2 − 5a − 3)
11. 3x(x − y)
31. (x + 1)(x + 2)(x + 3)
12. −6 y(6 − 2x y)
32. 5x 2 (x + 4)(x − 4)
13. 3m(−m2 + 2m − 40)
33. (a + b)3
14. 12x(2x 3 − x + 3)
34. (x − 2 y)3
15. 0.5b (0.2b − b + 4)
3 2
35. (c 2 − 2c + 3)(c 2 + 4c + 1)
16. 0.1c 4 (−4c 2 + 0.7c − 10)
36. (−2x 2 − x − 1)(x 2 − 5x + 1)
282
CHAPTER 5. POLYNOMIAL EXPRESSIONS 5. PRODUCTS OF POLYNOMIALS
Homework 5.5 Answers.
1. 32x 4 19. x 2 − 1.3x + 0.42
3. −x 13 21. 5x 2 + 5x y + 6x y 3 + 6 y 4
2 5 1 25
5. − t 23. − x 2 + x −2
15 3 6
7. −6x 3 y 25. x 3 + 1
9. −30x 9 27. x 3 + 2x + 3
11. 3x 2 − 3x y 29. 5x 4 − 3x 3 + 19x 2 − 12x − 4
13. −3m3 + 6m2 − 120m 31. x 3 + 6x 2 + 11x + 6
15. 0.1b5 − 0.5b4 + 2b3 33. a3 + 3a2 b + 3ab2 + b3
17. 2x 2 + 9x + 9 35. c 4 + 2c 3 − 4c 2 + 10c + 3
283
6. SPECIAL PRODUCTS CHAPTER 5. POLYNOMIAL EXPRESSIONS
6. Special Products
6.1. Difference of Squares.
THEOREM 6.1.1 (Difference of Squares Formula). For all real numbers A and B,
(A + B)(A − B) = A2 − B 2
PROOF.
(A + B)(A − B) = A(A − B) + B(A − B)
= A2 − AB + BA − B 2 −AB and BA are like terms
= A2 − B 2
This is a nice computational shortcut which has many algebraic applications. In this text we
will use it for factoring polynomial expressions and for simplifying radical expressions.
EXAMPLE 6.1.1. Simplify and state the result as a polynomial expression:
(2x + 3)(2x − 3)
SOLUTION: We will use the difference of squares formula (A + B)(A − B) = A2 − B 2 with
A = 2x and B = 3.
(2x + 3)(2x − 3) = (2x)2 − (3)2
= 22 x 2 − 9
= 4x 2 − 9
ANSWER: 4x 2 − 9
EXAMPLE 6.1.2. Simplify and state the result as a polynomial expression:
(x 3 − 5 y 6 )(x 3 + 5 y 6 )
SOLUTION: We will use the difference of squares formula (A + B)(A − B) = A2 − B 2 with
A = x 3 and B = 5 y 6 .
284
CHAPTER 5. POLYNOMIAL EXPRESSIONS 6. SPECIAL PRODUCTS
(x 3 − 5 y 6 )(x 3 + 5 y 6 ) = (x 3 )2 − (5 y 6 )2
= x 3·2 − 52 y 6·2
= x 6 − 25 y 12
ANSWER: x 6 − 25 y 12
EXAMPLE 6.1.3. Rewrite the difference of squares as a product of two binomials:
36 − x 2
SOLUTION: 36 = (6)2 and x 2 = (x)2 , so we can use the difference of squares formula
(A + B)(A − B) = A2 − B 2 with A = 6 and B = x.
36 − x 2 = (6 + x)(6 − x)
ANSWER: (6 + x)(6 − x)
EXAMPLE 6.1.4. Rewrite the difference of squares as a product of two binomials:
4a2 − 100b6
SOLUTION: 4a2 = (2a)2 and 100b6 = (10b3 )2 , so we can use the difference of squares
formula (A + B)(A − B) = A2 − B 2 with A = 2a and B = 10b3 .
4a2 − 100b6 = (2a)2 − (10b3 )2
= (2a + 10b3 )(2a − 10b3 )
ANSWER: (2a + 10b3 )(2a − 10b3 )
285
6. SPECIAL PRODUCTS CHAPTER 5. POLYNOMIAL EXPRESSIONS
6.2. Square of a Binomial.
THEOREM 6.2.1 (Square of a Binomial). For all real numbers A and B, the square of a
binomial sum can be computed using the following formula:
(A + B)2 = (A + B)(A + B)
= A2 + 2AB + B 2
The square of a binomial difference can be found using a similar formula:
(A − B)2 = (A − B)(A − B)
= A2 − 2AB + B 2
PROOF.
(A + B)(A + B) = A(A + B) + B(A + B)
= A2 + AB + BA + B 2 AB and BA are like terms
= A + 2AB + B
2 2
(A − B)(A − B) = A(A − B) − B(A − B)
= A2 − AB − BA + B 2 −AB and −BA are like terms
= A2 − 2AB + B 2
EXAMPLE 6.2.1. Simplify and state the result as a polynomial expression:
(2a + 4)2
SOLUTION: We will use the square of a binomial formula
(A + B)2 = A2 + 2AB + B 2
with A = 2a and B = 4:
(2a + 4)2 = (2a)2 + 2(2a)(4) + (4)2
= 22 a2 + 2 · 2 · 4 · a + 16
= 4a2 + 16a + 16
This is a polynomial in one variable, so we leave the answer in the standard form, listing
the terms in the descending order of their monomial degrees.
ANSWER: 4a2 + 16a + 16
286
CHAPTER 5. POLYNOMIAL EXPRESSIONS 6. SPECIAL PRODUCTS
EXAMPLE 6.2.2. Simplify and state the result as a polynomial expression:
(7x − 6 y 5 )2
SOLUTION: We will use the square of a binomial formula
(A − B)2 = A2 − 2AB + B 2
with A = 7x and B = 6 y 5 :
(7x − 6 y 5 )2 = (7x)2 − 2(7x)(6 y 5 ) + (6 y 5 )2
= 72 x 2 − 2 · 7 · 6 · x y 5 + 62 y 5·2
= 49x 2 − 84x y 5 + 36 y 10
ANSWER: 49x 2 − 84x y 5 + 36 y 10
There are other two sign patters for a binomial square, (−A + B)2 and (−A − B)2 , and they can
be reduced to the ones we already discussed.
EXAMPLE 6.2.3. Simplify and state the result as a polynomial expression: (−x + 10)2
SOLUTION:
(−x + 10)2 = (10 − x)2 same as (A − B)2
= (10)2 − 2(10)(x) + (x)2
= 100 − 20x + x 2
ANSWER: 100 − 20x + x 2
EXAMPLE 6.2.4. Simplify and state the result as a polynomial expression: (−5 − x y)2
SOLUTION:
(−5 − x y)2 = ((−5) + (−x y))2 same as (A + B)2
= (−5)2 + 2(−5)(−x y) + (−x y)2 all minuses cancel
= 25 + 10x y + x 2 y 2
ANSWER: 25 + 10x y + x 2 y 2
287
6. SPECIAL PRODUCTS CHAPTER 5. POLYNOMIAL EXPRESSIONS
Homework 5.6.
2
Use appropriate special product formulas 1 1
to simplify the given expression and state 16. + x
3 5
the answer as a polynomial expression. If a
2
polynomial is in one variable, list the terms 1 4 6
of the answer in the decreasing order of de- 17. x − x
4 5
gree.
2
2 4
1. (5a + 3)2 18. y − y3
3 3
2. (1 + 4x)2 19. (−7 − 20x)2
3. (x + 10)(x − 10) 20. (−2x − 0.4 y)2
4. (9 − a)(9 + a)
12 15 2
21. − + c
5 2
5. (4x − 2 y)2
2
5 6
6. (a2 − 3)2 22. − d + 6d 3
3
7. (2b + 5c)(2b − 5c)
23. ((2x + 1)(2x − 1))2
8. (p3 − 7)(p3 + 7)
24. ((0.5 y − 4x)(0.5 y + 4x))2
9. (2nb − 5b)2
10. (5 − 6x 2 y)2 Rewrite the difference of squares as a prod-
uct of two binomials.
11. (2x + x y 3 )2
25. 36x 2 − y 2
12. ( y 4 + x 2 )2
26. 4a2 − 9b2
13. (0.5x − 0.4)(0.5x + 0.4)
27. c 4 − d 10
14. (2.5x + 3.1 y)(2.5x − 3.1 y)
28. m6 − 16n8
1 7 2
15. 3 + x
2 29. 25k2 s2 − 0.01k4 s2
1 4 4
30. 49x 8 y 2 − x y
4
288
CHAPTER 5. POLYNOMIAL EXPRESSIONS 6. SPECIAL PRODUCTS
Homework 5.6 Answers.
1. 25a2 + 30a + 9 1 8 3 5 36 2
17. x − x + x
16 5 25
3. x 2 − 100
19. 49 + 280x + 400x 2
5. 16x − 16x y + 4 y
2 2
225 2 144
2 2 21. c − 36c +
7. 4b − 25c 4 25
9. 4n2 b2 − 20nb2 + 25b2 23. 16x 4 − 8x 2 + 1
11. 4x 2 + 4x 2 y 3 + x 2 y 6 25. (6x + y)(6x − y)
13. 0.25x 2 − 0.16
27. (c 2 + d 5 )(c 2 − d 5 )
1 14
15. 9 + 3x 7 + x 29. (5ks + 0.1k2 s)(5ks − 0.1k2 s)
4
289
7. QUOTIENTS OF POLYNOMIALS CHAPTER 5. POLYNOMIAL EXPRESSIONS
7. Quotients of Polynomials
7.1. Monomial Divisor.
THEOREM 7.1.1. A division of a polynomial by a monomial can be rewritten as a sum
of monomial quotients with a common denominator. For all monomials A and B and all
non-zero monomials C,
A+ B A B
= +
C C C
This result naturally generalizes to sums with more than two terms.
PROOF.
A+ B 1
= (A + B) ·
C C
1 1
= A· +B·
C C
A B
= +
C C
EXAMPLE 7.1.1. Simplify and state the result as a polynomial expression:
x 4 − 2x 3 + 6x 2
2x
SOLUTION: We rewrite the division as a sum of monomial quotients and then simplify the
resulting fractions by canceling common factors:
x 4 − 2x 3 + 6x 2 x 4 2x 3 6x 2
= − +
2x 2x 2x 2x
1 x4 x3 x2
= · − +3· lowest terms
2 x x x
1 3
= x − x 2 + 3x
2
This is a polynomial in one variable, so we leave the answer in the standard form, listing
the terms in the descending order of their monomial degrees.
1 3
ANSWER: x − x 2 + 3x
2
290
CHAPTER 5. POLYNOMIAL EXPRESSIONS 7. QUOTIENTS OF POLYNOMIALS
EXAMPLE 7.1.2. Simplify and state the result as a polynomial expression:
−10a5 b6 + 6a3 b4 + 5ab2
−5ab2
SOLUTION: We rewrite the division as a sum of monomial quotients and then simplify
the resulting fractions by canceling common factors. Notice that we compute the correct
signs for each term just as we write the sum of quotients:
−10a5 b6 + 6a3 b4 + 5a b2 10a5 b6 6a3 b4 5ab2
= − − cancel common factors
−5a b2 5ab2 5ab2 5ab2
6
= 2a4 b4 − a2 b2 − 1 in each fraction
5
6
ANSWER: 2a4 b4 − a2 b2 − 1
5
Unlike monomial products, not all monomial quotients are monomial. In general, a so-called
rational expression results, and we dedicate a whole chapter to them in this text. For now we
take a sneak peek at how this happens.
EXAMPLE 7.1.3. Simplify and state the result as a sum of simplified monomial quotients:
−p5 s − 7p2 s3 + p
−p2 s2
SOLUTION: We rewrite the division as a sum of monomial quotients and then cancel
common factors in each term.
−p5 s − 7p2 s3 + p p5 s 7p2 s3 p
= + −
−p2 s2 p2 s2 p2 s2 p2 s2
p3 1
= + 7s − 2
s ps
The terms p3 /s and −1/(ps2 ) contain variable reciprocals, so they are not monomial, but
proper rational expressions in simplified form.
p3 1
ANSWER: + 7s − 2
s ps
291
7. QUOTIENTS OF POLYNOMIALS CHAPTER 5. POLYNOMIAL EXPRESSIONS
7.2. Polynomial Divisor. There exists a division algorithm for polynomials, which is useful
for rewriting them as products of polynomial factors.
THEOREM 7.2.1. If A is a polynomial and B is a non-zero polynomial in one variable, and
the degree of A is at least as big as the degree of B, then
A R
=Q+
B B
where Q and R are polynomials and the degree of R is less than the degree of B. If the
polynomial remainder R happens to be zero, then A/B = Q is a polynomial quotient of
two polynomials.
The polynomial long division algorithm will be explained in examples. In each case, we will
state the result in the form Q + R/B, just as in the theorem (7.2.1).
EXAMPLE 7.2.1. Simplify using polynomial long division:
x 2 + 5x + 6
x +3
SOLUTION: We begin by setting up the long division. It resembles the integer long division,
but instead of columns with digits it has columns of monomial terms. Starting from the
right and going to the left, the powers of x are 0, 1, 2, and so on, as many as the dividend
monomial has.
x +3 x 2 + 5x + 6
Divide the leftmost monomial of the dividend by the leftmost monomial of the divisor:
x 2 /x = x. Write the result in the appropriate column of the answer space.
x
x +3 x 2 + 5x + 6
Take the opposite of the monomial you just found and multiply it by the divisor x + 3:
(−x)(x + 3) = −x 2 − 3x
Write the result below the dividend in appropriate columns.
x
x +3 x 2 + 5x + 6
− x 2 − 3x
292
CHAPTER 5. POLYNOMIAL EXPRESSIONS 7. QUOTIENTS OF POLYNOMIALS
Add the monomials above the line. No need to write zero for x 2 + (−x 2 ), since that place
always cancels. Carry down the next monomial term from the dividend.
x
x +3 x 2 + 5x + 6
− x 2 − 3x
2x + 6
Now we repeat the steps with 2x + 6 as the new dividend. Divide the leftmost monomial
of the dividend by the leftmost monomial of the divisor: 2x/x = 2, and write the result
in the answer space.
x +2
x +3 x + 5x + 6
2
− x 2 − 3x
2x + 6
Take the opposite of the monomial you just found and multiply it by the divisor x + 3:
(−2)(x + 3) = −2x − 6
Write the result below the dividend in appropriate columns.
x +2
x +3 x 2 + 5x + 6
− x 2 − 3x
2x + 6
− 2x − 6
Add the monomials above the line.
x +2
x +3 x + 5x + 6
2
− x 2 − 3x
2x + 6
− 2x − 6
0
There’s nothing more to carry down, so we have reached the answer. The quotient x + 2 is
written in the answer space. The remainder 0 is at the very bottom. We state the answer
as a quotient without a remainder.
ANSWER: x + 2
293
7. QUOTIENTS OF POLYNOMIALS CHAPTER 5. POLYNOMIAL EXPRESSIONS
EXAMPLE 7.2.2. Simplify using polynomial long division:
x 3 − 2x 2 − 4
x −3
SOLUTION: We begin by setting up the long division. Notice the empty space between the
terms −2x 2 and −4 of the dividend. This column is reserved for monomials of degree 1.
Some people prefer to fill the zero entries below the dividend with visible zeroes, but we
are just leaving them blank.
x −3 x 3 − 2x 2 −4
Divide the leftmost monomial of the dividend by the leftmost monomial of the divisor:
x 3 /x = x 2 . Write the result in the appropriate column of the answer space.
x2
x −3 x 3 − 2x 2 −4
Take the opposite of the monomial you just found and multiply it by the divisor x − 3:
(−x 2 )(x − 3) = −x 3 + 3x 2
Write the result below the dividend in appropriate columns.
x2
x −3 x 3 − 2x 2 −4
− x 3 + 3x 2
Add the monomials above the line. Carry down the next monomial term from the dividend.
The next term to be carried down is 0x, so we just carry down the empty space.
x2
x −3 x 3 − 2x 2 −4
− x 3 + 3x 2
x2
Repeat the steps with 2x + 6 as the new dividend. Divide the leftmost monomial of the
dividend by the leftmost monomial of the divisor: x 2 /x = x, and write the result in the
appropriate column of the answer space.
294
CHAPTER 5. POLYNOMIAL EXPRESSIONS 7. QUOTIENTS OF POLYNOMIALS
x2 + x
x −3 x 3 − 2x 2 −4
− x 3 + 3x 2
x2
Take the opposite of the monomial you just found and multiply it by the divisor x − 3:
(−x)(x − 3) = −x 2 + 3x
Write the result below the dividend in appropriate columns.
x2 + x
x −3 x 3 − 2x 2 −4
− x 3 + 3x 2
x2
− x 2 + 3x
Add the monomials above the line. Carry down the next monomial term from the dividend.
x2 + x
x −3 x 3 − 2x 2 −4
− x 3 + 3x 2
x2
− x 2 + 3x
3x − 4
Repeat the steps with 3x − 4 as the new dividend. Divide the leftmost monomial of the
dividend by the leftmost monomial of the divisor: 3x/x = 3, and write the result in the
answer space.
x2 + x + 3
x −3 x 3 − 2x 2 −4
− x 3 + 3x 2
x2
− x 2 + 3x
3x − 4
Take the opposite of the monomial you just found and multiply it by the divisor x − 3:
(−3)(x − 3) = −3x + 9
Write the result below the dividend in appropriate columns.
295
7. QUOTIENTS OF POLYNOMIALS CHAPTER 5. POLYNOMIAL EXPRESSIONS
x2 + x + 3
x −3 x 3 − 2x 2 −4
− x 3 + 3x 2
x2
− x 2 + 3x
3x − 4
− 3x + 9
Add the monomials above the line.
x2 + x + 3
x −3 x 3 − 2x 2 −4
− x 3 + 3x 2
x2
− x 2 + 3x
3x − 4
− 3x + 9
5
There’s nothing more to carry down, so we have reached the answer. The quotient x 2 +x+3
is written in the answer space. The remainder 5 is at the very bottom. We state the answer
in the form Q + R/B, where Q is the polynomial quotient, R is the polynomial remainder,
and R is the original polynomial divisor.
5
ANSWER: x 2 + x + 3 +
x −3
Checking these answers is possible, but not easy. One way to do that is to multiply the answer
by the divisor, simplify, and make sure we get the original dividend back:
5 5(x − 3)
x2 + x + 3 + (x − 3) = x 2 (x − 3) + x(x − 3) + 3(x − 3) +
x −3 x −3
= x 3 − 3x 2 + x 2 − 3x + 3x − 9 + 5
= x 3 + (−3 + 1)x 2 + (−3 + 3)x − 4
= x 3 − 2x 2 − 4
We will present a few more examples of polynomial long division without explaining the steps.
The procedure is exactly the same in each case. Readers can do these on their own, and then
compare the solutions in order to check the answer and to detect where the first mistake was
made.
296
CHAPTER 5. POLYNOMIAL EXPRESSIONS 7. QUOTIENTS OF POLYNOMIALS
6x 2 − 5x − 27
EXAMPLE 7.2.3. Simplify using polynomial long division:
2x − 5
SOLUTION: 3x + 5
2
2x − 5 6x − 5x − 27
− 6x 2 + 15x
10x − 27
− 10x + 25
−2
−2
ANSWER: 3x + 5 +
2x − 5
Checking the answer:
−2 −2(2x − 5)
3x + 5 + (2x − 5) = 3x(2x − 5) + 5(2x − 5) +
2x − 5 (2x − 5)
= 6x 2 − 15x + 10x − 25 + (−2)
= 6x 2 − 5x − 27
x 4 − 2x 3 + 3x − 4
EXAMPLE 7.2.4. Simplify using polynomial long division:
x2 − 5
SOLUTION: x 2 − 2x + 8
x2 − 5 x 4 − 2x 3 + 3x 2 −4
− x4 + 5x 2
− 2x 3 + 8x 2
2x 3 − 10x
8x 2 − 10x − 4
− 8x 2 + 40
− 10x + 36
−10x + 36
ANSWER: x 2 − 2x + 8 +
x2 − 5
297
7. QUOTIENTS OF POLYNOMIALS CHAPTER 5. POLYNOMIAL EXPRESSIONS
Homework 5.7.
Simplify the given expression and state the Simplify the expression using polynomial
result as a polynomial expression. If a poly- long division. State the answer as a poly-
nomial is in one variable, list the terms of nomial expression when the remainder is
the answer in the decreasing order of de- zero; otherwise use the form Q + R/B,
gree. where Q is the polynomial quotient, B is
the polynomial divisor, and the degree of
42x 6 − 24 the polynomial R is lower than the degree
1.
4 of B.
8a2 + 10a x 2 + 2x − 15
2. 11.
16 x +5
x − 2x 2 + x 7 x 2 − 8x + 12
3. 12.
x x −6
4 y 10 − 2 y 3 + y 2 4x 2 − 6x + 7
4. 13.
y2 −x − 3
20t 3 − 25t 2 + 10t 3 y2 + 5 y + 4
5. 14.
−5t −y + 5
14 y 5 + 28 y 4 − 70 y 3 x 2 − 10x − 20
6. 15.
−14 y x −5
−12x 2 y 2 + 4x 4 y 2 + 16x y 3 t 2 + 8t − 15
7. 16.
4x y t +6
9m2 n2 − 3m2 n − 6mn2 6 y 2 + 17 y + 8
8. 17.
−3mn 2y + 5
20a b2 c 3 − 14a2 b2 c 4 10a2 + 19a + 10
9. 18.
10ab2 c 2 2a + 3
17x 3 yz 5 + 16x 2 yz 6 t 3 + 64
10. 19.
2x 2 yz t +4
27 − x 3
20.
3− x
a3 − a2 + a − 1
21.
a2 + 1
8t 3 − 22t 2 − 5t + 12
22.
2t 2 − 7t + 4
298
CHAPTER 5. POLYNOMIAL EXPRESSIONS 7. QUOTIENTS OF POLYNOMIALS
Homework 5.7 Answers.
1. 10.5x 6 − 6 61
13. −4x + 18 +
−x − 3
3. x 6 − 2x + 1
−45
15. x − 5 +
x −5
5. −4t 2 + 5t − 2
3
7. −3x y + x y + 4 y
3 2 17. 3 y + 1 +
2y + 5
9. 2c − 1.4ac 2 19. t 2 − 4t + 16
11. x − 3 21. a − 1
299
8. NEGATIVE EXPONENT CHAPTER 5. POLYNOMIAL EXPRESSIONS
8. Negative Exponent
8.1. Negative Integer Exponent. Recall that by definition of negative integer exponent,
if b is any non-zero real number and n is a positive integer, then
n
1 1
b =
−n
= n
b b
Recall also that negative integer exponent obeys all of the exponential rules we’ve seen so far:
(1) The value of exponent with base 1 is 1
(2) The product rule and the quotient rule
(3) The power rule
(4) Distributivity over multiplication and division
In this section we will practice manipulating products and quotients involving negative ex-
ponent by rewriting them in the form where each variable base appears only once, and all
exponents are positive.
BASIC EXAMPLE 8.1.1. We start with some basic examples. By definition, raising a base
to a negative power amounts to raising the reciprocal of the base to the corresponding
positive power:
2
1 1
4 −2
= =
4 16
Next, observe what happens to the sign of the base. The reciprocal of the negative number
is negative, so the sign of the exponent has no influence on the sign of the base. What
does make a difference is the oddness of the exponent. An odd exponent of any sign will
preserve the sign of the base, and an even exponent will always produce a non-negative
result:
3
1 1 1 1 1
(−5) = −
−3
= − − − =−
5 5 5 5 125
4
1 1 1 1 1 1
(−3) = −
−4
= − − − − =
3 3 3 3 3 81
Rewriting variable exponents is even easier, since we do not have to compute their values:
2
1 1 a3
a b 3 −2
=a ·
3
=a ·
3
=
b b2 b2
300
CHAPTER 5. POLYNOMIAL EXPRESSIONS 8. NEGATIVE EXPONENT
In general, getting rid of the negative exponent within products and fractions is quite straight-
forward, thanks to the following fact.
THEOREM 8.1.1. Within a product or a fraction, a factor in the numerator with negative
exponent can be replaced by the same factor in the denominator with positive exponent.
For all real numbers a, all non-zero real numbers b and X , and all integers n,
aX −n a
=
b bX n
Conversely, a factor in the denominator with negative exponent can be replaced by the
same factor in the numerator with positive exponent.
a aX n
=
bX −n b
PROOF.
First part:
aX −n a
= · (X −n )
b b
n
a 1
= ·
b X
1
a
= ·
b Xn
a
=
bX n
Second part:
1
a a
= ·
bX −n b X −n
−n
a 1
= ·
b X
a
= · (X n )
b
aX n
=
b
301
8. NEGATIVE EXPONENT CHAPTER 5. POLYNOMIAL EXPRESSIONS
EXAMPLE 8.1.1. Simplify and state the answer as an expression with positive exponents:
x −17 y 5
x −13 y −6
SOLUTION: One way to deal with negative exponent factors is by replacing them with
corresponding positive exponent factors, using the theorem 8.1.1:
x −17 y 5 x 13 y 5 y 6
=
x −13 y −6 x 17
y 5+6
= common factor x 13 canceled
x4
y 11
=
x4
Another way to deal with this is by applying the product and the quotient rules directly to
the negative exponent. Here we are diving bases, so we will subtract the corresponding
exponents:
x −17 y 5
= x −17−(−13) y 5−(−6)
x −13 y −6
= x −4 y 11
y 11
= got rid of the negative exponent for the answer
x4
Both ways are correct and about as efficient, and choosing one over the other, or a mix of
the two, is more of a personal preference.
y 11
ANSWER:
x4
EXAMPLE 8.1.2. Simplify and state the answer as an expression with positive exponents:
(5x 7 x −4 )−2
SOLUTION: We will simplify inside the parentheses first since we can. Alternatively, we
could distribute the exponent over the product first.
(5x 7 x −4 )−2 = (5x 7+(−4) )−2
= (5x 3 )−2
302
CHAPTER 5. POLYNOMIAL EXPRESSIONS 8. NEGATIVE EXPONENT
Now we distribute exponent over the simplified product:
= 5−2 · (x 3 )−2
1
= · x 3·(−2) product rule
52
1
= · x −6
25
1
=
25x 6
1
ANSWER:
25x 6
EXAMPLE 8.1.3. Simplify and state the answer as an expression with positive exponents:
−10x 7 y 20
(5x 3 y 10 )2
SOLUTION: We have to distribute the exponent over the product in the denominator
before we can cancel anything:
−10x 7 y 20 −10x 7 y 20
=
(5x 3 y 10 )2 52 (x 3 )2 ( y 10 )2
−10x 7 y 20
= product rule
52 x 3·2 y 10·2
−10x 7 y 20
=
25x 6 y 20
2
= − x 7−6 y 20−20 quotient rule
5
2
= − x1 y0
5
2
= − x
5
2
ANSWER: − x
5
303
8. NEGATIVE EXPONENT CHAPTER 5. POLYNOMIAL EXPRESSIONS
EXAMPLE 8.1.4. Simplify and state the answer as an expression with positive exponents:
5 −3
x
2 y2
SOLUTION: Get rid of the negative exponent −3 by taking the reciprocal of its base:
5 −3 2 3
x 2y
=
2 y2 x5
And now distribute exponent over the fraction:
2 3
2y 23 ( y 2 )3
=
x5 (x 5 )3
8 y 2·3
=
x 5·3
8 y6
=
x 15
8 y6
ANSWER:
x 15
8.2. Scientific Notation. Scientific notation is a way of writing down extremely large and
extremely small numbers, and is useful in areas such as physics, engineering, and computer
science, to name a few. In fact, many calculators support some sort of scientific notation in or-
der to display very large and very small numbers on a tiny display only a dozen or so characters
wide.
A lot of quantities found in nature look bizarre, to say the least, when we express them in
familiar units suitable for things we do every day as humans. For example, the mass of our
Sun is approximately
1988550000000000000000000000000 kg
while the mass of an electron, which is one of the basic building blocks of the tangible world
around us, is approximately
0.000000000000000000000000000000910938356 kg
The difference between these two quantities is about 60 decimal places, and these are not even
the most extreme examples of ridiculously large/small numbers commonly occurring in nature.
Scientific notation allows writing these down in a concise form, making it easier to understand
the true magnitude of a number at a glance.
304
CHAPTER 5. POLYNOMIAL EXPRESSIONS 8. NEGATIVE EXPONENT
DEFINITION 8.2.1. A number is said to be in scientific notation if it is written in the form
D.ddd . . . × 10n
where D is a non-zero digit, ddd . . . are the rest of digits, × is a multiplication sign, and
n is an integer exponent. The factor 100 , corresponding to n = 0, does not have to be
shown.
BASIC EXAMPLE 8.2.1. Let’s write some numbers in scientific notation. Here’s the mass of
the Sun in kilograms:
1.98855 × 1030
The mass of an electron in kilograms:
9.10938356 × 10−31
Recall that multiplying a decimal representation by 10n amounts to shifting the decimal
point by n digits. If n is positive, the decimal point shifts to the right, and if n is negative,
the decimal point shifts to the left. Here are some other numbers in decimal notation,
which we rewrite in scientific notation:
31 = 3.1 × 101
0.6 = 6 × 10−1
402 = 4.02 × 102
0.073 = 7.3 × 10−2
0.0000101 = 1.01 × 10−5
12345678 = 1.2345678 × 107
1450.67 = 1.45067 × 103
EXAMPLE 8.2.1. Simplify the expression and state the answer in scientific notation:
16
10000
SOLUTION:
16
= 0.0016
10000
= 1.6 × 10−3
ANSWER: 1.6 × 10−3
305
8. NEGATIVE EXPONENT CHAPTER 5. POLYNOMIAL EXPRESSIONS
EXAMPLE 8.2.2. Simplify the expression and state the answer in scientific notation:
(3.4 × 105 )(5 × 107 )
SOLUTION: The order of multiplication in a product does not matter, so let’s multiply the
numbers and combine the exponential expressions first:
(3.4 × 105 )(5 × 107 ) = (3.4)(5)(105 )(107 )
= 17 · 105+7 product rule
= 17 · 1012
This is not yet scientific notation; we still need to express 17 as 1.7 times a power of 10:
17 · 1012 = (1.7 × 101 )(1012 ) put 17 in scientific notation
= 1.7 × 101+12 simplify powers of 10
= 1.7 × 1013
ANSWER: 1.7 × 1013
EXAMPLE 8.2.3. Simplify the expression and state the answer in scientific notation:
7.08 × 10−18
9.6 × 10−12
SOLUTION:
7.08 × 10−18 7.08 10−18
= ·
9.6 × 10−12 9.6 10−12
= 0.7375 · 10−18−(−12) quotient rule
= 0.7375 · 10 −6
= 7.375 · 10−1 · 10−6 put 0.7375 in scientific notation
= 7.375 · 10−1+(−6) product rule
= 7.375 × 10−7
ANSWER: 7.375 × 10−7
306
CHAPTER 5. POLYNOMIAL EXPRESSIONS 8. NEGATIVE EXPONENT
Homework 5.8.
Simplify the given expression and state the 20. (b−5 )−7
answer as an expression with positive ex-
ponents. 21. (4a7 )−3
1. 3−3 22. (7x 4 )−2
2. 10−4 23. (−2x 2 y −1 )10
3. (−3)−6 24. (−3a−9 b4 )−3
4. (−2)−4 15x −7
25.
10x −10
5. 8x −3
−12 y −5
26.
6. a b−5 2 y −8
4 −2
x4
7.
x −2 27.
3
Q
8.
2
−3
X −9 28.
−2 y2
b
9. 0
4 29. x4 + x2 + 1
−3
−2 30. 17x 3 y −7
1
10.
y
−3a3 b−5
11. 3−5 · 38 31.
−6a7 b−8
12. 511 · 5−7 3x −2 y 4
32.
13. 2x −14 · x −20 −12x y −7
14. − y 7 · y −10 33. (x 2 y −3 )(x y −5 )−2
15. (5x −2 y −3 )(2x −4 y) 34. (a−5 b4 )(a2 b−1 )−3
−4
16. (3x −5
y )(2x y )
−7 −2 −2x 5 y −6
35.
x −2
17. (a−5 )3
−3
3a−2 b4
18. (b ) 10 −5 36.
2b−1
19. (2−3 )−2
307
8. NEGATIVE EXPONENT CHAPTER 5. POLYNOMIAL EXPRESSIONS
Simplify the given expression and state the 54. 0.000000065
answer in decimal notation.
55. 50000200
3
37. 4.93 × 10
56. 4000000000
38. 8.14 × 104
39. 8.93 × 10−3 Simplify the given expression and state the
answer in scientific notation.
40. 7.27 × 10−4
57. (3 × 105 )(2 × 108 )
41. 3.007 × 10−6
58. (3.2 × 107 )(2.2 × 10−4 )
42. 1.0204 × 10−5
59. (3.8 × 109 )(6.5 × 10−2 )
43. 5.033 × 105
60. (7.3 × 107 )(8.5 × 10−5 )
44. 9.99887 × 106
8.5 × 108
61.
3.4 × 10−5
Write the given number in scientific nota- 5.7 × 10−2
tion. 62.
2.4 × 105
45. 15 7.0 × 103
63.
46. 123 5.0 × 108
7.5 × 10−3
47. 0.7 64.
8.0 × 10−6
48. 0.04 12
65.
4000
49. 37000000
13
50. 402000 66.
5000
51. 0.0584 658.7
67.
14
52. 0.000695
333
53. 0.0000007 68.
2500
308
CHAPTER 5. POLYNOMIAL EXPRESSIONS 8. NEGATIVE EXPONENT
Homework 5.8 Answers.
1 b3
1. 31.
27 2a4
1 33. y 7
3.
729
y 24
8 35.
5. 16x 28
x3
37. 4930
2
7. 4x
39. 0.00893
16
9. 2
b 41. 0.000003007
11. 27 43. 503300
2
13. 45. 1.5 × 10
x 34
10 47. 7 × 10−1
15. 6 2
x y
49. 3.7 × 107
1
17. 51. 5.84 × 10−2
a15
19. 64 53. 7.0 × 10−7
1 55. 5.00002 × 107
21.
64a21
57. 6 × 1013
1024x 20
23. 59. 2.47 × 108
y 10
3 3 61. 2.5 × 1013
25. x
2
63. 1.4 × 10−5
9
27. 65. 3.0 × 10−3
x8
29. 1 67. 4.705 × 101
309
8. PRACTICE TEST 5 CHAPTER 5. POLYNOMIAL EXPRESSIONS
Practice Test 5
1. Find an expression for the perimeter of 7. Multiply and simplify by combining the
the shown shape, and simplify it by com- like terms:
bining the like terms: (3x − 6)(4 − x + 6x 2 )
4 8. Simplify the expression by performing
x +1 the polynomial long division:
14x 3 + 7x 2 − 19x + 3
2x + 3
5x 2 − 3x 9. Simplify the expression by performing
x − 2x 2 the polynomial long division:
6x 3 − 2x − 5
x −2
6x
10. Simplify the expression and state the
answer with positive exponents only:
2. Simplify the expression:
(−5x −2 y)(−2x −3 y 2 )
(3k5 m0 )2 · (−k4 m3 )
11. Simplify the expression and state the
3. Simplify the expression: answer with positive exponents only:
−1 2 −2
a7 b2
3 x y
2
ab −5x −2 y 4
a6 b2
4. Find the degree of the polynomial: 12. Simplify the expression and state the
answer in decimal notation:
x 10 − 6x 9 y 2 + 7x 6 y 6 (1.5 × 10−4 )(1.7 × 107 )
5. Describe the polynomial degree and the 13. Simplify the expression and state the
number of terms in words: answer in scientific notation:
−4x 2 y + 7x y (4 × 109 )(6 × 104 )
6. Describe the polynomial degree and the 14. Simplify the expression and state the
number of terms in words: answer in scientific notation:
0.9
ax + at − i t 1800
310
CHAPTER 5. POLYNOMIAL EXPRESSIONS 8. PRACTICE TEST 5
Practice Test 5 Answers.
1. 3x 2 + 5x + 5 39
9. 6x 2 + 12x + 22 +
x −2
2. −9k14 m3
10 y 3
4 2
3. a b 10.
x5
4. 11 25 y 4
11.
5. cubic binomial x2
12. 2250
6. quadratic trinomial
7. 18x 3 − 39x 2 + 18x − 24 13. 2.4 × 1014
8. 7x 2 − 7x + 1 14. 5.0 × 10−4
311
CHAPTER 6
Factoring Polynomials
1. Greatest Common Factor
6 y
4
y = x 2 − 3x = x(x − 3)
2
x
−2 −1 1 2 3 4 5
−2
−4
1.1. Factoring Polynomials. Factoring a polynomial means rewriting it in a product form,
with as many polynomial factors as possible. This is useful for solving polynomial equations, as
well as for providing an expression which offers hints about the behavior of the corresponding
graph. Some polynomials can only be written as a product with a single factor, for example
(x + 1)
Other polynomials can be written with multiple factors:
(x 2 − 3x) = (x)(x − 3)
In this section we start exploring basic strategies for finding these factorizations.
DEFINITION 1.1.1. We call a polynomial irreducible over reals if it cannot be written as a
product of two or more non-constant polynomials with real coefficients.
We call a polynomial irreducible over integers or simply prime if it cannot be written as a
product of two or more non-constant polynomials with integer coefficients.
313
1. GREATEST COMMON FACTOR CHAPTER 6. FACTORING POLYNOMIALS
THEOREM 1.1.1 (Polynomial Factorization). Every non-constant polynomial expression
has a factorization: an equivalent form which is a product of polynomial expressions
which are irreducible over reals. This factorization is unique if we disregard the order of
multiplication and state polynomial factors in some kind of standard form.
Stated in a modern way, all polynomial structures are unique factorization domains.
THEOREM 1.1.2. All linear polynomials are irreducible over reals, and hence also prime.
BASIC EXAMPLE 1.1.1. A quadratic polynomial
a2 − b2
can be written as a product of two linear factors
(a + b)(a − b)
which are irreducible. The only other equivalent polynomial products with two factors are
(a − b)(a + b), (a − b)(b + a), and so on, differing only in the order of operations, which
is what we mean when we say that the factorization is essentially unique.
In the sections up ahead we will not worry too much about doing a perfect job and obtaining
the optimal factorization, but eventually we will list the most notorious types of irreducible
polynomials in order to put together a comprehensive factoring strategy.
1.2. Greatest Common Factor.
DEFINITION 1.2.1. The greatest common factor, or GCF for short, for non-zero integers a
and b is the largest positive integer c which is a factor of both a and b. In the absence
of common prime factors, the GCF is defined to be 1. The definition extends naturally to
larger collections of integers.
Numbers with GCF 1 are called relatively prime.
BASIC EXAMPLE 1.2.1. Some basic cases:
integers GCF
2, 3 1 no common prime factors
10, 30 10 the largest common integer factor is 10
4, 6 2 the only common factor is 2
6, 9, 15 3 the only common factor is 3
12, 18, 60 3 the largest common integer factor is 6
314
CHAPTER 6. FACTORING POLYNOMIALS 1. GREATEST COMMON FACTOR
EXAMPLE 1.2.1. Find the GCF for 16 and 20.
SOLUTION: For smaller numbers it’s easy enough to guess the GCF by simply staring at
them, and you may have figured out already that the GCF in this example is 4. Regardless,
here we will show how one can compute GCF even in difficult cases without guessing.
First, we represent the numbers as products of primes:
16 = 2·2·2·2
20 = 2·2·5
The GCF is a product of primes which is in common to all of these factorizations, in this
case 2 · 2 = 4.
ANSWER: 4
EXAMPLE 1.2.2. Find the GCF for 15 and 28.
SOLUTION: Represent the numbers as products of primes and detect all the ones they
have in common:
15 = 3·5
28 = 2·2·7
These factorizations have no primes in common, meaning 15 and 28 do not have a single
common prime factor, so the GCF is 1.
ANSWER: 1
EXAMPLE 1.2.3. Find the GCF for 96 and 66.
SOLUTION: Represent the numbers as products of primes and detect all the ones they
have in common:
96 = 2·2·2·2·2·3
66 = 2 · 3 · 11
The largest collection of prime factors in common is 2 · 3 = 6
ANSWER: 6
EXAMPLE 1.2.4. Find the GCF for 500 and 120.
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1. GREATEST COMMON FACTOR CHAPTER 6. FACTORING POLYNOMIALS
SOLUTION: Represent the numbers as products of primes and detect all the ones they
have in common:
500 = 2·2·5·5·5
120 = 2·2·2·3·5
The largest collection of prime factors in common is 2 · 2 · 5 = 20
ANSWER: 20
EXAMPLE 1.2.5. Find the GCF for 144, 60, and 90.
SOLUTION: Represent the numbers as products of primes and detect all the ones they
have in common:
144 = 2·2·2·2·3·3
60 = 2·2·3·5
90 = 2·3·3·5
The largest collection of prime factors in common is 2 · 3 = 6
ANSWER: 6
1.3. Monomial GCF.
DEFINITION 1.3.1. The greatest common factor, or GCF for short, for non-zero monomials
A and B is the monomial C such that its coefficient is the GCF of the coefficients of A and
B, and the exponent of each variable factor of C is no greater than the exponent of the
same variable in either A or B. The definition extends naturally to larger collections of
monomials.
Monomials with GCF 1 are called relatively prime.
EXAMPLE 1.3.1. Find the GCF for 4a2 b and 2a4 b.
SOLUTION: The GCF for coefficients is 2, the lowest exponent on a is 2, and the lowest
exponent on b is 1, so GCF is 2a2 b.
ANSWER: 2a2 b
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CHAPTER 6. FACTORING POLYNOMIALS 1. GREATEST COMMON FACTOR
EXAMPLE 1.3.2. Find the GCF for 10x 3 y 5 and 15x 4 y 2 z 7 .
SOLUTION: The GCF for coefficients is 5, the lowest exponent on x is 3, the lowest
exponent on y is 2, and the lowest exponent on z is taken to be 0, since the first monomial
does not have z at all. The overall GCF is 5x 3 y 2 .
ANSWER: 5x 3 y 2
EXAMPLE 1.3.3. Find the GCF for 14x 7 , 21x 5 , and 6x 4 .
SOLUTION: The GCF for coefficients is 1, and the lowest exponent on x is 4, so GCF is x 4 .
ANSWER: x 4
EXAMPLE 1.3.4. Rewrite the expression by factoring out the GCF:
12a2 − 6a
SOLUTION: We determine that the GCF is 6a, and use the distributive property:
12a2 − 6a = 6a(2a − 1)
We can check the answer by multiplying the polynomials:
6a(2a − 1) = 6a · 2a − 6a · 1 = 12a2 − 6a
ANSWER: 6a(2a − 1)
EXAMPLE 1.3.5. Rewrite the expression by factoring out the GCF:
80x 2 y + 40x y − 90x y 3
SOLUTION: The GCF for coefficients is 10, the lowest exponent on x is 1, and the lowest
exponent on y is 1, so the GCF is 10x y.
80x 2 y + 40x y − 90x y 3 = 10x y(8x + 4 − 9 y 2 )
317
1. GREATEST COMMON FACTOR CHAPTER 6. FACTORING POLYNOMIALS
We can check the answer by multiplying the polynomials:
10x y(8x + 4 − 9 y 2 ) = 10x y · 8x + 10x y · 4 − 10x y · 9 y 2
= 80x 2 y + 40x y − 90x y 3
ANSWER: 10x y(8x + 4 − 9 y 2 )
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CHAPTER 6. FACTORING POLYNOMIALS 1. GREATEST COMMON FACTOR
Homework 6.1.
Find the GCF for given expressions. 13. 2a2 − 2a − 8
1. 18, 24 14. 12x 2 + 6x − 30
2. 20, 32 15. 5t 2 + t
3. 12, 30, 60 16. 3x 3 − x 2
4. 14, 42, 70 17. −10 y 2 − 5 y
5. 14x, 7x y 2 18. −4a2 + 3a3
6. 25 y 2 , 16 y 2 z 19. 6a6 + 9a4 − 12a2
7. a2 b3 , 2ab4 20. −10x 2 − 15x 4 − 9x 3
8. 6x 3 y 3 , 4x y 2 21. 5x y 2 − 10x 2 y
9. 10x, 2x 2 , 14x 3 22. 20a2 b2 + 15ab
10. 5 y 2 , 25 y 3 , y 5 23. 12x 2 y + 6x y − 2x y 2
24. −4a2 − 2ab + 8b2
Factor out the GCF.
25. 3x 6 y 2 + 3x 5 y 3
11. 6x + 24
26. a4 y 5 − 17a6 y
12. 10x + 40
319
1. GREATEST COMMON FACTOR CHAPTER 6. FACTORING POLYNOMIALS
Homework 6.1 Answers.
1. 6 15. t(5t + 1)
3. 6 17. 5 y(−2 y − 1)
5. 7x
19. 3a2 (2a4 + 3a2 − 4)
7. a b3
21. 5x y( y − 2x)
9. 2x
11. 6(x + 4) 23. 2x y(6x + 3 − y)
13. 2(a2 − a − 4) 25. 3x 5 y 2 (x + y)
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CHAPTER 6. FACTORING POLYNOMIALS 2. GROUPING
2. Grouping
2.1. Factoring by Grouping. An occasional polynomial with 4 terms can be factored using
a technique knows as grouping, which will be explained in examples.
EXAMPLE 2.1.1. Factor by grouping: 6a + 3b + 10a2 + 5ab
SOLUTION: Represent the given polynomial with 4 terms as a sum of two binomials:
6a + 3b + 10a2 + 5ab = (6a + 3b) + (10a2 + 5ab)
For each binomial, factor out the GCF:
(6a + 3b) + (10a2 + 5ab) = 3(2a + b) + 5a(2a + b)
Notice that now we have a sum of two terms, and there is a common factor (2a + b). The
distributive property allows us to factor it out:
3(2a + b) + 5a(2a + b) = (3 + 5a)(2a + b)
ANSWER: (3 + 5a)(2a + b)
Sometimes it may be hard to see how we got from
3(2a + b) + 5a(2a + b)
to
(3 + 5a)(2a + b)
with one application of the distributive property, but we can break this process down into
several steps. To see these expressions are equivalent, it may be helpful to create a temporary
variable T and set it equal to 2a + b. Then we can write
3(2a + b) + 5a(2a + b) = 3(T ) + 5a(T ) substitute T for 2a + b
= (3 + 5a)T distributivity
= (3 + 5a)(2a + b) substitute 2a + b for T
EXAMPLE 2.1.2. Factor by grouping: x 5 − x 4 + x 3 − x 2
SOLUTION: Represent the given polynomial with 4 terms as a sum of two binomials:
x 5 − x 4 + x 3 − x 2 = (x 5 − x 4 ) + (x 3 − x 2 )
For each binomial, factor out the GCF:
(x 5 − x 4 ) + (x 3 − x 2 ) = x 4 (x − 1) + x 2 (x − 1)
321
2. GROUPING CHAPTER 6. FACTORING POLYNOMIALS
Notice that now we have a sum of two terms, and there is a common factor (x − 1). The
distributive property allows us to factor it out:
x 4 (x − 1) + x 2 (x − 1) = (x 4 + x 2 )(x − 1)
ANSWER: (x 4 + x 2 )(x − 1)
EXAMPLE 2.1.3. Factor by grouping: 5ab − b + 5a − 1
SOLUTION: Represent the given polynomial with 4 terms as a sum of two binomials:
5ab − b + 5a − 1 = (5ab − b) + (5a − 1)
For each binomial, factor out the GCF:
(5a b − b) + (5a − 1) = b(5a − 1) + 1(5a − 1)
Notice that we explicitly factored out the GCF 1 in the second group. This is merely a
precaution to make the application of the distributive property more clear:
b(5a − 1) + 1(5a − 1) = (b + 1)(5a − 1)
ANSWER: (b + 1)(5a − 1)
EXAMPLE 2.1.4. Factor by grouping: 16a3 − 12a2 − 12a + 9
SOLUTION: Represent the given polynomial with 4 terms as a sum of two binomials:
16a3 − 12a2 − 12a + 9 = (16a3 − 12a2 ) + (−12a + 9)
For each binomial, factor out the GCF:
(16a3 − 12a2 ) + (−12a + 9) = 4a2 (4a − 3) + 3(−4a + 3)
We have a sum of two terms, but there is no common factor. Fortunately, (4a − 3) is the
opposite of (−4a + 3), so we can obtain a common factor by factoring out −1:
4a2 (4a − 3) + 3(−4a + 3) = 4a2 (4a − 3) − 3(4a − 3)
The distributive property allows us to factor out (4a − 3):
4a2 (4a − 3) − 3(4a − 3) = (4a2 − 3)(4a − 3)
ANSWER: (4a2 − 3)(4a − 3)
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CHAPTER 6. FACTORING POLYNOMIALS 2. GROUPING
Homework 6.2.
Factor by grouping. 13. 35x 3 − 10x 2 − 56x + 16
1. x 3 − 2x 2 + 5x − 10 14. 14v 3 + 10v 2 − 7v − 5
2. z 3 − 3z 2 + 7z − 21 15. 6x 3 − 48x 2 + 5x − 40
3. 9n3 + 6n2 + 3n + 2 16. 28p3 + 21p2 + 20p + 15
4. 10x 3 + 25x 2 + 2x + 5 17. 15ab − 6a + 5b3 − 2b2
5. 4t 3 + 20t 2 + 3t + 15 18. 3mn − 8m + 15n − 40
6. 8a3 + 2a2 + 12a + 3 19. 5mn + 2m − 25n − 10
7. 7x 3 − 5x 2 + 7x − 5 20. 8x y + 56x − y − 7
8. 2t 3 − 12t 2 − t + 6 21. 24x y − 30 y 3 − 20x + 25 y 2
9. 5x 3 + 6x 2 − 10x − 12 22. 56ab − 49a − 16b + 14
10. 10x 3 + 8x 2 − 5x − 4 23. 10x y + 25x + 12 y + 30
11. x 3 − 3x 2 + 6 − 2x 24. 16x y − 6x 2 + 8 y − 3x
12. a3 + 8a2 + 2a + 16
323
2. GROUPING CHAPTER 6. FACTORING POLYNOMIALS
Homework 6.2 Answers.
1. (x − 2)(x 2 + 5) 13. (5x 2 − 8)(7x − 2)
3. (3n + 2)(3n2 + 1) 15. (6x 2 + 5)(x − 8)
5. (t + 5)(4t 2 + 3) 17. (3a + b2 )(5b − 2)
7. (7x − 5)(x 2 + 1) 19. (m − 5)(5n + 2)
9. (x 2 − 2)(5x + 6) 21. (6 y − 5)(4x − 5 y 2 )
11. (x − 3)(x 2 − 2) 23. (5x + 6)(2 y + 5)
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CHAPTER 6. FACTORING POLYNOMIALS 3. FACTORING X 2 + BX + C
3. Factoring x 2 + bx + c
3.1. Guessing the Coefficients.
THEOREM 3.1.1. If a quadratic trinomial x 2 + bx + c with integer coefficients b and c
factors over integers at all, then it can be written in the form
x 2 + bx + c = (x + m)(x + n)
where m and n are integers such that mn = c and m + n = b.
PROOF.
(x + m)(x + n) = x(x + n) + m(x + n)
= x 2 + x n + mx + mn
= x 2 + (m + n)x + mn
= x 2 + bx + c
EXAMPLE 3.1.1. Factor the trinomial x 2 + 5x + 6
SOLUTION: We need to guess coefficients m and n such that mn = 6 and m + n = 5. One
methodical way to do so is by checking all the possible ways to write 6 as a product of two
integer factors:
6 = 1·6
= 2·3
= (−1)(−6)
= (−2)(−3)
For each one of these, we can check whether m + n = 5. If it is, then we have found a
factorization.
If mn = 1 · 6 then m + n = 1 + 6 = 7 6= 5, no luck.
If mn = 2 · 3 then m + n = 2 + 3 = 5, success, so x 2 + 5x + 6 = (x + 2)(x + 3).
We can check the answer by multiplying the polynomials:
(x + 2)(x + 3) = x 2 + 3x + 2x + 2 · 3
= x 2 + 5x + 6
ANSWER: (x + 2)(x + 3)
325
3. FACTORING X 2 + BX + C CHAPTER 6. FACTORING POLYNOMIALS
EXAMPLE 3.1.2. Factor the trinomial x 2 + 2x − 35
SOLUTION: We need to guess coefficients m and n such that mn = −35 and m + n = 2.
−35 = 1(−35)
= 5(−7)
= 7(−5)
= 35(−1)
If mn = 1(−35) then m + n = 1 − 35 = −34 6= 2, no luck.
If mn = 5(−7) then m + n = 5 − 7 = −2 6= 2, no luck.
If mn = 7(−5) then m + n = 7 − 5 = 2, success, so x 2 + 2x − 35 = (x + 7)(x − 5).
Checking the answer: (x + 7)(x − 5) = x 2 − 5x + 7x − 35 = x 2 + 2x − 35.
ANSWER: (x + 7)(x − 5)
EXAMPLE 3.1.3. Factor the trinomial x 2 − 8x + 7
SOLUTION: We need to guess coefficients m and n such that mn = 7 and m + n = −8.
−7 = 1·7
= (−1)(−7)
If mn = 1 · 7 then m + n = 1 + 7 = 8 6= −8, no luck.
If mn = (−1)(−7) then m + n = −1 + (−7) = −8, success, so x 2 − 8x + 7 = (x − 1)(x − 7).
ANSWER: (x − 1)(x − 7)
EXAMPLE 3.1.4. Factor the trinomial h2 − 5h − 2
SOLUTION: We need to guess coefficients m and n such that mn = −2 and m + n = −5.
−2 = 1(−2)
= 2(−1)
If mn = 1(−2) then m + n = 1 − 2 = −1 6= −5, no luck.
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CHAPTER 6. FACTORING POLYNOMIALS 3. FACTORING X 2 + BX + C
If mn = 2(−1) then m + n = 2 − 1 = 1 6= −5, no luck.
There are essentially no other ways to have two integers with a product −2, we tried every
possible one. Since both addition and multiplication are commutative, checking the prod-
ucts (−1)2 and −2(1) would be redundant. We can conclude that the given polynomial is
irreducible over integers, or prime for short.
ANSWER: prime
3.2. Variations of the Pattern. It is possible to spot this binomial pattern in other places
as well, like quadratics with leading coefficient −1, or the ones with a constant coefficient ±1,
or multivariate polynomials.
EXAMPLE 3.2.1. Factor the trinomial −x 2 + 4x + 21
SOLUTION: This is not a trinomial of the form x 2 + bx + c because the leading coefficient
is −1 when it needs to be 1. But we can easily fix that by factoring out −1:
−x 2 + 4x + 21 = −(x 2 − 4x − 21)
Now we need to find two numbers m and n such that mn = −21 and m + n = −4. Let us
try m = 3 and n = −7:
−(x + 3)(x − 7) = −(x 2 − 7x + 3x − 21)
= −(x 2 − 4x − 21) looks right
= −x 2 + 4x + 21
Note that there are other ways to write this factorization, which look different, but are
essentially the same. They can be obtained by distributing the leading minus over either
sum:
−(x + 3)(x − 7) = (−x − 3)(x − 7) = (x + 3)(−x + 7)
ANSWER: −(x + 3)(x − 7)
EXAMPLE 3.2.2. Factor the trinomial −6x 2 − x + 1
SOLUTION: This is not really a polynomial of the form x 2 + bx + c, since the leading
coefficient is −6 when it needs to be 1. We can, however, notice that the constant term is
1, and if we expect an answer in the form
(mx + 1)(nx + 1)
327
3. FACTORING X 2 + BX + C CHAPTER 6. FACTORING POLYNOMIALS
then we are looking for m and n such that mn = −6 and m + n = −1.
If mn = 1(−6) then m + n = 1 − 6 = −5, no luck.
If mn = 2(−3) then m + n = 2 − 3 = −1, success, so −6x 2 − x + 1 = (2x + 1)(−3x + 1).
We can check the answer by multiplying the polynomials:
(2x + 1)(−3x + 1) = (2x)(−3x) + (2x)1 + 1(−3x) + 1 · 1
= −6x 2 + 2x − 3x + 1
= −6x 2 − x + 1
ANSWER: (2x + 1)(−3x + 1)
EXAMPLE 3.2.3. Factor the trinomial y 2 + 11 yz − 26z 2
SOLUTION: This is a trinomial with two variables, but it uses the same pattern for coef-
ficients as x 2 + bx + c, with the mixed term 11 yz playing the role of the linear term bx,
and the −26z 2 term playing the role of the constant c. So we are looking for m and n such
that mn = −26 and m + n = 11.
If mn = 1(−26) then m + n = 1 − 26 = −25 6= 11, no luck.
If mn = 2(−13) then m + n = 2 − 13 = −11 6= 11, no luck.
If mn = 13(−2) then m+n = 13−2 = 11, success, so y 2 +11 yz −26z 2 = ( y +13z)( y −2z).
ANSWER: ( y + 13z)( y − 2z)
3.3. Basic Factoring Strategy. By now we have seen 3 different techniques for factoring
a polynomial expression: factoring out the GCF, factoring by grouping, and factoring a simple
trinomial. Recall that the final factorization of any polynomial is essentially unique, so we
can apply these and other techniques incrementally and in any order we want. In most cases,
however, it is convenient to follow a simple strategy.
To take the advantage of all the factoring techniques we know so far,
(1) Factor out the GCF, unless it’s 1.
(2) If any of the remaining polynomial factors have four terms, try to factor them by
grouping.
(3) If any of the remaining polynomial factors are trinomials of the form x 2 + bx + c, try
to factor them by guessing integer coefficients.
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CHAPTER 6. FACTORING POLYNOMIALS 3. FACTORING X 2 + BX + C
EXAMPLE 3.3.1. Factor the trinomial x 3 + 3x 2 + 2x
SOLUTION: Factor out the GCF first:
x 3 + 3x 2 + 2x = x(x 2 + 3x + 2)
We attempt to factor the quadratic trinomial next. We need to replace it by an expression
(x + n)(x + m) where mn = 2 and m + n = 3. The ways to write 2 as a product of two
integers mn:
2 = 1·2
= (−1)(−2)
If mn = 1 · 2 then m + n = 1 + 2 = 3, success, so
x(x 2 + 3x + 2) = x(x + 2)(x + 3)
ANSWER: x(x + 2)(x + 3)
EXAMPLE 3.3.2. Factor the polynomial 10x 3 y − 2x 2 y + 30x y − 6 y
SOLUTION: Every term of this polynomial has factors 2 and y, so we start by factoring
out the GCF 2 y:
10x 3 y − 2x 2 y + 30x y − 6 y = 2 y(5x 3 − x 2 + 15x − 3)
We attempt to factor the polynomial with 4 terms next. The only appropriate technique
at our disposal is grouping. Notice that the GCF will have to be simply rewritten all the
way until the answer. Nothing else we factor can possibly affect it.
2 y(5x 3 − x 2 + 15x − 3) = 2 y (5x 3 − x 2 ) + (15x − 3) create binomial groups
= 2 y x 2 (5x − 1) + 3(5x − 1) find GCF for each binomial
= 2 y (x 2 + 3)(5x − 1) distributivity
ANSWER: 2 y(x 2 + 3)(5x − 1)
329
3. FACTORING X 2 + BX + C CHAPTER 6. FACTORING POLYNOMIALS
Homework 6.3.
Factor the polynomial expression. 16. 3 y 2 + 9 y − 84
1. x 2 − 8x + 16 17. x 3 − 6x 2 − 16x
2. a2 + 9a + 20 18. x 3 − x 2 − 20x
3. x 2 − 11x + 10 19. 5b2 + 35b − 150
4. y 2 − 8 y + 7 20. x 4 + x 3 − 56x 2
5. t 2 + 9t + 14 21. −b5 + b4 + 2b3
6. a2 + 14a + 40 22. −2x 2 + 4x + 70
7. b2 + 5b + 4 23. a2 + 5a + 9
8. z 2 + 8z + 7 24. x 2 − 7x + 18
9. d 2 + 7d + 10 25. −3x 3 + 63x 2 + 300x
10. x 2 + 8x + 15 26. 2x 2 − 42x − 144
11. a2 − 2a − 15 27. x 2 + 7x y + 10 y 2
12. b2 − b − 42 28. x 2 − 2x y − 3 y 2
13. x 2 − 4x − 45 29. b2 + 7bc + 7c 2
14. x 2 + 22x + 121 30. 5x 7 − 20x 6 − 25x 5
15. 2x 2 + 14x − 36
330
CHAPTER 6. FACTORING POLYNOMIALS 3. FACTORING X 2 + BX + C
Homework 6.3 Answers.
1. (x − 4)(x − 4) 17. x(x + 2)(x − 8)
3. (x − 1)(x − 10) 19. 5(b − 3)(b + 10)
5. (t + 7)(t + 2)
21. −b3 (b − 2)(b + 1)
7. (b + 1)(b + 4)
23. prime
9. (d + 2)(d + 5)
25. −3x(x − 25)(x + 4)
11. (a − 5)(a + 3)
13. (x + 5)(x − 9) 27. (x + 5 y)(x + 2 y)
15. 2(x − 2)(x + 9) 29. prime
331
4. FACTORING AX 2 + BX + C CHAPTER 6. FACTORING POLYNOMIALS
4. Factoring ax 2 + bx + c
4.1. Guessing the Coefficients.
THEOREM 4.1.1. If a quadratic trinomial ax 2 + bx + c with integer coefficients a, b and c
factors over integers at all, then it can be written in the form
a x 2 + bx + c = (g x + m)(hx + n)
where g, h, m, n are integers such that gh = a, mn = c, and gn + mh = b.
PROOF.
(g x + m)(hx + n) = g x(hx + n) + m(hx + n)
= g xhx + g x n + mhx + mn
= (gh)x 2 + (gn + mh)x + mn
= ax 2 + bx + c
When we try to factor
ax 2 + bx + c
we would like to make sure we checked every possible combination of integer coefficients g,
h, m, n such that gh = a and mn = c, but we want to avoid redundant checking of equivalent
combinations of coefficients. To do so, we will use a specific pattern. We will also assume that
a is positive: if it is not, we can easily make it positive by factoring out −1, for example
−3x 2 + 5x + 2 = −(3x 2 − 5x − 2)
THEOREM 4.1.2. In order to check whether a trinomial with integer coefficients
ax 2 + bx + c
with a > 0, factors over integers as
(g x + m)(hx + n)
it is sufficient to examine every pair of positive integers g and h such that gh = a and
g ≤ h, and for each one of these, every possible integer pair m and n with mn = c.
There are many other ways to check all combinations of four coefficients systematically, and
none of them are particularly better than the one presented in this text. It is important to
understand that experienced human algebraists rarely go through a list of possibilities to factor
trinomials. Instead, after doing dozens and dozens of basic examples, they train the brain to
recognize these patterns instantly, and combinations of coefficients which just feel right often
turn out to be the solutions. The reader should expect to learn (eventually) to try only a few
combinations “at random”, check whether the factorization works each time, and “stumble” on
the correct answer after only a few attempts.
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CHAPTER 6. FACTORING POLYNOMIALS 4. FACTORING AX 2 + BX + C
4.2. Factoring Trinomials.
EXAMPLE 4.2.1. Factor the trinomial 2x 2 + 7x + 3
SOLUTION: We need to guess 4 coefficients g, h, m, n such that gh = 2, mn = 3, and
g n + mh = 7. We will do so by going over all of the ways to represent coefficients 2 and
3 as products of two integers. The combinations are presented in the table that follows.
Note that for g and h we only try 1 and 2 respectively, because 1 · 2 is the only way to get
gh = 2 with positive g ≤ h. But for m and n we try every possible assignment which gives
us mn = 7. The row with the correct value of gn + mh = 7 is highlighted.
g h m n gn + mh
1 2 1 3 5
1 2 3 1 7
1 2 −1 −3 −5
1 2 −3 −1 −7
The second row yields the correct coefficients, so the answer is
(g x + m)(hx + n) = (x + 3)(2x + 1)
The answer can be checked by multiplying and combining like terms:
(x + 3)(2x + 1) = x(2x) + x · 1 + 3(2x) + 3 · 1
= 2x 2 + x + 6x + 3
= 2x 2 + 7x + 3
ANSWER: (x + 3)(2x + 1)
EXAMPLE 4.2.2. Factor the trinomial 8x 2 + 22x + 15
SOLUTION: We need to guess 4 coefficients g, h, m, n such that gh = 8, mn = 15, and
g n + mh = 22. We will do so by going over all of the ways to represent coefficients 8 and
15 as products of two integers. The combinations are presented in the table that follows.
Note that for g · h we only try two pairs: 1 · 8 and 2 · 4, because these are the only ways to
get gh = 8 with positive g ≤ h. But for m and n we try every possible assignment which
gives us mn = 15.
333
4. FACTORING AX 2 + BX + C CHAPTER 6. FACTORING POLYNOMIALS
g h m n gn + mh
1 8 1 15 23
1 8 3 5 29
1 8 5 3 43
1 8 15 1 122
1 8 −1 −15 −23
1 8 −3 −5 −29
1 8 −5 −3 −43
1 8 −15 −1 −122
2 4 1 15 34
2 4 3 5 22
2 4 5 3 26
2 4 15 1 62
2 4 −1 −15 −34
2 4 −3 −5 −22
2 4 −5 −3 −26
2 4 −15 −1 −62
The highlighted row with gn + mh = 22 provides the coefficients for the answer
(g x + m)(hx + n) = (2x + 3)(4x + 5)
ANSWER: (2x + 3)(4x + 5)
EXAMPLE 4.2.3. Factor the trinomial 5x 2 − 27x + 10
SOLUTION: We need to guess 4 coefficients g, h, m, n such that gh = 5, mn = 10, and
g n + mh = −27. We will do so by going over all of the ways to represent coefficients 5 and
10 as products of two integers. The combinations are presented in the table that follows.
Note that for g · h we only try one pair: 1 · 5, because this is the only way to get gh = 5
with positive g ≤ h. But for m and n we try every possible assignment which gives us
mn = 10.
334
CHAPTER 6. FACTORING POLYNOMIALS 4. FACTORING AX 2 + BX + C
g h m n gn + mh
1 5 1 10 15
1 5 2 5 15
1 5 5 2 27
1 5 10 1 52
1 5 −1 −10 −15
1 5 −2 −5 −15
1 5 −5 −2 −27
1 5 −10 −1 −15
The highlighted row gn + mh = −27 shows the correct coefficients, so the answer is
(g x + m)(hx + n) = (5x − 2)(x − 5)
ANSWER: (5x − 2)(x − 5)
335
4. FACTORING AX 2 + BX + C CHAPTER 6. FACTORING POLYNOMIALS
Homework 6.4.
1. 2x 2 − 13x − 7 14. 49x 2 − 42x + 9
2. 2x 2 + 7x − 4 15. −20 y 2 + 25 y − 5
3. 9x 2 − 9x + 2 16. −10a2 + 8a + 18
4. 5a2 − 19a − 4 17. 6z 2 + 21z + 15
5. 3x 2 + 19x + 6 18. 12a2 + 68a − 24
6. 35x 2 + 34x + 8 19. 16t 2 − 23t + 7
7. 8t 2 + 6t − 9 20. 9t 2 − 14t + 5
8. −4x 2 − 12x − 5 21. 9x 2 + 18x + 5
9. 1 − 4x + 3x 2 22. 16x 2 + 32x + 7
10. 9 − 21x − 18x 2 23. 18x 3 + 33x 2 + 9x
11. 2x 2 − 6x − 17 24. 6a3 − 4a2 − 10a
12. 15 + c − 2c 2 25. 14x 4 − 19x 3 − 3x 2
13. 25x 2 − 40x + 16 26. 70 y 4 − 68 y 3 + 16 y 2
336
CHAPTER 6. FACTORING POLYNOMIALS 4. FACTORING AX 2 + BX + C
Homework 6.4 Answers.
1. (2x + 1)(x − 7) 15. −5( y − 1)(4 y − 1)
3. (3x − 1)(3x − 2) 17. 3(x + 1)(2x + 5)
5. (3x + 1)(x + 6)
19. (16t − 7)(t − 1)
7. (2t + 3)(4t − 3)
21. (3x + 1)(3x + 5)
9. (3x − 1)(x − 1)
11. prime 23. 3x(3x + 1)(2x + 3)
13. (5x − 4)2 25. x 2 (2x − 3)(7x + 1)
337
5. FACTORING SPECIAL PRODUCTS CHAPTER 6. FACTORING POLYNOMIALS
5. Factoring Special Products
5.1. Difference of Squares. Recall the difference of squares formula:
A2 − B 2 = (A + B)(A − B)
for any two reals A and B.
EXAMPLE 5.1.1. Factor the special product expression 25x 2 − 9 y 2
SOLUTION: This can actually be done by guessing coefficients for a factored form of the
trinomial 25x 2 + 0x y − 9 y 2 , but being able to detect a difference of squares saves a lot of
time and effort:
25x 2 − 9 y 2 = (5x)2 − (3 y)2
= (5x + 3 y)(5x − 3 y)
ANSWER: (5x + 3 y)(5x − 3 y)
EXAMPLE 5.1.2. Factor the special product expression 200x 6 − 18
SOLUTION: 200 and 18 are not squares of integers, but may be a special product will
become evident if we factor out GCF first:
200x 6 − 18 = 2(100x 6 − 9)
= 2 (10x 3 )2 − (3)2
= 2 (10x 3 + 3)(10x 3 − 3)
ANSWER: 2(10x 3 + 3)(10x 3 − 3)
It is worthy of note that a sum of squares is irreducible over reals.
THEOREM 5.1.1. If X and Y are monomials with relatively prime coefficients and no vari-
ables in common, then the binomial X 2 + Y 2 is irreducible over reals.
BASIC EXAMPLE 5.1.1. It may be tempting to write that x 2 +4 is equivalent to (x +2)(x −2)
or perhaps (x +2)2 , but neither statement is true. Just like any other sum of squares, x 2 +4
is prime.
338
CHAPTER 6. FACTORING POLYNOMIALS 5. FACTORING SPECIAL PRODUCTS
EXAMPLE 5.1.3. Factor the expression 27x 7 + 3x 5 .
SOLUTION: Factor out GCF first.
27x 7 + 3x 5 = 3x 5 (9x 2 + 1)
(9x 2 + 1) is a sum of squares, which is prime, so we cannot factor the polynomial any
further.
ANSWER: 3x 5 (9x 2 + 1)
EXAMPLE 5.1.4. Factor the expression a4 − b4
SOLUTION: Factor as a difference of squares:
a4 − b4 = (a2 )2 − (b2 )2
= (a2 + b2 )(a2 − b2 )
The factor (a2 + b2 ) is a sum of squares, which is prime. The factor (a2 − b2 ) though is a
difference of squares again, so we can factor it further:
(a2 + b2 )(a2 − b2 ) = (a2 + b2 ) ((a + b)(a − b))
ANSWER: (a2 + b2 )(a + b)(a − b)
339
5. FACTORING SPECIAL PRODUCTS CHAPTER 6. FACTORING POLYNOMIALS
5.2. Square of a Binomial. Recall the binomial square formulas:
A2 + 2AB + B 2 = (A + B)2
A2 − 2AB + B 2 = (A − B)2
for any two reals A and B.
EXAMPLE 5.2.1. Factor the special product expression 64x 2 + 16x + 1
SOLUTION: This can be done by guessing the coefficients, but detecting the square of a
binomial saves some time. Here we notice that 64x 2 = (8x)2 and 1 = (1)2 , while 16x is
indeed 2(8)(x).
64x 2 + 16x + 1 = (8x)2 + 2(8)(x) + (1)2
= (8x + 1)2
ANSWER: (8x + 1)2
EXAMPLE 5.2.2. Factor the special product expression 25 y 2 − 60 yz + 36z 2
SOLUTION: This can be done by guessing the coefficients, but detecting the square of a
binomial saves some time. Here we notice that 25 y 2 = (5 y)2 , and 36z 2 = (6z)2 , while
−60 yz is indeed −2(5 y)(6x).
25 y 2 − 60 yz + 36z 2 = (5 y)2 − 2(5 y)(6x) + (6z)2
= (5 y − 6z)2
ANSWER: (5 y − 6z)2
340
CHAPTER 6. FACTORING POLYNOMIALS 5. FACTORING SPECIAL PRODUCTS
Homework 6.5.
Factor the given expression. 16. 8 y 2 − 98
1. x 2 − 14x + 49 17. x 2 − 5x y + 9 y 2
2. x 2 − 10x + 25 18. a2 − 9ab − 10b2
3. 4x 2 + 8x + 4 19. 5x 4 + 125x 2
4. 9 y 2 + 12 y + 4 20. ab2 + 9a
5. x 2 − 100 21. 25 y 2 − 4
6. 25x 2 − 1 22. 80x 2 − 45
7. 3n3 + 60n2 + 300n 23. 81x 4 − 625
8. x 5 + 24x 4 + 144x 3 24. x 4 y 4 − 1
9. 20a2 + 100a + 125 25. x 8 − 256
10. 27x 2 + 36x + 12 26. y 16 − 1
11. 2x 2 + 28x + 98 27. x 2 + 16x y + 64 y 2
12. 16 y 2 − 24 y + 9 28. 81 y 2 − 18 yz + z 2
13. 6x 2 − 24 29. 32x 2 − 80x y + 50 y 2
14. 5 y 2 − 5 30. −36c 2 − 96cd − 64d 2
15. 200t 2 − 8
341
5. FACTORING SPECIAL PRODUCTS CHAPTER 6. FACTORING POLYNOMIALS
Homework 6.5 Answers.
1. (x − 7)2 17. prime
3. 4(x + 1)2 19. 5x 2 (x 2 + 25)
5. (x + 10)(x − 10)
21. (5 y + 2)(5 y − 2)
7. 3n(n + 10) 2
23. (9x 2 + 25)(3x + 5)(3x − 5)
9. 5(2x + 5) 2
25. (x 4 + 16)(x 2 + 4)(x + 2)(x − 2)
11. 2(x + 7) 2
13. 6(x + 2)(x − 2) 27. (x + 8 y)2
15. 8(5t 2 + 1)(5t 2 − 1) 29. 2(4x − 5 y 2 )
342
CHAPTER 6. FACTORING POLYNOMIALS 6. GENERAL FACTORING STRATEGY
6. General Factoring Strategy
6.1. Irreducible Polynomials. In this section we will combine all of the polynomial fac-
toring techniques presented so far. We will also describe which polynomials are irreducible.
Knowing which polynomials cannot be factored any further will gives us an assurance that our
answers are in fact in the fully factored form.
(1) By theorem 1.1.2, all linear polynomials are prime.
(2) Some prime trinomials such as x 2 + x + 1 can be detected by trying out every feasible
combination of coefficients in factorizations. See theorems 3.1.1 and 4.1.1.
(3) By theorem 5.1.1, sums of relatively prime monomial squares X 2 + Y 2 are irreducible
over reals and therefore also prime.
BASIC EXAMPLE 6.1.1. In this text we will usually recognize a fully factored polynomial as
a product of linear and irreducible quadratic factors, something like
5x 7 (x + 1)3 (x 2 + 4)
We use exponential notation for brevity, but if we think of this as a product of polynomial
factors, then x 7 as a linear factor x raised to the 7th power, (x + 1)3 is a linear factor
x + 1 raised to the 3rd power, and x 2 + 4 is an irreducible quadratic factor raised to the
1st power.
6.2. Factoring Strategy. By now we have seen several different techniques for factoring a
polynomial expression: factoring out the GCF, factoring by grouping, factoring a trinomial by
guessing the coefficients, and the special product shortcuts. Recall that the final factorization
of any polynomial is essentially unique, so we can apply these techniques incrementally and in
any order we want. In most cases, however, it is convenient to try them in a specific order.
To take the advantage of all the factoring techniques we know so far,
(1) Factor out the GCF, unless it’s 1.
(2) If a polynomial factor has four terms, try to factor it by grouping.
(3) If a polynomial factor has two terms, try to factor it as a difference of squares A2 − B 2 .
(4) If a polynomial factor is a trinomial of the form x 2 + bx + c, try to factor it by guessing
two integer coefficients for a product of two binomials.
(5) If a polynomial factor is a trinomial of the form ax 2 + bx + c, try to factor it as a
binomial square, and if that fails, try to guess four integer coefficients for a product
of two binomials.
One may have to apply these techniques repeatedly, until only prime polynomial factors remain.
343
6. GENERAL FACTORING STRATEGY CHAPTER 6. FACTORING POLYNOMIALS
EXAMPLE 6.2.1. Factor the polynomial completely:
x2 + x
SOLUTION: x 2 + x = x(x + 1), which is a product of a linear monomial and a linear
binomial, so every factor is irreducible.
ANSWER: x(x + 1)
EXAMPLE 6.2.2. Factor the polynomial completely:
3x 3 − 3x 2 + x − 1
SOLUTION: Factoring by grouping yields
3x 3 − 3x 2 + x − 1 = 3x 2 (x − 1) + 1(x − 1)
= (3x 2 + 1)(x − 1)
The factor 3x 2 +1 is an irreducible quadratic binomial, and x−1 is linear, so also irreducible
over reals.
ANSWER: (3x 2 + 1)(x − 1)
EXAMPLE 6.2.3. Factor the polynomial completely:
−m6 + 2m5 + 35m4
SOLUTION: The GCF is m4 , and we also notice that factoring out −1 will create an easy
trinomial pattern:
−m6 + 2m5 + 35m4 = −m4 (m2 − 2m − 35)
= −m4 (m + 5)(m − 7)
We think of m4 as a linear factor m raised to 4th power, so every factor is linear, and hence
prime, and we are done.
ANSWER: −m4 (m + 5)(m − 7)
344
CHAPTER 6. FACTORING POLYNOMIALS 6. GENERAL FACTORING STRATEGY
EXAMPLE 6.2.4. Factor the polynomial completely:
6x 3 y 5 − 21x 4 y 5 + 3x 3 y 6
SOLUTION: We start by factoring out the GCF:
6x 3 y 5 − 21x 4 y 5 + 3x 3 y 6 = 3x 3 y 5 (2 − 7x + y)
This is a product of a monomial and a linear trinomial, so we are done.
ANSWER: 3x 3 y 5 (2 − 7x + y)
EXAMPLE 6.2.5. Factor the polynomial completely:
a3 − 5a2 − 4a + 20
SOLUTION: The GCF is 1, and there are four terms, so let’s try grouping:
a3 − 5a2 − 4a + 20 = a2 (a − 5) − 4(a − 5)
= (a2 − 4)(a − 5)
a2 − 4 is not linear, so we should check whether we can factor it more, and in fact we can
as a difference of squares:
(a2 − 4)(a − 5) = (a2 − 22 )(a − 5)
= ((a + 2)(a − 2)) (a − 5)
ANSWER: (a + 2)(a − 2)(a − 5)
345
6. GENERAL FACTORING STRATEGY CHAPTER 6. FACTORING POLYNOMIALS
Homework 6.6.
Factor the given polynomial completely. 16. z 5 − 2z 4 + 5z 3 − 10z 2
1. 5 y 2 − 80 17. 5 y 5 − 80 y
2. 10 y 2 − 490 18. 7z 4 − 112
3. 18 y 4 − 12 y 3 19. x 2 − 5x y + 8 y 2
4. x 2 + x − 12 20. 25z 2 + 10z x + x 2
5. 6x 3 − 9x 2 + 2x − 3 21. 16t 2 + 28t − 30
6. x 2 + 5x + x y + 5 y 22. 36x 2 + 24x − 45
7. 6x 3 + x 2 − 5x 23. −6x 2 − 7x + 1
8. 25 y 3 − 30 y 2 + 9 y 24. 16x 4 + y 4
9. 3x 4 + 6x 2 25. 4w3 + 20w2 − 4w − 20
10. 4 y 2 + 36 26. u4 + 3u3 − 16u2 − 48u
11. 4z 2 w + 13zw + 10w 27. 4x 2 + y 2 − 4x y
12. 5x 2 − 30x + 10 28. 11a4 − 11b4
13. x 2 y 4 + 4x y 4 − 32 y 4 29. −360 y 2 + x 2 + 2x y
14. −60 + 52 y − 8 y 2 30. 3x 3 + 17x 2 y − 6x y 2
15. x 4 + 6x 3 − 6x 2 − 36x
346
CHAPTER 6. FACTORING POLYNOMIALS 6. GENERAL FACTORING STRATEGY
Homework 6.6 Answers.
1. 5( y + 4)( y − 4) 17. 5 y( y 2 + 4)( y + 2)( y − 2)
3. 6 y 3 (3 y − 2) 19. prime
5. (2x − 3)(3x 2 + 1)
21. 2(2t + 5)(4t − 3)
7. x(6x − 5)(x + 1)
23. prime
9. 3x 2 (x 2 + 2)
25. 4(w + 5)(w + 1)(w − 1)
11. w(4z + 5)(z + 2)
13. y 4 (x − 4)(x + 8) 27. (2x − y)2
15. x(x 2 − 6)(x + 6) 29. (x + 20 y)(x − 18 y)
347
7. SOLVING EQUATIONS BY FACTORING CHAPTER 6. FACTORING POLYNOMIALS
7. Solving Equations by Factoring
7.1. Zero Product Property.
THEOREM 7.1.1. If a product of several numbers is zero, then one of these numbers must
be zero. Conversely, if one of the factors in a product is zero, then the value of the product
is zero.
Formally, for all real numbers a and b, ab = 0 if and only if a = 0 or b = 0. This statement
generalizes naturally to products with any number of factors.
THEOREM 7.1.2 (Zero Product in Equations). If P and Q are polynomials, then the solution
set of the equation PQ = 0 consists of all solutions of P = 0 together with all solutions of
Q = 0.
EXAMPLE 7.1.1. Solve the equation
x 2 − 3x + 2 = 0
SOLUTION: Factoring the expression on the left yields
(x − 1)(x − 2) = 0
Here (x − 1) and (x − 2) are the polynomial factors P and Q respectively in the theorem
7.1.2.
Solving the equation x − 1 = 0 gives us x = 1.
Solving the equation x − 2 = 0 gives us x = 2.
So the solutions to the quadratic equation are 1 and 2.
Let’s check the answers. If x = 1, then
x 2 − 3x + 2 = 12 − 3 · 1 + 2 = 0
If x = 2, then
x 2 − 3x + 2 = 22 − 3 · 2 + 2 = 0
Multiple solutions are common for quadratic equations, so we will state answers using the
roster notation.
ANSWER: {1, 2}
348
CHAPTER 6. FACTORING POLYNOMIALS 7. SOLVING EQUATIONS BY FACTORING
7.2. Equation Solving Strategy. We are already familiar with the procedure for solving
linear polynomial equations, and now we can tackle all the rest of them. Faced with a poly-
nomial equation, we will first try to determine whether it is linear or not. If it is linear, then
we will solve it by isolating the variable. If it is a quadratic or a higher degree equation, then
we will arrange it so that all the non-zero terms are on the same side, factor the polynomial
completely, and take advantage of the zero product property.
EXAMPLE 7.2.1. Solve the equation
2x 2 + x = 1
SOLUTION: This equation appears to be quadratic, so we will add −1 to both sides in
order to make the right side zero, then factor by guessing the coefficients.
2x 2 + x = 1
2x + x − 1
2
= 0
(2x − 1)(x + 1) = 0
The left side is a product, and the right side is zero, so we can apply the zero product
property. Solving the equation 2x − 1 = 0 gives us
2x − 1 = 0
2x = 1
x = 0.5
Solving the equation x + 1 = 0 gives us x = −1.
ANSWER: {0.5, −1}
EXAMPLE 7.2.2. Solve the equation
8x 2 − 18 = 0
SOLUTION: All the non-zero terms of this quadratic equation are already on the left side,
so let’s try factoring it completely. Instead of factoring out 2 we opt to multiply both sides
by 0.5, canceling it altogether.
8x 2 − 18 = 0
2
0.5(8x − 18) = 0.5(0)
2
4x − 9 = 0 this is a difference of squares
(2x)2 − 32 = 0
(2x + 3)(2x − 3) = 0
349
7. SOLVING EQUATIONS BY FACTORING CHAPTER 6. FACTORING POLYNOMIALS
The left side is a product, and the right side is zero, so we can apply the zero product
property. Solving the equation 2x + 3 = 0 gives us x = −1.5.
Solving the equation 2x − 3 = 0 gives us x = 1.5.
ANSWER: {1.5, −1.5}
EXAMPLE 7.2.3. Solve the equation
14x 2 = −16x
SOLUTION: We will add 16x to both sides to make the right side zero, and try to factor.
Notice that we opt to divide both sides of the equation by the common factor 2 rather than
to factor it out, because dividing both sides by a non-zero number makes an equivalent
equation.
14x 2 = −16x
14x 2 + 16x = 0 divide both sides by 2
14x 2 + 16x 0
=
2 2
14x 2 16x
+ = 0 distributivity
2 2
7x 2 + 8x = 0 the GCF is x
x(7x + 8) = 0 distributivity
Notice we are not allowed to divide both sides of the equation by x. Unlike the non-zero
quantity 2, x is an unknown quantity which may turn out to be zero.
The expression x(7x + 8) has two polynomial factors: x and 7x + 8. By the zero product
property x = 0 is a solution, and solving 7x + 8 = 0 gives us
7x + 8 = 0
7x = −8
x = −8/7
ANSWER: {0, −8/7}
350
CHAPTER 6. FACTORING POLYNOMIALS 7. SOLVING EQUATIONS BY FACTORING
EXAMPLE 7.2.4. Solve the equation
4 y 3 = 7 y 2 + 15 y
SOLUTION: This looks like a cubic equation, so we will subtract (7 y 2 +15 y) on both sides,
and then factor completely.
4 y3 = 7 y 2 + 15 y
4 y 3 − 7 y 2 − 15 y = 0 the GCF is y
y(4 y 2 − 7 y − 15) = 0 factor by guessing coefficients
y(4 y + 5)( y − 3) = 0
By the zero product property, y = 0 is a solution;
solving 4 y + 5 = 0 yields y = −1.25;
and solving y − 3 = 0 yields y = 3.
As always the case with any equation, we can check the answers by substituting them and
checking that the equation holds.
If y = 0 then
4 · 03 = y · 02 + 15 · 0
0 = 0
If y = 3 then
4 · 33 = 7 · 32 + 15 · 3
108 = 63 + 45
108 = 108
Finally, if y = −1.25 then
4(−1.25)3 = 7(−1.25)2 + 15(−1.25)
−7.8125 = 10.9375 − 18.75
−7.8125 = −7.8125
ANSWER: {−1.25, 0, 3}
351
7. SOLVING EQUATIONS BY FACTORING CHAPTER 6. FACTORING POLYNOMIALS
Homework 6.7.
Solve the given equation by using the zero 15. y 2 + 4 y − 21 = 0
product property.
16. y 2 = 7 y + 18
1. (x + 3)(x − 10) = 0
17. x 2 − 11x + 18 = 0
2. (x − 2)(x + 9) = 0
18. x 2 − 8x + 15 = 0
3. (x − 1)(x − 8) = 0
19. y 2 − 6.5 y = 0
4. (x + 4)(x + 3) = 0
20. 4z 2 − 12z = 0
5. 3x(x + 4) = 0
2 2 1
21. x + x =0
6. 2 y( y − 6) = 0 3 3
7. (2x + 3)(4x − 5) = 0 22. 3z 2 + 4z = 0
8. (5x − 10)(6x + 10) = 0 23. 2x 2 = 72
1
24. 6 y 2 − 54 = 0
9. x x +
2
=0
2
25. 10x + x 2 + 25 = 0
2
3
10. 4 y y − =0 26. 6x + 9 + x 2 = 0
3
11. x(4x + 0.8)(x − 0.1)2 = 0 27. (x + 1)(x − 7) = −16
12. y 2 (3 y − 6)( y + 1.7) = 0 28. (x + 2)(7 − x) = 18
29. (3x + 5)(x + 3) = 7
Solve the given equation by factoring. 30. (5x + 4)(x − 1) = 2
13. x 2 + 7x + 6 = 0 31. 81x 2 − 6 = 19
14. x 2 + 6x + 5 = 0 32. 14x 2 − 3 = 42x − 3
352
CHAPTER 6. FACTORING POLYNOMIALS 7. SOLVING EQUATIONS BY FACTORING
Homework 6.7 Answers.
1. {−3, 10} 17. {2, 9}
3. {1, 8} 19. {0, 6.5}
1
§ ª
5. {−4, 0} 21. − , 0
2
3 5
§ ª
7. − , 23. {−6, 6}
2 4
25. {−5}
1
§ ª
9. − , 0
2 27. {3}
11. {−0.2, 0, 0.1} §
2
ª
29. −4, −
3
13. {−6, −1}
5 5
§ ª
31. − ,
15. {−7, 3} 9 9
353
8. FACTORING APPLICATIONS CHAPTER 6. FACTORING POLYNOMIALS
8. Factoring Applications
8.1. Applications to Areas. Recall that the area of a rectangle with width w and length l
is the product of its dimensions.
A= w·l h
w
Recall also that the area of a triangle with base b and height h is one half of their product.
bh h
A=
2
b
EXAMPLE 8.1.1. The length of a rectangular carpet is 1 foot greater than the width. Find
the dimensions of the carpet if its surface area is 12 square feet.
SOLUTION: Let w and l be the width and the length in feet, respectively. We can translate
“the length is 1 foot greater than the width” as
l = w+1
We can also write the equation for the area:
wl = 12
This is not a linear system, but we can solve it by substitution. The first equation is already
solved for l, so we substitute (w + 1) for l in the area equation and solve for w:
wl = 12
w(w + 1) = 12
w2 + w = 12
354
CHAPTER 6. FACTORING POLYNOMIALS 8. FACTORING APPLICATIONS
This is a quadratic equation, which we will solve by factoring.
w2 + w = 12 put non-zero terms on one side
w2 + w − 12 = 0 trinomial pattern x 2 + bx + c
(w − 3)(w + 4) = 0
By the zero product property, the solutions are w = 3 and w = −4. The width cannot be
negative, so the solution −4 is not applicable. The only useful solution is 3. To find the
length, we substitute 3 for w in the equation solved for l:
l = w+1
l = (3) + 1
l = 4
ANSWER:
w and l are width and length respectively, in feet
l = w+1
wl = 12
solution: w = 3, l =4
EXAMPLE 8.1.2. The height of a triangular sail is 1 meter less than the twice the length of
its base. Find the dimensions of the sail if its surface area is 14 square meters.
SOLUTION: Let b and h be the base and the height of the sail in meters. The sentence
“the height of a triangular sail is 1 meter less than the twice the length of its base” can be
translated as
h = 2b − 1
We can also write the equation for the area:
bh/2 = 14
The first equation is already solved for h, so we can substitute (2b − 1) for h in the area
equation and solve for b:
bh
= 14 area equation
2
b(2b − 1)
= 14 substitution
2
b(2b − 1) = 28 multiplied both sides by 2
355
8. FACTORING APPLICATIONS CHAPTER 6. FACTORING POLYNOMIALS
With the last step, we multiplied both sides by 2 to get rid of the fraction. The result looks
like a quadratic equation, so we solve it by factoring:
2b2 − b = 28 distributed on the left side
2
2b − b − 28 = 0 trinomial pattern ax 2 + bx + c
(2b + 7)(b − 4) = 0 factored by guessing coefficients
By the zero product property, either
2b + 7 = 0
2b = −7
b = −7/2
or
b−4 = 0
b = 4
The negative solution b = −7/2 is not applicable because it cannot denote a length, so the
only useful solution is b = 4. To find the height, substitute 4 for b in the equation solved
for h:
h = 2b − 1
h = 2(4) − 1
h = 7
ANSWER:
b and h are base and height respectively, in meters
h = 2b − 1
bh/2 = 14
solution: b = 4, h=7
EXAMPLE 8.1.3. The height of a triangular tile is 40% greater than its base. Find the
dimensions of the tile if its area is 17.5 square inches.
SOLUTION: Let b and h be the base and the height of the triangle in inches. The sentence
“the height of a triangular tile is 40% greater than its base” is a statement about percent
increase, and can be translated as
h = b + 0.4b see definition 4.2.1 in chapter 2
h = 1.4b combined like terms
356
CHAPTER 6. FACTORING POLYNOMIALS 8. FACTORING APPLICATIONS
And the statement about the area can be translated as
bh
= 17.5
2
Using the first equation, which is solved for h, we can substitute 1.4b for h into the second
equation, and solve for b:
bh
= 17.5
2
b(1.4b)
= 17.5 substituted 1.4b for h
2
To get rid of the fraction, we multiply both sides by 2:
b(1.4b) = 35
1.4b2 = 35 simplify left side
b 2
= 25 divide both sides by 1.4
This looks like a quadratic equation, so we solve it by factoring:
b2 − 25 = 0 subtracted 25 on both sides
(b + 5)(b − 5) = 0 difference of squares b2 − 52
By the zero product property, the solutions are −5 and 5. Negative length does not make
sense, so the negative solution is not applicable, and the only useful solution is b = 5. We
can find the other dimension using the equation solved for h:
h = 1.4b
h = 1.4 · 5
h = 7
ANSWER:
b and h are base and height respectively, in inches
h = b + 0.4b
bh/2 = 17.5
solution: b = 5, h=7
357
8. FACTORING APPLICATIONS CHAPTER 6. FACTORING POLYNOMIALS
Homework 6.8.
1. The width of a rectangular rug is 4 times 5. The width of a rectangular laptop
less than its length. Find the dimensions of screen is 7 inches less than its length. Find
the rug if its area is 36 square feet. the dimensions of the screen if its area is
144 square inches.
2. The length of a rectangular room is 6
yards greater than its width. Find the di- 6. The width of a rectangular mural is 10%
mensions of the room if its area is 55 square greater than its length. Find the dimen-
yards. sions of the mural if its area is 110 square
meters.
3. The base of a triangle is 2 meters longer
than twice its height. Find the base and 7. Dmitriy buys a gold earring shaped like
the height if the area of the triangle is 12 a thin flat triangle. Find the dimensions of
square meters. the earring if the base of the triangle is 25%
shorter than its height, and the area of the
4. The base of a triangle is 2 units longer triangle is 24 mm2 .
than the height. Find the base and the
height if the area of the triangle is 12 8. The height of a triangular piece of fab-
square units. ric is 3 times greater than its base. Find the
dimensions of the triangle if its area is 13.5
square inches.
358
CHAPTER 6. FACTORING POLYNOMIALS 8. FACTORING APPLICATIONS
Homework 6.8 Answers.
1. 5.
w and l are width and length respectively, w and l are width and length respectively,
in feet in inches
w = l/4 w= l −7
wl = 36 wl = 144
solution: w = 3, l = 12 solution: w = 9, l = 16
3. 7.
b and h are base and height respectively, in b and h are base and height respectively, in
meters mm
b = 2h + 2 b = h − 0.25h
bh/2 = 12 bh/2 = 24
solution: b = 8, h=3 solution: b = 6, h=8
359
8. PRACTICE TEST 6 CHAPTER 6. FACTORING POLYNOMIALS
Practice Test 6
Factor each of the following expressions Solve each of the following equations using
completely: factoring:
1. 48x 7 y 3 + 42x 4 y 4 11. x 2 + 100 = 20x
2. 6x 4 − 42x 3 − 6x 2 12. 25x 2 − 9 = 0
3. x 2 − 10x + 24 13. 3x 2 − 3x = 36
4. 18x 2 + 63x 2 a − 24x − 84x a 14. 5x 3 = 45x
5. 5x 2 − 17x + 6 15. 18x 2 − 3x = 6
6. 4 y 2 + 10 y − 6 16. x(2x − 1) = 3
7. 1 − 16x 4 17. x 3 − 9x 2 + 2x − 18 = 0
18. The length of a rectangular carpet is 5
Solve each of the following equations using times greater than the width, and the area
the zero product property: of the carpet is 45 square feet. Find the di-
mensions of the carpet.
8. x(x + 4) = 0
19. A piece of stained glass is shaped like a
9. 10(x − 6)(3x + 1) = 0 triangle, with its height 4 cm shorter than
its base. Find the base and the height of
10. −7x 2 (x 2 + 1)(x − 1) = 0 the triangle if its area is 48 cm2 .
360
CHAPTER 6. FACTORING POLYNOMIALS 8. PRACTICE TEST 6
Practice Test 6 Answers.
1. 6x 4 y 3 (8x 3 + 7 y) 14. {0, 3, −3}
2. 6x 2 (x 2 − 7x − 1) 15. {2/3, −1/2}
3. (x − 4)(x − 6)
16. {3/2, −1}
4. 3x(3x − 4)(2 + 7a)
17. {9}
5. (5x − 2)(x − 3)
18.
6. 2(2 y − 1)( y + 3) l and w are length and width respectively,
in feet
7. (1 + 4x 2 )(1 + 2x)(1 − 2x)
lw = 45
8. {0, −4} l = 5w
solution: l = 15, w=3
9. {6, −1/3}
10. {0, 1} 19.
b and h are base and height respectively, in
11. {10} cm
bh/2 = 48
12. {−3/5, 3/5}
h= b−4
13. {4, −3} solution: b = 12, h=8
361
CHAPTER 7
Rational Expressions
1. Simplifying Rational Expressions
4
y
3
2 1
y=
x
1
x
−4 −3 −2 −1 1 2 3 4
−1
−2
−3
−4
1.1. Definition.
DEFINITION 1.1.1. A rational expression is either a polynomial expression or a quotient of
a polynomial and a non-zero polynomial.
BASIC EXAMPLE 1.1.1. Here are some rational expressions:
5− x 1 1 + x2 x 2 + 2x + 1 x + yw
x2 − x
4 x x −2 x +1 x y − w5
On the other hand
1 1
+
x x
is not a rational expression, since it is neither a polynomial nor a quotient of polynomials.
363
1. SIMPLIFYING RATIONAL EXPRESSIONS CHAPTER 7. RATIONAL EXPRESSIONS
1.2. Variable Restrictions. Unlike polynomials, which can always be evaluated, rational
expressions will occasionally fail to have a defined value. As the most basic example, the
expression 1/x can be evaluated for every real number x except for x = 0. Because of that,
the process of simplifying rational expressions may well produce expressions which are not
equivalent, but almost equivalent, with only a finite number of disagreements.
BASIC EXAMPLE 1.2.1. Expressions
(x + 2)(x − 1)
and x +2
x −1
are not equivalent even though they agree on the value for almost all real numbers x. But
when x = 1,
x +2=1+2=3
whereas
(x + 2)(x − 1) (1 + 2)(1 − 1) 3 · 0
= =
x −1 1−1 0
which is undefined.
In this text we will simplify rational expressions by canceling common polynomial factors, even
though in many cases it will result in non-equivalent expressions. Our excuse will be that we
can find exactly which inputs lead to disagreements.
EXAMPLE 1.2.1. Find all variable values which make the given expression undefined:
x +1
x −5
SOLUTION: A rational expression is a quotient of two polynomials. The numerator is
always defined, and the denominator is always defined. The only operation which may
possibly fail is the division: the fraction bar. It will fail just in case if the denominator is
equal to zero. So we can make the denominator equal to zero, and then solve for x. We
ignore the numerator completely.
x −5 = 0
x = 5
The only number which makes this expression undefined is x = 5.
ANSWER: {5}
EXAMPLE 1.2.2. Find all variable values which make the given expression undefined:
x +4
x 2 − 16
364
CHAPTER 7. RATIONAL EXPRESSIONS 1. SIMPLIFYING RATIONAL EXPRESSIONS
SOLUTION: We can make the denominator equal to zero, and then solve for x. We ignore
the numerator completely, even though it happens to have the factor (x + 4), which can
also be found in the denominator.
x 2 − 16 = 0
(x + 4)(x − 4) = 0
By the zero product property, x is either −4 or 4.
ANSWER: {−4, 4}
EXAMPLE 1.2.3. Find all variable values which make the given expression undefined:
1 1
+
x 7
x 2
−
6 9
SOLUTION: There are two variable denominators in this expression, and if either one
happens to be 0, then the whole expression is undefined. One variable denominator is
x from 1/x up on top, so x = 0 will make the expression undefined. The other variable
denominator is
x 2
−
6 9
so we make it equal to zero and solve for x:
x 2
− = 0
6 9
x 2
=
6 9
4
x =
3
ANSWER: {0, 4/3}
365
1. SIMPLIFYING RATIONAL EXPRESSIONS CHAPTER 7. RATIONAL EXPRESSIONS
1.3. Canceling Common Polynomial Factors. We can simplify rational expressions by
fully factoring both numerator and denominator, and then canceling common polynomial fac-
tors. In answers, we will write simplified rational expressions as quotients of polynomial fac-
torizations. This way we can be sure nothing else can be canceled.
EXAMPLE 1.3.1. Simplify and state the answer as a polynomial:
a2 − 1
a+1
SOLUTION:
a2 − 1 (a + 1)(a − 1)
=
a+1 (a + 1)
= a−1
ANSWER: a − 1
EXAMPLE 1.3.2. Simplify and state the answer as a quotient of polynomial factorizations:
6x − 3
4x − 4x + 1
2
SOLUTION:
6x − 3 3(2x − 1)
=
4x 2 − 4x + 1 (2x − 1)(2x − 1)
3
=
2x − 1
3
ANSWER:
2x − 1
EXAMPLE 1.3.3. Simplify and state the answer as a quotient of polynomial factorizations:
4a2 − 5ab
30b − 24a
SOLUTION:
4a2 − 5ab a(4a − 5b)
=
30b − 24a 6(5b − 4a)
366
CHAPTER 7. RATIONAL EXPRESSIONS 1. SIMPLIFYING RATIONAL EXPRESSIONS
The factors (4a − 5b) and (5b − 4a) are not identical, so they cannot be canceled outright.
But they are opposites of each other, so the quotient is equivalent to −1 whenever defined.
Algebraically, we can make them look exactly the same if we factor out −1.
a(4a − 5b) a(−1)(−4a + 5b)
=
6(5b − 4a) 6(5b − 4a)
a(−1)(5b − 4a)
= now they are identical and can be canceled
6(5b − 4a)
−a
=
6
−a
ANSWER:
6
EXAMPLE 1.3.4. Simplify and state the answer as a quotient of polynomial factorizations:
x 3 + 2x 2 + x + 2
x4 − 1
SOLUTION: The numerator has 4 terms, so we will attempt grouping; the denominator
is a difference of squares. We factor polynomials completely before rewriting the rational
expression.
x 3 + 2x 2 + x + 2 = x 2 (x + 2) + 1(x + 2)
= (x 2 + 1)(x + 2)
x 2 + 1 is an irreducible quadratic, and x + 2 is linear, so the numerator is fully factored.
2
x 4 − 1 = (x 2 + 1)(x 2 − 1) difference of squares x 2 − 12
= (x 2 + 1)(x + 1)(x − 1) difference of squares x 2 − 12
x 2 + 1 is an irreducible quadratic, and the other two factors are linear, so the denominator
is fully factored. Now we can rewrite the rational expression in a fully factored form and
cancel common polynomial factors:
x 3 + 2x 2 + x + 2 (x 2 + 1)(x + 2)
=
x4 − 1 (x 2 + 1)(x + 1)(x − 1)
(x + 2)
=
(x + 1)(x − 1)
x +2
ANSWER:
(x + 1)(x − 1)
367
1. SIMPLIFYING RATIONAL EXPRESSIONS CHAPTER 7. RATIONAL EXPRESSIONS
Homework 7.1.
Find all variable values which make the 21k
14.
given expression undefined. 24k3
3k2 + 30k 32x 3 y
1. 15.
k + 10 8x y 2
15n2 90x 2
2. 16.
10n + 25 20x 2 y
27p 18m − 24
3. 17.
18p2 − 36p 60
x + 10 20
4. 18.
8x 2 + 80x 4p + 2
10m2 + 8m x +1
5. 19.
10m x 2 + 8x + 7
10x + 16 32x 2
6. 20.
6x + 20 28x 2 + 28
b2 + 12b + 32 n2 + 4n − 12
7. 21.
b2 − 4b − 32 n2 − 7n + 10
10 y 2 + 30 y b2 + 14b + 48
8. 22.
35 y 2 − 5 y b2 + 15b + 56
w2 − 1 9v + 54
9. 23.
w2 + 1 v2 − 4v − 60
16x 2 + 25 30x − 90
10. 24.
16x 2 − 25 50x + 40
12x 2 − 42x
25.
30x 2 − 42x
Simplify the given expression and state the
answer as a quotient of polynomial factor- 6a − 10a2
izations. 26.
10a + 4a2
21x 2 k2 − 12k + 32
11. 27.
18x k2 − 64
12n
12. 9p + 18
4n2 28.
p2 + 4p + 4
24a
13.
40a2
368
CHAPTER 7. RATIONAL EXPRESSIONS 1. SIMPLIFYING RATIONAL EXPRESSIONS
9n2 + 89n − 10 n2 − 2n + 1
29. 33.
9n + 90 6n + 6
3x 2 − 29x + 40 7a2 − 26a − 45
30. 34.
5x 2 − 30x − 80 6a2 − 34a + 20
2x 2 − 10x + 8 56x − 48
31. 35.
3x 2 − 7x + 4 24x 2 + 56x + 32
7n2 − 32n + 16 4k3 − 2k2 − 2k
32. 36.
4n − 16 9k3 − 18k2 + 9k
369
1. SIMPLIFYING RATIONAL EXPRESSIONS CHAPTER 7. RATIONAL EXPRESSIONS
Homework 7.1 Answers.
1. {−10} n+6
21.
n−5
3. {0, 2}
9
23.
5. {0} v − 10
7. {−4, 8} 2x − 7
25.
5x − 7
9. ∅
k−4
7x 27.
11. k+8
6
9n − 1
3 29.
13. 9
5a
2(x − 4)
4x 2 31.
15. 3x − 4
y
(n − 1)2
3m − 4 33.
17. 6(n + 1)
10
1 7x − 6
19. 35.
x +7 (3x + 4)(x + 1)
370
CHAPTER 7. RATIONAL EXPRESSIONS 2. PRODUCTS OF RATIONAL EXPRESSIONS
2. Products of Rational Expressions
2.1. Multiplying Rational Expressions. We can multiply rational expressions just like we
would multiply fractions: the numerator of the product is the product of numerators, and
the denominator of the product is the product of denominators. Rather than to carry out the
multiplication of polynomials, we will factor them, so that we can cancel common polynomial
factors and leave the answer in a simplified form.
EXAMPLE 2.1.1. Simplify and state the answer as a quotient of polynomial factorizations:
25x 2 24 y 4
·
9 y 8 55x 7
SOLUTION: We begin by multiplying across and then we cancel common integer and
variable factors.
25x 2 24 y 4 25 · 24 · x 2 y 4
· =
9 y 8 55x 7 9 · 55 · y 8 x 7
(5 · 5)(2 · 2 · 2 · 3)
= cancel variables, factor integers
(3 · 3)(5 · 11) · x 5 y 4
5·2·2·2
= canceled common prime factors
3 · 11 · x 5 y 4
40
=
33x 5 y 4
40
ANSWER:
33x 5 y 4
EXAMPLE 2.1.2. Simplify and state the answer as a quotient of polynomial factorizations:
x2 − 4 x +1
· 2
x + 3 x + 3x + 2
SOLUTION: We can multiply across, then factor all the polynomials, and cancel the com-
mon polynomial factors. Note that the invisible parentheses of the fraction notation be-
come visible once we rewrite the expression as a single fraction.
x2 − 4 x +1 (x 2 − 4)(x + 1)
· 2 =
x + 3 x + 3x + 2 (x + 3)(x 2 + 3x + 2)
371
2. PRODUCTS OF RATIONAL EXPRESSIONS CHAPTER 7. RATIONAL EXPRESSIONS
The difference of squares x 2 − 4 factors as
x 2 − 4 = (x + 2)(x − 2)
while the trinomial x 2 + 3x + 2 factors as
x 2 + 3x + 2 = (x + 1)(x + 2)
Now we can rewrite the original fraction in a fully factored form, and cancel common
polynomial factors.
(x 2 − 4)(x + 1) (x + 2)(x − 2)(x + 1)
=
(x + 3)(x 2 + 3x + 2) (x + 3)(x + 1)(x + 2)
(x − 2)
= common polynomial factors canceled
(x + 3)
x −2
ANSWER:
x +3
2.2. Dividing Rational Expressions.
EXAMPLE 2.2.1. Simplify and state the answer as a quotient of polynomial factorizations:
14 21
÷
25 10
SOLUTION: Recall that division amounts to multiplying by the reciprocal of the divisor:
14 21 14 10 14 · 10
÷ = · =
25 10 25 21 25 · 21
Instead of finding these large products, we opt to factor the integers completely so that it
is easier to cancel all the common factors.
14 · 10 (2 · 7)(2 · 5)
=
25 · 21 (5 · 5)(3 · 7)
2·2
=
5·3
4
=
15
ANSWER: 4/15
372
CHAPTER 7. RATIONAL EXPRESSIONS 2. PRODUCTS OF RATIONAL EXPRESSIONS
EXAMPLE 2.2.2. Simplify and state the answer as a polynomial or a quotient of polynomial
factorizations:
x 2 + 8x + 16 x + 4
÷
x x2
SOLUTION: First we rewrite the division as a multiplication by the reciprocal, and then
multiply fractions as usual:
x 2 + 8x + 16 x + 4 x 2 + 8x + 16 x2
÷ = · when we multiply sums
x x2 x x +4
(x 2 + 8x + 16)x 2
= fraction parentheses become visible
x(x + 4)
Now we can factor the polynomials completely and cancel all common factors. x 2 +8x +16
is a square of a binomial, and factors as (x + 4)2 , so we can write
(x 2 + 8x + 16)x 2 (x + 4)2 x 2
=
x(x + 4) x(x + 4)
= (x + 4)x common factors x and (x + 4) cancel
For the sake of consistency, we will state polynomial answers in the standard form.
ANSWER: x 2 + 4x
EXAMPLE 2.2.3. Simplify and state the answer as a polynomial or a quotient of polynomial
factorizations:
3x 2 − 75 5 − x
÷
5+ x 12
SOLUTION: First we rewrite the division as a multiplication by the reciprocal, and then
multiply fractions as usual:
3x 2 − 75 5 − x 3x 2 − 75 12
÷ = ·
5+ x 12 5+ x 5− x
(3x 2 − 75)12
=
(5 + x)(5 − x)
3x 2 − 75 has GCF 3 and then factors as a difference of squares:
3x 2 − 75 = 3(x 2 − 25)
= 3(x 2 − 52 )
= 3(x + 5)(x − 5)
373
2. PRODUCTS OF RATIONAL EXPRESSIONS CHAPTER 7. RATIONAL EXPRESSIONS
So in a fully factored form, the rational expression looks like this:
(3x 2 − 75)12 3(x + 5)(x − 5)12
=
(5 + x)(5 − x) (5 + x)(5 − x)
36(x − 5)
= common factor (x + 5) canceled
5− x
Note that we canceled x + 5 with 5 + x because they are equivalent, but we can not cancel
x −5 with 5− x as easily because they are opposites, and dividing a number by its opposite
results in −1.
36(x − 5) 36(x − 5)
=
5− x (−1)(−5 + x)
36(x − 5)
=
(−1)(x − 5)
36
= common factor (x − 5) canceled
−1
= −36
ANSWER: −36
374
CHAPTER 7. RATIONAL EXPRESSIONS 2. PRODUCTS OF RATIONAL EXPRESSIONS
Homework 7.2.
Simplify and state the answer as a polyno- 8k 1
15. ÷
mial or a quotient of polynomial factoriza- 24k2 − 40k 15k − 25
tions.
p−8 1
2 16. ÷
8x 9 p2 − 12p + 32 p − 10
1. ·
9 2
6
20n 7 17. (n − 8) ·
2. · 10n − 80
3n 5n
3x − 6
5x 2 5 18. · (x + 3)
3. ÷ 12x − 24
4 6
x 2 − 7x + 10 x 2 − x − 20
10p 8 19. ÷
4. ÷ x −2 x + 10
5 10
21v 2 + 16v − 16 35v − 20
7(m − 6) 5m(7m − 5) 20. ÷
5. · 3v + 4 v−9
m−6 7(7m − 5)
b+2
7r (r − 6)
2 21. · (5b − 3)
6. · 40b2 − 24b
7r(r + 10) r −6
2n2 − 12n − 54 1
7 n−2 22. ·
7. ÷ n+7 2n + 6
10(n + 3) (n + 3)(n − 2)
x 2 + 11x + 24 x 2 + 5x − 24
23. ÷
6x(x + 4) 6x(x − 6) 6x 3 + 18x 2 6x 3 + 6x 2
8. ÷
x −3 (x − 3)(x − 6)
n−7 n2 − 13n + 42
24. ÷
25n + 25 4 6n − 12 12 − 6n
9. ·
5 30n + 30
k−7 7k2 − 28k
v−1 4 25. ·
10. · 2 k2 − k − 12 8k2 − 56k
4 v − 11v + 10
x 2 − 12x + 32 7x 2 + 14x
9 b−5 26. ·
11. ÷ 2 x 2 − 6x − 16 7x 2 + 21x
b2 − b − 12 b − b − 12
n−7 9n + 54
x − 10 7 27. ÷
12. ÷ n2 − 2n − 35 10n + 50
35x + 21 35x + 21
27a + 36 6a + 8
x 2 − 6x − 7 x + 5 28. ÷
13. · 9a + 63 2
x +5 x −7
x2 − 1 x2 − 4 x2 + x − 2
1 8a + 80 29. · 2 ÷
14. · 2x − 4 x − x − 2 3x − 6
a−6 8
x 2 + 3x − 10 8x + 20 2x 2 − x − 3
30. ÷ ·
x 2 + 6x + 5 6x + 15 2x 2 + x − 6
375
2. PRODUCTS OF RATIONAL EXPRESSIONS CHAPTER 7. RATIONAL EXPRESSIONS
Homework 7.2 Answers.
1. 4x 2 3
17.
5
3x 2
3. x + 10
2 19.
x +4
5. 5m
b+2
21.
7 8
7.
10 x +1
23.
2 x −3
9.
3 7
25.
8(k + 3)
9
11.
b−5 10
27.
9(n + 6)
13. x + 1
3
15. 5 29.
2
376
CHAPTER 7. RATIONAL EXPRESSIONS 3. LCD FOR RATIONAL EXPRESSIONS
3. LCD for Rational Expressions
3.1. LCM for Polynomials.
DEFINITION 3.1.1. A common multiple for a collection of polynomials P, Q, R, . . . with
integer coefficients is a polynomial which is divisible by P, Q, R, . . . without a remainder,
using the polynomial long division.
The least common multiple (or LCM) for a collection of polynomials with integer coeffi-
cients is a common multiple of the least possible polynomial degree and with coefficients
as low as possible.
The lowest common denominator (or LCD) for a collection of rational expressions is the
LCM of their denominators.
THEOREM 3.1.1. To find the least common multiple for a collection of polynomials with
integer coefficients, we can factor each polynomial completely, and then take the product
with each prime factor and each irreducible polynomial factor to the highest degree we
found. Note that the integer factor of the polynomial LCM is the LCM for the collection
of individual integer factors.
EXAMPLE 3.1.1. Find the LCM for given polynomial expressions and state it in a fully
factored form.
x 4 , x, x 7
SOLUTION: These polynomials are already fully factored. The only factor here is x, and
its highest degree is 7, so LCM is x 7 .
ANSWER:
EXAMPLE 3.1.2. Find the LCM for given polynomial expressions and state it in a fully
factored form.
4m2 , 6mn, 2n2
SOLUTION: These polynomials are already fully factored. The LCM for integer factors 4,
6, and 2 is 12, the highest degree of m is 2, and the highest degree of n is also 2. So the
LCM is 12m2 n2 .
ANSWER: 12m2 n2
377
3. LCD FOR RATIONAL EXPRESSIONS CHAPTER 7. RATIONAL EXPRESSIONS
EXAMPLE 3.1.3. Find the LCM for given polynomial expressions and state it in a fully
factored form.
25x 2 , 5x 2 − 5, 10x 2 − 20x + 10
SOLUTION: We factor the polynomials first:
25x 2 already factored
5x 2 − 5 = 5(x 2 − 1) GCF
= 5(x + 1)(x − 1) difference of squares
10x 2 − 20x + 10 = 10(x 2 − 2x + 1) GCF
= 10(x − 1)2 special product pattern
The LCM for the integer factors 5, 25, and 10 is 50. The highest degree of (x − 1) is 2,
the highest degree of x is 2, and the highest degree of (x + 1) is 1. Hence the polynomial
LC M is 50x 2 (x + 1)(x − 1)2 .
ANSWER: 50x 2 (x + 1)(x − 1)2
3.2. LCD for Rational Expressions.
EXAMPLE 3.2.1. Find the LCD for rational expressions
1 1
,
2x x3
SOLUTION: These denominators are already fully factored. The highest degree of the
factor 2 is 1, and the highest degree of the factor x is 3, so LCD = 21 · x 3 .
ANSWER: 2x 3
EXAMPLE 3.2.2. Find the LCD for rational expressions
1 1 1
, ,
25x 2 y 20x 3 y 2 2x w7
SOLUTION: In the examples above we found the integer factor for the polynomial LCM
by taking the LCM of individual integer factors. In more complicated situations we may
have to find integer factorizations in order to compute the LCM.
378
CHAPTER 7. RATIONAL EXPRESSIONS 3. LCD FOR RATIONAL EXPRESSIONS
We factor the denominators first:
25x 2 y = 52 · x 2 y
20x 3 y 2 = 22 · 5 · x 3 y 2
2x w7 already factored
The highest degree of 2 is 2, the highest degree of 5 is 2, the highest degree of x is 3, the
highest degree of y is 2, and the highest degree of w is 7, so LCD = 22 · 52 · x 3 y 2 w7
ANSWER: 100x 3 y 2 w7
EXAMPLE 3.2.3. Find the LCD for rational expressions
x −5 7
, 2
4x − 4 6x − 6
SOLUTION: Recall that for the purpose of finding the LCD, we are free to ignore the
numerators completely. We factor the denominators first:
4x − 4 = 22 (x − 1)
6x 2 − 6 = 2 · 3(x 2 − 1)
= 6(x + 1)(x − 1)
The highest degree of 2 is 2, the highest degree of 3 is 1, the highest degree of (x − 1) is
1, and the highest degree of (x + 1) is 1, so LCD = 22 · 3 · (x + 1)(x − 1).
ANSWER: 12(x + 1)(x − 1)
EXAMPLE 3.2.4. Find the LCD for rational expressions
x4 19x
,
x2 − x − 6 x 2 y − 6x y + 9 y
SOLUTION: Recall that for the purpose of finding the LCD, we are free to ignore the
numerators completely. We factor the denominators first:
x2 − x − 6 = (x − 3)(x + 2)
x 2 y − 6x y + 9 y = y(x 2 − 6x + 9) GCF
= y(x − 3)(x − 3) special product pattern
= y(x − 3)2
379
3. LCD FOR RATIONAL EXPRESSIONS CHAPTER 7. RATIONAL EXPRESSIONS
The highest degree of y is 1, the highest degree of (x − 3) is 2, and the highest degree of
(x + 2) is 1, so LCD = y(x − 3)2 (x + 2)
ANSWER: y(x − 3)2 (x + 2)
EXAMPLE 3.2.5. Find the LCD for rational expressions
1 1
,
x − 6x + 9
2 3x − x 2
SOLUTION: We factor the denominators first:
x 2 − 6x + 9 = (x − 3)2
3x − x 2 = x(3 − x)
The highest degree of x is 1, and the highest degree of (x −3) is 2. (3− x) is not equivalent
to (3 − x), but we can make it equivalent by factoring out −1. In the interest of keeping
the overall degree of the LCD as low as possible, we treat the factor (3 − x) as if it was
(x − 3) with degree 1. Hence the LCD is x(x − 3)2 .
ANSWER: x(x − 3)2
380
CHAPTER 7. RATIONAL EXPRESSIONS 3. LCD FOR RATIONAL EXPRESSIONS
Homework 7.3.
Find the least common multiple for each Find the least common denominator for
collection of polynomial expressions and each collection of rational expressions and
state it in a fully factored form: state it in a fully factored form:
1. 2a3 , 6a4 b2 , 4a3 b5 1 1 1
11. , ,
a+1 (a − 1)2 a2 −1
2. 5x 2 y, 25x 3 y 5 z
1 1 1
12. , ,
3. x 2 − 3x, x − 3, x x −2 (x + 2)2 x2 −4
4. 4x − 8, x − 2, 4 2n − 1 n+1
13. ,
2n2 + n − 1 2n2 + 3n − 2
5. x + 2, x −4
4 m
14. ,
6. x, x − 7, x +1 m2 − 2m − 3 2m2 + 3m + 1
7. x 2 − 25, x +5 1 2 3
15. , ,
t −3 t +3 t2 − 9
8. x 2 − 9, x 2 − 6x + 9
5 10
16. ,
9. x + 3x + 2,
2
x + 5x + 6
2
a−5 a2 − 10a + 25
10. x 2 −7x +10, x 2 −2x −15, x 2 + x −6 5x 2x
17. ,
x2 − 9 x 2 + 11x + 24
2x 4x
18. ,
x2−4 x2 + 5x + 6
x +3 x −1
19. ,
6x 3 − 24x 2 + 18x 4x 5 − 24x 4 + 20x 3
1 1
20. ,
9 y3 − 9 y 2 − 18 y 6 y5 − 24 y 4 + 24 y 3
381
3. LCD FOR RATIONAL EXPRESSIONS CHAPTER 7. RATIONAL EXPRESSIONS
Homework 7.3 Answers.
1. 12a4 b5 11. (a + 1)(a − 1)2
3. x(x − 3) 13. (2n − 1)(n + 1)(n + 2)
5. (x + 2)(x − 4) 15. (t + 3)(t − 3)
7. (x + 5)(x − 5) 17. (x + 3)(x − 3)(x + 8)
9. (x + 1)(x + 2)(x + 3) 19. 12x 3 (x − 1)(x − 3)(x − 5)
382
CHAPTER 7. RATIONAL EXPRESSIONS 4. SUMS OF RATIONAL EXPRESSIONS
4. Sums of Rational Expressions
4.1. Sums with a Common Denominator. When simplifying a sum with a common de-
nominator, we take the sum of numerators over the common denominator, and then simplify
the resulting rational expression.
EXAMPLE 4.1.1. Simplify and state the answer as a quotient of polynomial factorizations:
4x 4
+
x2 − 1 x2 − 1
SOLUTION: The denominators are the same, so we can add the numerators over the
common denominator, then factor the numerator, and cancel common polynomial factors,
if any.
4x 4 4x + 4
2
+ 2 =
x −1 x −1 x2 − 1
4(x + 1)
=
(x + 1)(x − 1)
4
=
x −1
4
ANSWER:
x −1
EXAMPLE 4.1.2. Simplify and state the answer as a quotient of polynomial factorizations:
4x − 1 2x − 9
−
x +4 x +4
SOLUTION: The denominators are the same, so we can subtract the numerators over the
common denominator, combine like terms, factor the numerator, and cancel any common
polynomial factors we find. Notice that we are subtructing a sum, so we have to make the
fraction parentheses visible.
4x − 1 2x − 9 4x − 1 − (2x − 9)
− = subtract the entire numerator
x +4 x +4 x +4
4x − 1 − 2x + 9
= add the sum of opposites
x +4
2x + 8
= combine like terms
x +4
383
4. SUMS OF RATIONAL EXPRESSIONS CHAPTER 7. RATIONAL EXPRESSIONS
Now we will factor polynomials completely and cancel identical polynomial factors:
2x + 8 2(x + 4)
=
x +4 x +4
= 2
ANSWER: 2
4.2. General Sums. The most economical way to add rational expressions with different
denominators is by rewriting them all with the LCD.
EXAMPLE 4.2.1. Simplify and state the answer as a quotient of polynomial factorizations:
3 4
−
10x 15x 2
SOLUTION: LCD = 30x 2 , so before we can subtract, we need to modify each fraction to
have this denominator.
3 4 3 3x 4 2
− = · − ·
10x 15x 2 10x 3x 15x 2 2
9x 8
= 2
−
30x 30x 2
Now that the denominators are the same, we can subtract the numerators:
9x 8 9x − 8
2
− 2
=
30x 30x 30x 2
Both the numerator and the denominator are fully factored, so we are done.
9x − 8
ANSWER:
30x 2
384
CHAPTER 7. RATIONAL EXPRESSIONS 4. SUMS OF RATIONAL EXPRESSIONS
EXAMPLE 4.2.2. Simplify and state the answer as a quotient of polynomial factorizations:
x 1
+
x −1 x
SOLUTION: LCD = x(x − 1), so before we can add, we need to modify each fraction to
have this denominator.
x 1 x x 1 x −1
+ = · + · when multiplying sums
x −1 x x −1 x x x −1
x2 x −1
= + show fraction parentheses
x(x − 1) x(x − 1)
Now that the denominators are the same, we can add the numerators:
x2 x −1 x 2 + (x − 1)
+ =
x(x − 1) x(x − 1) x(x − 1)
x2 + x − 1
=
x(x − 1)
The polynomial in the numerator is prime, so everything is fully factored, and we are done.
x2 + x − 1
ANSWER:
x(x − 1)
EXAMPLE 4.2.3. Simplify and state the answer as a quotient of polynomial factorizations:
2 x
−
x 2 + 3x + 2 x 2 + 5x + 6
SOLUTION: To find the LCD, we factor the denominators:
x 2 + 3x + 2 = (x + 1)(x + 2)
x 2 + 5x + 6 = (x + 2)(x + 3)
So LCD = (x + 1)(x + 2)(x + 3), and before we can subtract fractions, we need to rewrite
them with this denominator. To do so, we need to multiply the first fraction by
x +3
x +3
and the second fraction by
x +1
x +1
385
4. SUMS OF RATIONAL EXPRESSIONS CHAPTER 7. RATIONAL EXPRESSIONS
The result looks like this:
2 x 2 x
− 2 = −
x + 3x + 2 x + 5x + 6
2 (x + 1)(x + 2) (x + 2)(x + 3)
2 x +3 x x +1
= · − ·
(x + 1)(x + 2) x + 3 (x + 2)(x + 3) x + 1
2(x + 3) x(x + 1)
= −
(x + 1)(x + 2)(x + 3) (x + 2)(x + 3)(x + 1)
Now that the denominators are equivalent, we can subtract the numerators:
2(x + 3) − x(x + 1)
=
(x + 1)(x + 2)(x + 3)
Finally, we will simplify the numerator by combining like terms, factoring it, and then
canceling common polynomial factors, if any:
2x + 6 − x 2 − x
=
(x + 1)(x + 2)(x + 3)
−x 2 + x + 6
=
(x + 1)(x + 2)(x + 3)
If we factor out −1 in the numerator, then we can detect an easy trinomial pattern and
factor it as a product of two binomials:
−(x 2 − x − 6)
=
(x + 1)(x + 2)(x + 3)
−(x − 3)(x + 2)
=
(x + 1)(x + 2)(x + 3)
−(x − 3)
=
(x + 1)(x + 3)
−x + 3
ANSWER:
(x + 1)(x + 3)
386
CHAPTER 7. RATIONAL EXPRESSIONS 4. SUMS OF RATIONAL EXPRESSIONS
EXAMPLE 4.2.4. Simplify and state the answer as a quotient of polynomial factorizations:
4a 7a
+
a−2 2−a
SOLUTION: Since a − 2 and 2 − a are opposites of each other, we treat them as the same
polynomial factor, and the LCD is (a − 2). Before we can add, though, we need to factor
out −1 in one of the denominators, so that they become equivalent.
4a 7a 4a 7a
+ = +
a−2 2−a a − 2 −1(a − 2)
Because −1 is its own reciprocal, we can move the factor −1 into the numerator, making
the denominators the same, and allowing us to perform the addition:
4a 7a 4a −1(7a)
+ = +
a − 2 −1(a − 2) a−2 a−2
4a + (−7a)
=
a−2
−3a
=
a−2
Everything is fully factored, so we are done.
−3a
ANSWER:
a−2
387
4. SUMS OF RATIONAL EXPRESSIONS CHAPTER 7. RATIONAL EXPRESSIONS
Homework 7.4.
Perform the summation, simplify, and state 2 2
15. +
the answer as a quotient of polynomial fac- x −1 x +1
torizations.
2z 3z
2 4 16. −
1. + z−1 z+1
a+3 a+3
2 3
2 17. +
x 6x − 8 x − 5 4x
2. −
x −2 x −2
8x 3
18. −
t + 4t 2t − 7
2
x2−4 x +2
3. +
t −1 t −1 4x x
19. +
a2 + 3a 4 x 2 − 25 x + 5
4. − 2
a + 5a − 6 a + 5a − 6
2
t 5
20. −
2x 2 + 3 x 2 − 5x + 9 t − 3 4t − 12
5. −
x 2 − 6x + 5 x 2 − 6x + 5 2 4
21. +
3 4 x + 3 (x + 3)2
6. + 2
x x 2 4
22. −
5 5 5x 2 + 5x 3x + 3
7. −
6r 8r 3a 9a
23. +
6 3 4a − 20 6a − 30
8. 2
+ 2
xy x y x x
24. +
x −5 5− x
8 5
9. 3
+ 2 t y
9t 6t 25. −
y−t y+t
x +5 x −3
10. +
8 12 x x −5
26. +
x −5 x
a+2 a−4
11. −
2 4 2x 4
27. − 2
x2− 1 x + 2x − 3
2a − 1 5a + 1
12. +
3a2 9a x −1 x +5
28. + 2
x2 + 3x + 2 x + 4x + 3
x − 1 2x + 3
13. −
4x x x +1 x +6
29. + 2
x2 − 2x − 35 x + 7x + 10
a+2 a−4
14. −
2 4 3x + 2 x
30. +
3x + 6 4 − x 2
388
CHAPTER 7. RATIONAL EXPRESSIONS 4. SUMS OF RATIONAL EXPRESSIONS
4 − a2 a − 2 2x 3
31. 2 − 34. − 2
a −9 3−a x2− 1 x + 5x + 4
4y 2 2 x 7
32. − − 35. −
2
y −1 y y +1 x 2 + 15x + 56 x 2 + 13x + 42
x 2 2x 5
33. − 2 36. + 2
x + 5x + 6 x + 3x + 2
2 x2−9 x + x −6
389
4. SUMS OF RATIONAL EXPRESSIONS CHAPTER 7. RATIONAL EXPRESSIONS
Homework 7.4 Answers.
6 2(x + 5)
1. 21.
a+3 (x + 3)2
3. t + 7
9a
x +6 23.
5. 4(a − 5)
x −5
5 t 2 + 2t y − y 2
7. 25.
24r ( y + t)( y − t)
15t + 16
9. 2(x + 2)
18t 3 27.
(x + 1)(x + 3)
a+8
11.
4 2(x − 4)
29.
−7x − 13 (x − 7)(x + 2)
13.
4x
a−2
4x 31.
15. (a + 3)(a − 3)
x2−1
11x − 15 x −3
17. 33.
4x(x − 5) (x + 3)(x + 1)
x(x − 1) x −8
19. 35.
(x + 5)(x − 5) (x + 8)(x + 6)
390
CHAPTER 7. RATIONAL EXPRESSIONS 5. COMPLEX FRACTIONS
5. Complex Fractions
5.1. Definition.
DEFINITION 5.1.1. Within this text, complex fractions are all the expressions which can be
built up by combining polynomials with arithmetic operations +, −, ·, and ÷
Many other sources use a narrow definition which only admits fractions of fractions
and/or mixed numbers.
The reader should also be careful not to confuse complex fractions with fractions which
involve complex numbers, which we briefly mention later in the text.
BASIC EXAMPLE 5.1.1. Here are some traditional complex fractions:
1 x 1 34
÷ ,
2 5 6 73
And here are even more complex fractions which are built up as quotients of sums of
rational expressions:
3 1 x y
− +
x y 2 y z 1
, , ,
1 ab 1 b 1+ 1
−1 − 1
x a z 1+
2
5.2. Simplifying Complex Fractions.
THEOREM 5.2.1. For every complex fraction we can find a rational expression which is
almost equivalent. That is, the value of the complex fraction will be equal to the value of
a corresponding rational expression whenever both values are defined.
PROOF. By induction of arithmetic expressions. The most basic type of complex fraction
is a polynomial, which is a rational expression. A sum, a product, or a quotient of rational
expressions can be rewritten as a rational expression in simplified form, and they will have the
same values, except at finitely many points where one of them is undefined.
We can simplify every complex fraction by rewriting every sum as a rational expression, and
then simplifying products and quotients of rational expressions by canceling common polyno-
mial factors. These steps may have to be repeated several times. For answers, we will factor
the rational expressions and state them as quotients of polynomial factorizations.
391
5. COMPLEX FRACTIONS CHAPTER 7. RATIONAL EXPRESSIONS
EXAMPLE 5.2.1. Simplify and state the answer as a quotient of polynomial factorizations:
x
1
1+
x
SOLUTION: We start by rewriting the sum in the denominator as a rational expression. In
order to do that, we have to find the LCD for 1 and 1/x and then add fraction as usual.
x x 1
= the LCD for 1 and is x
1 x 1 x
1+ +
x x x
x
=
x +1
x
x +1
Now we can divide x by as usual:
x
x x +1
= x÷ division is multiplication by the reciprocal
x +1
x
x
x
= x·
x +1
x2
=
x +1
x2
ANSWER:
x +1
EXAMPLE 5.2.2. Simplify and state the answer as a quotient of polynomial factorizations:
1 x
+
2 4
1 x
−
4 2
SOLUTION: To avoid rewriting a giant fraction, we will simplify this expression in three
stages. First we will simplify the sums, and then divide the resulting rational expressions.
392
CHAPTER 7. RATIONAL EXPRESSIONS 5. COMPLEX FRACTIONS
Simplify the numerator:
1 x 2 x
+ = + the LCD is 4
2 4 4 4
2+ x
=
4
Simplify the denominator:
1 x 1 2x
− = − the LCD is 4
4 2 4 4
1 − 2x
=
4
Divide numerator by denominator:
2 + x 1 − 2x 2+ x 4
÷ = · multiply by reciprocal of divisor
4 4 4 1 − 2x
(2 + x)4
= show fraction parentheses
4(1 − 2x)
2+ x
= canceled common factor 4
1 − 2x
2+ x
ANSWER:
1 − 2x
EXAMPLE 5.2.3. Simplify and state the answer as a quotient of polynomial factorizations:
2
1+
4
3+
x
4
SOLUTION: We start by simplifying the sum of rational expressions 3 +
x
4 3x 4
3+ = + the LCD is x
x x x
3x + 4
=
x
393
5. COMPLEX FRACTIONS CHAPTER 7. RATIONAL EXPRESSIONS
3x + 4
Now we can compute the quotient of rational expressions 2 and in the usual way:
x
3x + 4 x 2x
2÷ =2· =
x 3x + 4 3x + 4
2x
Finally, we simplify the sum of rational expressions 1 + in the usual way:
3x + 4
2x 3x + 4 2x
1+ = + the LCD is 3x + 4
3x + 4 3x + 4 3x + 4
3x + 4 + (2x)
=
3x + 4
5x + 4
=
3x + 4
5x + 4
ANSWER:
3x + 4
EXAMPLE 5.2.4. Simplify and state the answer as a quotient of polynomial factorizations:
x −1 + x −2
x −3 − x −2
SOLUTION: We can convert negative exponents into rational expressions and then simplify
the resulting complex fraction just as before:
1 1
+ 2
x +x
−1 −2
x x
=
−3
x −x −2 1 1
3
− 2
x x
Simplify the numerator:
1 1 x 1
+ 2 = + the LCD is x 2
x x x2 x2
x +1
=
x2
Simplify the denominator:
1 1 1 x
3
− 2 = 3
− 3 the LCD is x 3
x x x x
1− x
=
x3
394
CHAPTER 7. RATIONAL EXPRESSIONS 5. COMPLEX FRACTIONS
Divide numerator by denominator:
x +1 1− x x +1 x3
÷ = ·
x2 x3 x2 1− x
(x + 1)x 3 x3
= =x
x 2 (1 − x) x2
(x + 1)x
=
1− x
x(x + 1)
ANSWER:
1− x
395
5. COMPLEX FRACTIONS CHAPTER 7. RATIONAL EXPRESSIONS
Homework 7.5.
Simplify and state the answer as a quotient of −5
−3
polynomial factorizations: b−5
10.
10
1 +6
1+ b−5
x
1. x 1
1 −
1− 2 x +1 x
x 11.
x 1
1 +
−1 x +1 x
y2
2. 2a 3
1 −
1+ a−1 a
y 12.
−6
−4
a−2 a−1
3.
4
−a 3
a
x
13.
25 9
−a
4.
a x2
5+a x
1 1 3x − 2
− 14.
a2 a x
5.
1 1 2
9x − 4
2
+
a a
1 1
1 1 −
+ a−h a
15.
b 2 h
6.
4
2
b −1 1 1
−
x +h x
16.
4 h
2−
x +2
7. x −1 + y −1
10 17. 2
5−
x − y2
x +2
xy
12
4+
2x − 3 x − x −1
8.
15 18.
5+ x + x −1
2x − 3
2
3 x −1+
+2 x −4
2a − 3 19.
9. 6
−6 x +3+
−4 x −4
2a − 3
396
CHAPTER 7. RATIONAL EXPRESSIONS 5. COMPLEX FRACTIONS
18 x −1 x +1
x −5− −
x +2 x +1 x −1
20. 22.
6 x −1 x +1
x +7+ +
x +2 x +1 x −1
2 5
−
b b+3
21.
3 3
+
b b+3
397
5. COMPLEX FRACTIONS CHAPTER 7. RATIONAL EXPRESSIONS
Homework 7.5 Answers.
x x
1. 13.
x −1 3
−a
3. 1
a+2 15.
a(a − h)
−a + 1
5.
a+1 1
17.
2 x−y
7.
5
x −2
1 19.
9. − x +2
2
x2 − x − 1 −b + 2
11. 21.
x2 + x + 1 2b + 3
398
CHAPTER 7. RATIONAL EXPRESSIONS 6. EQUATIONS WITH RATIONAL EXPRESSIONS
6. Equations with Rational Expressions
6.1. Extraneous Solutions.
DEFINITION 6.1.1 (Extraneous Solution). Some techniques for solving equations will yield
numbers which are not solutions. This is a reasonable compromise as long as these tech-
niques never miss a legitimate solution. In this case, we can solve an equation by making a
list of all possible solutions, and then checking each one. The ones that make the equation
true are the proper solutions, while the rest are called extraneous solutions. It is important
to understand that strictly speaking they are not solutions at all.
When solving equations involving rational expressions, we will often simplify them by cancel-
ing polynomial factors with their reciprocals, for example
(x + 1)(x − 2) x +1
=
x(x − 2) x
Recall that these expressions are not exactly equivalent, since the one on the left is not defined
for x = 2. In fact, if x = 2, then the reciprocal of x − 2 is not defined, and nothing can
be canceled. So when we do this within an equation, we make a tacit assumption that the
polynomial factor x − 2 6= 0, which is equivalent to x 6= 2. In other words, 2 can not be a
solution.
THEOREM 6.1.1. An equation involving rational expressions can be solved by multiplying
both sides by the polynomial LCD, canceling common polynomial factors, and then solving
the resulting polynomial equation. Some or all solutions of the polynomial equation may
be extraneous. To find out which ones, it is sufficient to determine which values of the
variable make the expressions within the original equation undefined: if they come up as
solutions, they must be extraneous, while everything else will work.
A more generic (and somewhat more tedious) way to detect an extraneous solution is to
use it in the original equation and show that the equation fails to be satisfied.
399
6. EQUATIONS WITH RATIONAL EXPRESSIONS CHAPTER 7. RATIONAL EXPRESSIONS
EXAMPLE 6.1.1. Solve the equation
x x −2
x− =
x −1 x −1
SOLUTION: Looking at denominators, if x = 1 then x − 1 = 0, and expressions involving
denominator x − 1 will have undefined values, so 1 can not be a solution.
Our technique is to multiply both sides of the equation by the LCD, which is x − 1, and
then simplify fractions:
x x −2
x− = original equation in factored form
x −1 x −1
x −2
x
x− (x − 1) = (x − 1) both sides multiplied by LCD
x −1 x −1
As we distribute, we end up multiplying each term of the equation by LCD, which amounts
to multiplying the numerator of each term by LCD:
x(x − 1) (x − 2)(x − 1)
x(x − 1) − = distributivity
(x − 1) (x − 1)
x(x − 1) − x = x −2 common polynomial factors canceled
This looks like a quadratic equation, so we will solve it by factoring:
x2 − x − x = x −2 combine like terms
2
x − 2x = x −2 subtract x and add 2 on both sides
x − 3x + 2
2
= 0 factor
(x − 1)(x − 2) = 0
By the zero product property, solutions are 1 and 2. Solutions of this quadratic equation
are the possible solutions of the original equation involving rational expressions, but some
of them may be extraneous. We figured out early on that 1 makes expressions undefined,
so it is extraneous, and the only solution is 2.
Alternatively, we can check the answers in the usual way and discard the ones that do not
work. When we substitute 1 for x in the original equation, we end up dividing by zero,
which is not allowed:
(1) (1) − 2
(1) − =
(1) − 1 (1) − 1
ANSWER: {2}
400
CHAPTER 7. RATIONAL EXPRESSIONS 6. EQUATIONS WITH RATIONAL EXPRESSIONS
EXAMPLE 6.1.2. Solve the equation
1 1
= 2
2x 2 + 3x + 1 x −1
SOLUTION: Factor denominators and find the LCD:
2x 2 + 3x + 1 = (2x + 1)(x + 1)
2
x −1 = (x + 1)(x − 1)
Recall that LCD is the product of prime polynomial factors, each with the highest exponent
we found, and each of these factors has the highest degree 1, so
LCD = (x + 1)(x − 1)(2x + 1)
If either of these factors becomes zero, some expressions within the original equation
become undefined. We can make a list of forbidden solutions by solving the equation
(x + 1)(x − 1)(2x + 1) = 0
By the zero product property,
• If x + 1 = 0 then x = −1
• If x − 1 = 0 then x = 1
• If 2x + 1 = 0 then x = −1/2
So if any of these numbers come up as solutions, we will know they are extraneous. Now
we can multiply both sides of the equation by the LCD and get rid of fractions by canceling
common polynomial factors:
1 1
= original equation in factored form
(2x + 1)(x + 1) (x + 1)(x − 1)
Multiply both sides by LCD to get rid of fractions:
1 1
(x + 1)(x − 1)(2x + 1) = (x + 1)(x − 1)(2x + 1)
(2x + 1)(x + 1) (x + 1)(x − 1)
Note that denominators do not change, and what we do amounts to multiplying the nu-
merator of each term in the equation by LCD:
(x + 1)(x − 1)(2x + 1) (x + 1)(x − 1)(2x + 1)
= cancel common factors
(2x + 1)(x + 1) (x + 1)(x − 1)
x −1 = 2x + 1 this is a linear equation
−2 = x x is isolated on the right
401
6. EQUATIONS WITH RATIONAL EXPRESSIONS CHAPTER 7. RATIONAL EXPRESSIONS
The solution −2 of the linear equation is not among the forbidden solutions, so it should
work. Alternatively, it can be checked in the original equation involving rational expres-
sions:
1 1
=
2(−2) + 3(−2) + 1
2 (−2)2 − 1
1 1
=
2 · 4 + (−6) + 1 4−1
1 1
=
8−6+1 3
1 1
=
3 3
The equation holds true, so −2 is the only solution.
ANSWER: {−2}
EXAMPLE 6.1.3. Solve the equation
24
x+ = −9
x −2
SOLUTION: The LCD is x − 2, so the only forbidden solution is x = 2. We multiply both
sides by the LCD, distribute as needed, and then simplify resulting rational expressions:
24
x+ (x − 2) = (−9) (x − 2) both sides multiplied by LCD
x −2
24(x − 2)
x(x − 2) + = −9x + 18 distributivity
x −2
x 2 − 2x + 24 = −9x + 18 factor (x − 2) canceled
This is a quadratic equation, which we can solve by moving all the non-zero terms to one
side, factoring, and applying the zero product property:
x 2 − 2x + 24 + 9x − 18 = −9x + 18 + 9x − 18 added 9x − 18 to both sides
x 2 + 7x + 6 = 0 combined like terms
(x + 1)(x + 6) = 0
By the zero product property, solutions of this quadratic equation are −1 and −6. The
only forbidden solution is 2, so both −1 and −6 should work. Alternatively, we can check
whether they are solutions for the original equation.
402
CHAPTER 7. RATIONAL EXPRESSIONS 6. EQUATIONS WITH RATIONAL EXPRESSIONS
If x = −1 then the left side of the original equation evaluates to
24 24
x+ = (−1) +
x −2 (−1) − 2
24
= −1 +
−3
= −1 + (−8)
= −9 this is the right side, so the equation holds true
If x = −6 then the left side of the original equation evaluates to
24 24
x+ = (−6) +
x −2 (−6) − 2
24
= −6 +
−8
= −6 + (−3)
= −9 this is the right side, so the equation holds true
ANSWER: {−6, −1}
403
6. EQUATIONS WITH RATIONAL EXPRESSIONS CHAPTER 7. RATIONAL EXPRESSIONS
Homework 7.6.
Solve the equation. 4x 4 1
16. − =
2x − 6 5x − 15 2
x 6
1. − =0 −3 x
6 x 17. =
x +3 x +3
3 4 1
2. = − 3 x
x x 5 18. =
2x − 6 2x − 6
4
3. a + = −5 4− x 12
a 19. =
1− x 3− x
3
4. y + = −4 7 1 3
y 20. + =
3− x 2 4− x
y +3 6
5. = 7 1 y −2
y −3 y −3 21. − =
y −3 2 y −4
x 18
6. = 2 6
2 x 22. − =1
3− x 8− x
1 1
7. 3x − − =0 1 1 3x + 8
2 x 23. − = 2
x +2 2− x x −4
4
8. x + 1 = x +2 1 3x − 3
x +1 24. − =
3x − 1 x 3x 2 − x
20 5x
9. x + = −2 x +1 x −1 5
x −4 x −4 25. − =
x −1 x +1 6
x2 + 6 x − 2
10. + = 2x x −1 x +2 3
x −1 x −1 26. + =
x −3 x +3 4
6 2x
11. x + = 3 2x + 1
x −3 x −3 27. + =1
2x + 1 2x − 1
x −4 12
12. = +1 3x − 5 5x − 1 x − 4
x −1 3− x 28. + − =2
5x − 5 7x − 7 1 − x
2x 4x + 5 3
13. = − 3 5 5x
3x − 4 6x − 1 3x − 4 29. + = 2
x −3 x +2 x − x −6
6x + 5 2 3x
14. − = 2 2 1 x
2
2x − 2x 1 − x 2 x −1 30. + = 2
x −2 x +4 x + 2x − 8
3m 7 3
15. − = 5 30
2m − 5 3m + 1 2 31. − 2 =1
y −3 y −9
404
CHAPTER 7. RATIONAL EXPRESSIONS 6. EQUATIONS WITH RATIONAL EXPRESSIONS
1 1 1 4 2t 12
32. + = 2 34. + = 2
x +3 x −3 x −9 t −3 t −3 t − 6t + 9
5 3x 4
33. + = 2
x −2 x −2 x − 4x + 4
405
6. EQUATIONS WITH RATIONAL EXPRESSIONS CHAPTER 7. RATIONAL EXPRESSIONS
Homework 7.6 Answers.
1. {−6, 6} 19. {−5, 0}
16
§ ª
3. {−4, −1} 21. ,5
3
5. ∅
23. {−8}
1 2
§ ª
7. − , 1
§ ª
2 3 25. − , 5
5
9. {3}
1
§ ª
27.
11. {2} 10
29. ∅
13. {−1}
31. {2}
15. {−5}
7
§ ª
33. −2,
17. ∅ 3
406
CHAPTER 7. RATIONAL EXPRESSIONS 6. PRACTICE TEST 7
Practice Test 7
Find all variable values which make the 1 1
9. + 2
given expression undefined: 4x 2− 9 4x + 12x + 9
x +3
1. 2 2
x 2 + 5x + 6 −
x 3x
10.
y5 1 5
2. −
y 3 − 10 y 2 + 25 y x 6x
1
1+
Find the lowest common denominator for x
11.
the given expressions and state it in a fully 1
1− 2
factored form: x
1 2x 4
3. 1
6 x +1 2x + 2 12.
1
1+
1 2 3 1+ 1
4. x
x2 2
x −1 x + 2x + 1
2
1 1 1
5.
6x 5 4x 7 8x 6
7 y 1 Solve the equation:
6.
10x + 5x
2 50x 2 2
4x − 1 5 1 1
13. − =
x 3 x
Simplify the expression and state the an- 1 3
swer in a fully factored form: 14. 2 − =
x2 + x x +1
x 2 + x − 6 x 2 + 10x + 21 x 2 5
7. ÷ 15. + 2 =
x2 − x − 2 x +1 x + 2 x + 5x + 6 x +3
x + 3 1 − 2x x 3 7x
8. − 16. + = 2
x − 4 8 − 2x x −5 x +2 x − 3x − 10
407
6. PRACTICE TEST 7 CHAPTER 7. RATIONAL EXPRESSIONS
Practice Test 7 Answers.
1. {−2, −3} 4x
9.
(2x − 3)(2x + 3)2
2. {0, 5}
10. 8
3. 6(x + 1) x
11.
x −1
4. x 2 (x + 1)2 (x − 1)
x +1
7 12.
5. 24x 2x + 1
6. 50x 2 (2x + 1)(2x − 1) 13. {12}
1 14. {−1/2, 1}
7.
x +7
15. {4}
7
8.
2(x − 4) 16. {−3}
408
CHAPTER 8
Radicals
1. Introduction to Radicals
y
9
y = x2
8
7
6
5
4 p
y= x
3
2
1
x
1 2 3 4 5 6 7 8 9
1.1. Square Root and Radicals.
DEFINITION 1.1.1 (Principal Square Root). The principal square root of a non-negative
real number b is the non-negative number y such that y 2 = b. It is written like so:
p
b= y
The fancy sign is known as radical or radix and the expression b is called the radicand of
the root.
The radical is actually just a traditional notation for exponent:
1
p
b = (b) 2
This is perhaps the best way to understand the principal square root, and the reason why it
obeys the properties of exponential expressions.
The principal square root of a negative number is not real, but complex.
409
1. INTRODUCTION TO RADICALS CHAPTER 8. RADICALS
BASIC EXAMPLE 1.1.1. Here are some easy-to-find square roots.
p
The principal square root of 16 is 4 because 42 = 16, so we can write 16 = 4.
p
100 = 10 because 102 = 100.
p
0 = 0 because 02 = 0.
p
0.25 = 0.5 because 0.52 = 0.25.
v 2
t4 2 2 4
= because =
9 3 3 9
p
2 = 1.41421 . . . is irrational, so it cannot be written as a fraction of integers, and we
can only list finitely many digits of its decimal expansion here.
THEOREM 1.1.1. Applying the principal square root to an exponential expression with
base b ≥ 0 amounts to dividing the exponent by 2. For all non-negative real numbers a
and all positive even integers k,
p k
a k = (a) 2
The same is true for exponentiating the principal square root:
p k k
a = (a) 2
This is true for odd integers as well, as will be shown in the rational exponent section.
BASIC EXAMPLE 1.1.2. When taking roots of exponential expressions, we can check the
results by definition.
p
x 14 = x 14/2 = x 7 , which can be checked by definition:
(x 7 )2 = x 7·2 = x 14
Æ
y −6 = y −6/2 = y −3 = 1/ y 3 , which can be checked by definition:
2
1 1 1
= 3 2 = 6 = y −6
y 3 (y ) y
Exponentiating roots works similarly:
p 4 p −20 1
5 = 54/2 = 52 = 25 z = z −20/2 = z −10 = 10
z
410
CHAPTER 8. RADICALS 1. INTRODUCTION TO RADICALS
1.2. Principal nth Root. The nth root is a generalization of the square root. Just like the
square root cancels the square for x ≥ 0
p
x2 = x
the nth root will cancel the nth power
p
xn = x
n
DEFINITION 1.2.1 (Principal nth Root). Given a positive even integer n, the principal nth
root of a non-negative real number b is the non-negative number y such that y n = b.
Given a positive odd integer n, the principal nth root of a real number b is the number y
such that y n = b.
The principal nth root of b is written like so:
p
n
b= y
The fancy sign is known as radical or radix, n is called the index and b is called the radicand
of the root.
Just as was the case with the square root, the radical is actually just a traditional notation for
exponent:
1
p n
b = (b) n
p
Note
p that the principal square root b is the principal nth root for n = 2, and could be written
2
as b, but 2 is traditionally not shown.
BASIC EXAMPLE 1.2.1. Here are some easy-to-find nth roots.
p
3
The principal 3rd root of 125 is 5 because 53 = 125, so we can write 125 = 5
p
5
0 = 0 because 05 = 0
p
3
8 = 2 because 23 = 8
p
4
0.0081 = 0.3 because 0.34 = 0.0081
p
17
1 = 1 because 117 = 1
v
3 3
t 27 3 3 27
= because =
125 5 5 125
411
1. INTRODUCTION TO RADICALS CHAPTER 8. RADICALS
p p p
BASIC EXAMPLE 1.2.2. Roots with even indices, such as the square root , 4 , 6 , and so
on, were only defined for non-negative radicands, but roots with odd index can be applied
to negative radicands as well.
p3
−1000 = −10 because (−10)3 = −1000.
p5
−32 = −2 because (−2)5 = −32
p7
−1 = −1 because (−1)7 = −1.
THEOREM 1.2.1. Applying the principal nth root to an exponential expression with base
b ≥ 0 amounts to dividing the exponent by n.
p
n k
b k = (b) n
Similarly for exponentiating the principal nth root:
p
n
k k
b = (b) n
BASIC EXAMPLE 1.2.3. When taking roots of exponential expressions, we can check the
results by definition.
p6
x 12 = x 12/6 = x 2 , which can be checked by definition:
(x 2 )6 = x 2·6 = x 12
p 1
y −15 = y −15/3 = y −5 =
3
, which can be checked by definition:
y5
3
1 1
=
y5 ( y 5 )3
1
=
y 15
= y −15
Exponentiating roots works similarly:
p
4 20
3 = 320/4 = 35 = 243
p −35 1
= z −35/5 = z −7 = 7
5
z
z
412
CHAPTER 8. RADICALS 1. INTRODUCTION TO RADICALS
EXAMPLE 1.2.1. Simplify the radical expression, assuming that all variable radicands and
bases are non-negative: Æ3
(a − 5)18
SOLUTION: Æ
3
(a − 5)18 = (a − 5)18/3 = (a − 5)6
ANSWER: (a − 5)6
EXAMPLE 1.2.2. Simplify the radical expression, assuming that all variable radicands and
bases are non-negative:
p4
−24
x +7
SOLUTION: The radicand is a sum x + 7, so we have to introduce parentheses when we
rewrite the radical as an exponent with base x + 7:
p4
−24 1
x +7 = (x + 7)−24/4 = (x + 7)−6 =
(x + 7)6
1
ANSWER:
(x + 7)6
413
1. INTRODUCTION TO RADICALS CHAPTER 8. RADICALS
Homework 8.1.
6
Simplify the square root.
p
19. a+6
p
1. 81 p 10
p 20. b+1
2. 25
Æ
p 21. (x − 7)2
3. 49
Æ
22. ( y − 13)2
p
4. 16
p Æ
5. 0 23. (m + k)14
p Æ
6. 1 24. (2a − b)4
Simplify the radical expression.
p
3
7. 27 Simplify the radical expression, assuming
p
3
that all variable radicands and bases are
8. 64 non-negative.
p
3
9. −1000
p3
25. x 30
p
3
10. −125
p3
26. y 12
p
3
11. −1 p5
27. a20
p
5
12. 0 p5
28. b10
p
5
13. 32 p3
18
29. a
p
4
14. 81 p 25
5
30. u
Simplify the square root, assuming that
p
4
31. x 2000
all variable radicands and bases are non-
negative.
p
4
32. u4
p p
15. x 12 33. x −12
p
16. y 18
p
34. b−16
p 8 p −20
17. a 35. ( 4 y)
p 4 p
4
−4
18. b 36. k
414
CHAPTER 8. RADICALS 1. INTRODUCTION TO RADICALS
Homework 8.1 Answers.
1. 9 21. x − 7
3. 7 23. (m + k)7
5. 0
25. x 10
7. 3
27. a4
9. −10
29. a6
11. −1
13. 2 31. x 500
15. x 6 1
33.
x6
17. a4
1
35.
19. (a + 6) 3
y5
415
2. PRODUCTS WITH SQUARE ROOTS CHAPTER 8. RADICALS
2. Products with Square Roots
2.1. Distributivity Over Multiplication.
THEOREM 2.1.1. Principal square root distributes over multiplication. For all non-negative
real numbers x and y
p p p
xy = x· y
This property extends naturally to a product with any number of factors.
PROOF. Using the exponential representation of the root and the properties of the exponent:
p 1 1 1 p p
x y = (x y) 2 = (x) 2 ( y) 2 = x · y
We are assuming all variable bases and radicands to be non-negative throughout this chapter
becausepseveral very nice properties fail otherwise. In particular, the distributive property
p p
x y = x· y works for positive radicands, but allowing negative radicands leads to nonsense
like
p
1 = 1
Æ
= (−1)(−1)
p p
= −1 −1 illegitimate distributivity
= −1
The
p mistake above is in the third equals sign: there is nothing wrong with the complex number
−1, it is just that the distributive property does not apply to negative radicands.
p
EXAMPLE 2.1.1. Simplify the expression 8
SOLUTION: When simplifying expressions involving square roots, it is customary to sim-
plify
p the radicand by taking square rootspof factors which happen to be perfect squares.
So 8 will be written in the answer as 2 2 because
p p
8 = 4·2
p p
= 4· 2
p
= 2· 2
p
ANSWER: 2 2
416
CHAPTER 8. RADICALS 2. PRODUCTS WITH SQUARE ROOTS
p
EXAMPLE 2.1.2. Simplify the expression 75
SOLUTION: We simplify the radicand by finding the largest integer factor which happens
to be a perfect square, and taking its square root separately by distributivity:
p p p p p
75 = 25 · 3 = 25 · 3 = 5 · 3
p
ANSWER: 5 3
EXAMPLE 2.1.3. Simplify the expression, assuming that all variable bases and radicands
are non-negative: p
x3
SOLUTION: We simplify the radicand by finding the largest even integer exponent of x,
and taking its square root separately by distributivity:
p p
x3 = x2 · x
p p
= x2 · x
p
= x· x
p
ANSWER: x x
EXAMPLE 2.1.4. Simplify the expression, assuming that all variable bases and radicands
are non-negative: p
x 17
SOLUTION: The largest even exponent below 17 is 16:
p p
x 17 = x 16 · x
p p
= x 16 · x
p
= x 16/2 · x simplified square root
p
= x8 · x
p
ANSWER: x 8 x
417
2. PRODUCTS WITH SQUARE ROOTS CHAPTER 8. RADICALS
EXAMPLE 2.1.5. Simplify the expression, assuming that all variable bases and radicands
are non-negative: p p
2x 3 · 8x
SOLUTION: Here it helps to simplify the product under the square root before simplifying
the radicand:
p p p
2x 3 · 8x = 2x 3 · 8x distributivity
p
= 16x 4
p p
= 16 · x 4 distributivity the other way
= 4 · x 4/2 simplify square roots
= 4 · x2
ANSWER: 4x 2
EXAMPLE 2.1.6. Simplify the expression, assuming that all variable bases and radicands
are non-negative: p p p
2 6 3+5 2
SOLUTION: It is important not to confuse the distributivity of multiplication over addition
with the distributivity of the principal root over multiplication.
p p p p p p p
2 6 3+5 2 = 2 6· 3+2 6·5 2 distributivity of · over +
p p p
= 2 6 · 3 + 10 6 · 2 distributivity of over ·
p p
= 2 18 + 10 12
Now we simplify the radicands:
p p p p
2 18 + 10 12 = 2 9 · 2 + 10 4 · 3 finding perfect square factors
p p p p p
= 2 9 · 2 + 10 4 · 3 distributivity of over ·
p p
= 2 · 3 · 2 + 10 · 2 · 3 simplified square roots
p p
= 6 2 + 20 3
p p
ANSWER: 6 2 + 20 3
418
CHAPTER 8. RADICALS 2. PRODUCTS WITH SQUARE ROOTS
EXAMPLE 2.1.7. Simplify the expression, assuming that all variable bases and radicands
are non-negative: p p
−5x x 3 y 4 · (−2) x y 5
SOLUTION: We will combine the two radicands into one using distributivity, and then
simplify the result. First, we change the order of multiplication:
p p p p
−5x x 3 y 4 · (−2) x y 5 = (−5)(−2) · x · x 3 y 4 · x y 5
p p
= 10x x 3 y 4 · x y 5 distributivity of over ·
p
= 10x x 3+1 y 4+5 properties of exponent
p
= 10x x 4 y 9
Now we can simplify the radicand:
p p
10x x 4 y 9 = 10x x 4 · y 8 · y find largest even power of y
p p p
= 10x x 4 · y 8 · y distributivity
p
= 10x · x 4/2 · y 8/2 · y simplify square roots
p
= 10x · x 2 · y 4 · y
p
= 10x 1+2 · y 4 y properties of exponent
p
= 10x 3 · y 4 · y
p
ANSWER: 10x 3 y 4 y
419
2. PRODUCTS WITH SQUARE ROOTS CHAPTER 8. RADICALS
Homework 8.2.
p p
Simplify the expression, assuming that 12. −5 10 · 15
all variable bases and radicands are non- p p
negative. 13. 12m · 15m
p p p
1. 12 14. 5 y 3 · (−5) 10 y 2
p p p
2. 18 15. 6( 2 + 2)
p p p p
3. 27 16. 10( 5 + 2)
p p p
4. 60 17. 10b · 50b
p p p
5. 72 18. 6a · 18a7
p
6. 80
p p
19. x2 y3 · x y4
p
7. 20x 5
p p
20. a3 b2 · ab
8.
p
32 y 7 p p
21. 5 10(5n + 2)
p p p p
9. a10 b13 22. 15( 5 − 3 3x)
p
10. 49x 49
p p
23. 6x y · 12x 2 y 5
p p
11. 3 5 · (−4) 16
p p
24. 5ab5 · 20a2 b3
420
CHAPTER 8. RADICALS 2. PRODUCTS WITH SQUARE ROOTS
Homework 8.2 Answers.
p p
1. 2 3 13. 6m 5
p p p
3. 3 3 15. 2 3 + 2 6
p p
5. 6 2 17. 10b 5
p p
7. 2x 2 5x 19. x y 3 x y
p p p
9. a5 b6 b 21. 25n 10 + 10 5
p p
11. −48 5 23. 6x y 3 2x
421
3. QUOTIENTS WITH SQUARE ROOTS CHAPTER 8. RADICALS
3. Quotients with Square Roots
3.1. Distributivity Over Division.
THEOREM 3.1.1. Principal square root distributes over division. For all non-negative real
numbers a and all positive reals c
s p
a a
= p
c c
This results generalizes naturally to fractions with any number of factors.
PROOF. Using the exponential representation of the root and the properties of the exponent:
12 1 p
(x) 2
v
tx x x
= = 1
=p
y y ( y) 2 y
As we simplify quotients with radicals, we will state all answers in a form where radicals are
confined to numerators, and radicands are fully simplified.
DEFINITION 3.1.1. We say that the denominator is rationalized in a fraction if radicals
appear in the numerator only, and radicands are simplified.
EXAMPLE 3.1.1. Simplify and rewrite with rationalized denominator:
1
p
3
SOLUTION: The radicand is simplified, but it is in the wrong place. To get rid of the radical
in the monomial
p p denominator, we can multiply the expression by the radical over itself, in
this case by 3/ 3:
p
1 1 3
p = p ·p
3 3 3
p
1· 3 p p p p
= p p but 3 · 3 = 3 · 3 = 32 = 3
3· 3
p
3
=
3
p
3
ANSWER:
3
422
CHAPTER 8. RADICALS 3. QUOTIENTS WITH SQUARE ROOTS
EXAMPLE 3.1.2. Simplify and rewrite with rationalized denominator:
2x
p
6x
SOLUTION: As before, we can rationalize the denominator by multiplying both numerator
and denominator by the radical we are trying to eliminate:
p
2x 2x 6x
p = p ·p
6x 6x 6x
p
2x · 6x
= p p
6x · 6x
p
2x 6x p
= p distributivity of over ·
(6x)2
p
2x 6x
= common factors 2 and x cancel
6x
p
6x
=
3
p
6x
ANSWER:
3
EXAMPLE 3.1.3. Simplify and rewrite with rationalized denominator:
p
20x 5
p
5x 3
SOLUTION: Here it makes more sense to apply distributivity and simplify the fraction
under the radical sign before rationalizing anything:
p v
20x 5 t 20x 5
p
p = 3
distributivity of over ÷
5x 3 5x
p
= 4x 2
= 2x
ANSWER: 2x
423
3. QUOTIENTS WITH SQUARE ROOTS CHAPTER 8. RADICALS
Homework 8.3.
v
Simplify and rewrite with rationalized de- t 10 y 9
nominator. 14.
18 y 5
p
500 v
1. p t 4x 3
5 15.
50x
p
72
2. p
v
t 100
2 16.
p 49
50 p
3. p 36
2 17. p
p 6
40 p
4. p 18
10 18. p
p 32
55
5. p p
63 y 3
5 19. p
p 7y
18
6. p p
3 8x
20. p
p 2x
5
7. p p
20 9b
21. p
p 3b
2
8. p p
18 48x 5
22. p
v 3x 3
t4
9. 5
25 23. p
v 7
t9
10. 2
49 24. p
3
p
v
t 2a5
11. 25
50a 25. p
8
v
t 25 p
16
12. 26. p
64 27
v p
t 6x 7 3
13. 27. p
32x 50
424
CHAPTER 8. RADICALS 3. QUOTIENTS WITH SQUARE ROOTS
p p
5 27x 5
28. p 33. p
7 3x
p p
3a 21a9
29. p 34. p
32 7a3
p p
2a 35x 13
30. p 35. p
45 5x 5
p p
6 7b5
31. p 36. p
5 28b
p
5
32. p
18
425
3. QUOTIENTS WITH SQUARE ROOTS CHAPTER 8. RADICALS
Homework 8.3 Answers.
1. 10 19. 3 y
p
3. 5 21. 3
p p
5. 11 5 7
23.
7
1
7. p
2 5 2
25.
4
2
9. p
5 6
27.
2 10
a
11. p
5 6a
29.
p 3 8
x 3
13. p
4 30
31.
p 5
x 2
15.
5 33. 3x 2
p p
17. 6 35. x 4 7
426
CHAPTER 8. RADICALS 4. SUMS WITH SQUARE ROOTS
4. Sums with Square Roots
4.1. Radical Like Terms. Adding terms with square roots is possible using the distributiv-
ity of multiplication over addition.
DEFINITION 4.1.1. Terms involving radicals are like (or similar) if they have the same
variable factors with the same respective exponents, and the same radical factors with
the same respective radicands.
BASIC EXAMPLE 4.1.1. Some pairs of like terms:
p p p p
x 7, 4x 7 x x, −6x x
p p p p
x, 5 x x 2 y 2ab, 10x 2 y 2ab
BASIC EXAMPLE 4.1.2. Some pairs of terms which are not like:
p p
y y, y are not like because they have different exponents for y variable.
p p
2x, 5x are not like because they have different radicands.
p p
8, 3 2 are not like because they have different radicands, but they will be like if we
simplify the radicands:
p p p p p
8= 4·2= 4· 2=2 2
p p
2 2 and 3 2 are indeed like.
EXAMPLE 4.1.1. Simplify the expression
p p p
4 6−5 2− 6
p p
SOLUTION: The terms 4 6 and − 6 are like, so we can combine them:
p p p p p p
4 6−5 2− 6 = 4 6− 6−5 2 terms can be added in any order
p p
= (4 − 1) 6 − 5 2 distributivity
p p
= 3 6−5 2
The remaining terms are not like, so we cannot simplify this expression any further.
p p
ANSWER: 3 6−5 2
427
4. SUMS WITH SQUARE ROOTS CHAPTER 8. RADICALS
EXAMPLE 4.1.2. Simplify the expression
p p
8 18 − 7 2
SOLUTION: These terms are not like, but the radicand 18 can be simplified:
p p p p
8 18 − 7 2 = 8 9 · 2 − 7 2
p p p
= 8 9 2−7 2
p p
= 8·3 2−7 2
p p
= 24 2 − 7 2
These terms are like, so we can combine them:
p p p
24 2 − 7 2 = (24 − 7) 2
p
= 17 2
p
ANSWER: 17 2
EXAMPLE 4.1.3. Simplify the expression
p p
(3 + 7)(2 − 5 7)
SOLUTION: We begin by multiplying the sums using the standard binomial pattern, where
we multiply each term of the first sum by each term of the second sum, and then add all
of these products together:
p p p p p p
(3 + 7)(2 − 5 7) = 3 · 2 − 3 · 5 7 + 7 · 2 − 7 · 5 7
p p p p
= 6 − 15 7 + 2 7 − 5 7 7 simplified products
p p p 2
= 6 − 15 7 + 2 7 − 5 · 7 because 7 =7
p p
= 6 − 15 7 + 2 7 − 35
p p
We have two pairs of like terms here: the two integers as well as −15 7 and 2 7.
p p p p
6 − 15 7 + 2 7 − 35 = 6 − 35 − 15 7 + 2 7
p
= −29 + (−15 + 2) 7
p
= −29 − 13 7
p
ANSWER: −29 − 13 7
428
CHAPTER 8. RADICALS 4. SUMS WITH SQUARE ROOTS
4.2. Rationalizing Sums. A sum with two terms can be rationalized if it is multiplied by
its conjugate, which is the sum of the first term and the opposite of the second term.
DEFINITION 4.2.1. For all real numbers A, B, C, D, the sums
p p
A B+C D
and p p
A B−C D
are the radical conjugates (or simply conjugates) of each other.
THEOREM 4.2.1 (Rationalizing the Denominator). In a fraction, a sum with two terms in
the numerator or the denominator can be rationalized by multiplying both the numerator
and the denominator by the conjugate of that sum.
PROOF. We will
p provepthis statement for a sum of
p square
p roots in the denominator. If the
denominator is a b + c d, then its conjugate is a b − c d.
p p
1 1 a b−c d
p p = p p · p p
a b+c d a b+c d a b−c d
p p
a b−c d
= p p p p
(a b + c d)(a b − c d)
Notice that denominators are sums, and we had to show the invisible fraction parentheses
when we multiplied the denominators. Notice also that the denominator now is a difference
of squares, and can be simplified:
p p p p
a b−c d a b−c d
p p p p = p 2 p 2
(a b + c d)(a b − c d) a b − c d
p p
a b−c d
= p 2 p 2
a2 b − c2 d
p p
a b−c d
=
a2 b − c 2 d
The new denominator is free of square roots now, even though they moved into the numerator.
The case with the difference of square roots, as well as the cases with rationalizing the numer-
ator have similar proofs.
429
4. SUMS WITH SQUARE ROOTS CHAPTER 8. RADICALS
EXAMPLE 4.2.1. Rationalize the denominator:
6
p
3+ 5
p p
SOLUTION: The conjugate of 3 + 5 is 3 − 5:
p
6 6 3− 5
p = p · p
3+ 5 3+ 5 3− 5
p
6(3 − 5)
= p p difference of squares in the denominator
(3 + 5)(3 − 5)
p
6(3 − 5)
= p 2
32 − 5
p
6(3 − 5)
= simplify the denominator
9−5
p
6(3 − 5)
= cancel common factor 2
4
p
3(3 − 5)
=
2
p
3(3 − 5)
ANSWER:
2
EXAMPLE 4.2.2. Rationalize the denominator:
5x
p p
x− y
p p p p
SOLUTION: The conjugate of x− y is x+ y:
p p
5x 5x x+ y
p p = p p ·p p
x− y x− y x+ y
p p
5x( x + y)
= p p p p
( x − y)( x + y)
430
CHAPTER 8. RADICALS 4. SUMS WITH SQUARE ROOTS
The denominator is now a difference of squares, so we can simplify it:
p p p p
5x( x + y) 5x( x + y)
p p p p = p 2 p 2
( x − y)( x + y) x − y
p p
5x( x + y)
=
x−y
p p
5x( x + y)
ANSWER:
x−y
431
4. SUMS WITH SQUARE ROOTS CHAPTER 8. RADICALS
Homework 8.4.
p p
Simplify the expression. 23. 4x x− x3
p p p p
1. 2 5 + 2 5 + 2 5
p
24. 2 y y + 3 y3
p p p p p
2. −3 6 − 3 3 − 2 3 25. ( 5 − 5)(2 5 − 1)
p p p p p
3. −3 2 + 3 5 + 3 5 26. (−2 + 3)(−5 + 2 3)
p p p
p p
4. −2 6 − 3 − 3 6
p p
27. ( 2 + 6)(2 6 − 3 2)
p p p p
p p p
5. −2 6 − 2 6 − 6 28. (4 5 − 20)( 20 − 5)
p p p
6. −3 3 + 2 3 − 2 3
p p p
7. 3 5 + 3 5 + 2 5 Rationalize the denominator and simplify.
p p p
8. − 5 + 2 3 − 2 3 4
29. p
p p p 3+ 5
9. 2 2 − 3 18 − 2
−4
p p p 30. p
10. − 54 − 3 6 + 3 27 4−4 2
p p p 1
11. −3 6 − 12 + 4 3 31. p
p p p 2+1
12. − 5 − 5 − 2 45
3
p p p 32. p
13. 3 2 + 2 8 − 3 18 6−2
p p p
14. 2 20 + 2 20 − 3 2
33. p
p p p 5+ 2
15. 3 18 − 2 − 3 2
p p 3
p 34. p
16. −3 27 + 2 3 − 12 4−3 3
p p p p
17. −3 6 − 3 6 − 3 + 3 6 5
35. p p
p p p p 3+4 5
18. −2 2 − 2 + 3 8 + 3 6
5
p p p p 36. p p
19. −2 18 − 3 8 − 20 + 2 20 2 3− 2
p p p p
20. −3 18 − 8 + 2 8 + 2 8 x−y
37. p p
p p x− y
21. 300x − 3x
p p 2
38. p p
22. 2x + 50x 2 5+2 3
432
CHAPTER 8. RADICALS 4. SUMS WITH SQUARE ROOTS
Homework 8.4 Answers.
p p
1. 6 5 21. 9 3x
p p p
3. −3 2 + 6 5 23. 3x x
p p
5. −5 6 25. 15 − 11 5
p p
7. 8 5 27. 6 − 2 3
p p
9. −8 2 29. 3 − 5
p p p
11. −3 6 + 2 3 31. 2−1
p p
13. −2 2 2(5 − 2)
33.
p 23
15. 5 2 p p
p −5( 3 − 4 5)
p 35.
17. −3 6 − 3 77
p p p p
19. −12 2 + 2 5 37. x + y
433
5. EQUATIONS WITH SQUARE ROOTS CHAPTER 8. RADICALS
5. Equations with Square Roots
5.1. Squaring Both Sides. Some equations involving principal square roots can be solved
by squaring both sides. This operation does not always produce equivalent equations, and will
create extraneous solutions every now and then.
THEOREM 5.1.1. If A = B, then the solution set of the equation A2 = B 2 includes all of the
solutions of A = B, but not necessarily the other way around. This means we can solve
the original equation by solving A2 = B 2 , and then discarding extraneous solutions, if any.
BASIC EXAMPLE 5.1.1. It is easy to see that squaring both sides may create additional
solutions. The solution set of the equation
x =1
is {1}, while the solution set of
(x)2 = (1)2
which is equivalent to
x2 = 1
is {1, −1}.
EXAMPLE 5.1.1. Determine which of the given numbers are solutions for the equation:
p
x −5= x +7
2, 9
SOLUTION: If x = 2 then
p
2−5 = 2+7
−3 = 3
This is false, so 2 is not a solution, but an extraneous solution.
If x = 9 then
p
9−5 = 9+7
4 = 4
This is true, so 9 is a solution.
ANSWER: {9}
434
CHAPTER 8. RADICALS 5. EQUATIONS WITH SQUARE ROOTS
p
EXAMPLE 5.1.2. Solve the equation x −6−3=0
SOLUTION: We want to solve the equation for the term with the radical, or else squaring
both sides will be pointless.
p
x −6−3 = 0
p
x −6 = 3 added 3 to both sides
p 2
x −6 = (3)2
x −6 = 9
This looks like a linear equation, which we solve by isolating the variable:
x −6 = 9
x = 9+6 added 6 to both sides
x = 15
The solution checks out:
p p
15 − 6 − 3 = 9−3=3−3=0
ANSWER: {15}
p
EXAMPLE 5.1.3. Solve the equation 6 + 2 x − 4 = 0
SOLUTION: We start by solving for the term with the radical:
p
6+2 x −4 = 0 subtract 6 on both sides
p
2 x − 4 = −6 divide both sides by 2 to simplify
p
x − 4 = −3 square both sides
p 2
x −4 = (−3)2
x −4 = 9 add 4 to both sides
x = 13
Check the solution:
p
6 + 2 13 − 4 = 0
p
6+2 9 = 0
6+2·3 = 0
12 = 0
435
5. EQUATIONS WITH SQUARE ROOTS CHAPTER 8. RADICALS
This is false, so 13 is an extraneous solution, and since there are no others, the solution
set is empty.
ANSWER: ∅
In the previous example, we could have saved ourselves some time if we used the fact that the
principal square root is never negative. We could have concluded there are no solutions from
p
2 x − 4 = −6
which is impossible to satisfy, since the left side is non-negative, while the right side is negative.
THEOREM 5.1.2. The principal square root is never negative. For any real number b,
p
• if b is positive,pthen b > 0
p
• if b = 0, then b = 0 = 0
• if b is negative, then the principal square root of b is not real, but complex.
EXAMPLE 5.1.4. Solve the equation
p
x2 − 4 + 4 = 1
SOLUTION: We start by solving for the term with the radical:
p
x2 − 4 + 4 = 1
p
x 2 − 4 = −3
We notice that the left side is non-negative, while the right side is negative, so there are
no solutions possible.
ANSWER: ∅
EXAMPLE 5.1.5. Solve the equation
p p
3 − 4x − −7x − 6 = 0
SOLUTION: The left side is a sum of two terms, and squaring a sum should be avoided
when possible. But if we isolate the radicals, one on each side of the equation, then we
436
CHAPTER 8. RADICALS 5. EQUATIONS WITH SQUARE ROOTS
can get rid of both of them at the same time:
p p
3 − 4x − −7x − 6 = 0
p p p
3 − 4x = −7x − 6 add −7x − 6 to both sides
p 2 p 2
3 − 4x = −7x − 6 square both sides
3 − 4x = −7x − 6 squares cancel square roots
3x = −9 isolate the term with x on the left
x = −3
Check whether −3 is a solution:
p p Æ Æ
3 − 4x − −7x − 6 = 3 − 4(−3) − −7(−3) − 6
p p
= 3 + 12 − 21 − 6
p p
= 15 − 15
= 0
The equation holds, so −3 is the only solution.
ANSWER: {−3}
EXAMPLE 5.1.6. Solve the equation
p
−8 = x − −2x − 16
SOLUTION: Isolate the radical on the left before squaring both sides:
p
−8 = x − −2x − 16
p
−2x − 16 = x + 8
Square both sides now, but note that the right side is a sum of two terms, and will have to
be squared as a binomial:
p 2
−2x − 16 = (x + 8)2
−2x − 16 = x 2 + 16x + 64
This is a quadratic equation, so we will move all non-zero terms to one side, factor, and
solve using the zero factor property:
−2x − 16 + 2x + 16 = x 2 + 16x + 64 + 2x + 16
0 = x 2 + 18x + 80
0 = (x + 8)(x + 10)
437
5. EQUATIONS WITH SQUARE ROOTS CHAPTER 8. RADICALS
By the zero product property, this quadratic equation has solutions x = −8 and x = −10.
We have to check them before concluding they are also solutions for the equation with
radicals.
If x = −8 then the right side of the original equation is
p Æ
x − −2x − 16 = (−8) − −2(−8) − 16
p
= −8 − 16 − 16
p
= −8 − 0
= −8
This is equal to the left side, so −8 is a solution.
If x = −10 then the right side of the original equation is
p Æ
x − −2x − 16 = (−10) − −2(−10) − 16
p
= −10 − 20 − 16
p
= −10 − 4
= −10 − 2
= −12
This is not equal to the left side, so −10 is not a solution, although it is an extraneous
solution.
ANSWER: {−8}
438
CHAPTER 8. RADICALS 5. EQUATIONS WITH SQUARE ROOTS
Homework 8.5.
p
Solve the equation. 14. y + 18 = y − 2
p p
1. x = 7 15. x +4=4 x +1
p p
2. x − 7 = 2 16. 1+2 a−1= a
p
3. x − 3 = 9
p
17. x2 + 6 = x − 3
p p
4. 4 − x = 0 18. 5x + 21 = x + 3
p p
5. 3x + 1 = 8 19. x =1+ 1− x
p p
6. 5x − 1 = 7 20. x = x +5+7
p p p
7. 4 x = −3 21. 3 x = x + 32
p p p
8. 1 − x = 2 22. 2 x − 7 = 5x
p p p
9. 8x + 3 = 6x + 7 23. x +5= x −1
p p p p
10. 7b − 9 = b + 3 24. 5x + 3 = 2x − 1
p p
11. 3a + 1 = a − 3 25. 1 + x = 1 + 5x
p p
12. b − 7 = b − 5 26. x +2−2= x
p
13. x − 9 = x − 3
439
5. EQUATIONS WITH SQUARE ROOTS CHAPTER 8. RADICALS
Homework 8.5 Answers.
1. {49} 15. {0, 8}
3. {144} 17. ∅
5. {21}
19. {1}
7. ∅
21. {4}
9. {2}
11. {8} 23. {4}
13. {12} 25. {0, 3}
440
CHAPTER 8. RADICALS 6. APPLICATIONS TO RIGHT TRIANGLES
6. Applications to Right Triangles
6.1. Distance Formula.
DEFINITION 6.1.1 (Euclidean Distance). On a coordinate plane, the distance d between
points with coordinates (x 1 , y1 ) and (x 2 , y2 ) is
Æ
d= (x 2 − x 1 )2 + ( y2 − y1 )2
y2
d
| y2 − y1 |
y
y1
|x 2 − x 1 |
x1 x2
x
This definition is in full agreement with the Pythagorean theorem. Applied to the high-
lighted right triangle pictured above, the theorem states
d 2 = (x 2 − x 1 )2 + ( y2 − y1 )2
If we take the principal square root of both sides of the equation, we will get our definition
of the distance.
Given a right triangle with two known sides, we can use the Pythagorean theorem to solve for
the length of the remaining side.
EXAMPLE 6.1.1. Given the lengths of two
sides of a pictured right triangle with
legs a and b and hypotenuse c, find the
c
length of the remaining side. a
a = 8, b=4
b
441
6. APPLICATIONS TO RIGHT TRIANGLES CHAPTER 8. RADICALS
SOLUTION: We state the Pythagorean theorem for this triangle and solve for the unknown
variable:
a2 + b2 = c2
82 + 42 = c2 substitute known lengths
80 = c2 take principal square root of both sides
p
80 = c
p
It is true that there is another solution to this quadratic equation, namely − 80, but
since there are no negative lengths, the negative solution is not applicable and can be
disregarded. We will state the answer with simplified radicand:
p
c = 80
p
c = 16 · 5
p p
c = 16 · 5
p
c = 4· 5
p
ANSWER: 4 5
EXAMPLE 6.1.2. Find the distance between the points with coordinates
(4, 7) and (−5, 4)
SOLUTION: Let (x 1 , y1 ) = (4, 7), (x 2 , y2 ) = (−5, 4), and use the distance formula:
Æ
d = (−5 − 4)2 + (4 − 7)2 simplify sums before squaring
Æ
= (−9)2 + (−3)2
p
= 81 + 9
p
= 90
Simplify the radicand for the answer:
p p
90 = 9 · 10
p p p
= 9 · 10 distributivity of over ·
p
= 3 · 10
p
ANSWER: 3 10
442
CHAPTER 8. RADICALS 6. APPLICATIONS TO RIGHT TRIANGLES
Homework 8.6.
7. a = 18, c = 30
p
8. b = 1, c = 2
c p p
a 9. a = 2, b = 6
10. a = 5, b = 5
b
Find the distance between the given points.
Given the lengths of two sides of a pictured
straight triangle with legs a and b and hy- 11. (2, 3) and (6, 10)
potenuse c, find the length of the remain-
ing side. 12. (0, 3) and (4, 0)
1. a = 13, b = 5 13. (−3, 2) and (−1, 5)
2. a = 9, c = 15 14. (−2, 4) and (−8, −4)
p
3. b = 1, c = 10 15. (0, −5) and (−12, 0)
p
4. a = 1, c = 3 16. (2, −7) and (1, −4)
p
5. c = 10, b = 5 3 17. (−2, −8) and (6, 7)
6. a = 24, b = 10 18. (1, 0) and (7, 3)
443
6. APPLICATIONS TO RIGHT TRIANGLES CHAPTER 8. RADICALS
Homework 8.6 Answers.
1. c = 13
p
11. 65
3. a = 3 p
13. 13
5. a = 5
15. 13
7. b = 26
p
9. c = 2 2 17. 17
444
CHAPTER 8. RADICALS 7. RATIONAL EXPONENT
7. Rational Exponent
3 y
2
y = (x) 3
2
1
x
−3 −2 −1 1 2 3
−1
7.1. Rational Exponent Definition.
DEFINITION 7.1.1 (Integer Reciprocal Exponent). If b is a real number, and n is a positive
integer, then the exponential expression (b)1/n is defined as follows:
1
(b) n = y, where yn = b
If n is even, then the positive y is chosen. Notationally, these are equivalent:
1 p
n
(b) =
n
b
BASIC EXAMPLE 7.1.1. Some easy-to-find exponents:
1
(25) 2 = 5 because 52 = 25.
1
(16) 4 = 2 because 24 = 16.
1
(−8) 3 = −2 because (−2)3 = −8.
1
(0.25) 2 = 0.5 because (0.5)2 = 0.25.
1
(−0.00001) 5 = −0.1 because (−0.1)5 = −0.00001
445
7. RATIONAL EXPONENT CHAPTER 8. RADICALS
1
EXAMPLE 7.1.1. Find (225)− 2
SOLUTION: All properties of exponent still apply, so
1
1 2
− 12
(225) =
225
1
= 1
(225) 2
1
=
15
1
ANSWER:
15
DEFINITION 7.1.2 (Rational Exponent). If b is a non-negative real number, and m/n is a
quotient of an integer and a positive integer in lowest terms, then the exponential expres-
m
sion (b) n is defined as follows:
m
1
m
(b) = (b)
n n
Note that all of these are equivalent:
m
1
m p
n
m p
1 n
(b) = (b)
n n = ((b) ) =m
b = bm
n
Several of the integer exponent properties stop working when we allow the base b to
be negative, so we are assuming throughout this chapter that all variable bases are non-
negative.
THEOREM 7.1.1. Assuming that all bases are non-negative, the rational exponent obeys
all the properties of the integer exponent.
Recall the properties of the integer exponent we have seen so far:
• Distributivity over multiplication:
(ab)n = a n b n
• Distributivity over division:
a n an
=
b bn
446
CHAPTER 8. RADICALS 7. RATIONAL EXPONENT
• The product rule:
b m b n = b m+n
• The quotient rule:
bm
= b m−n
bn
• The power rule:
(b m )n = b mn
• Negative exponent properties:
n
1 1
b −n
= = n
b b
EXAMPLE 7.1.2. Simplify and state the answer with positive exponents only:
6
13
a
SOLUTION:
13 1
a 6
= (a)6· 3
= a2
ANSWER: a2
EXAMPLE 7.1.3. Simplify and state the answer with positive exponents only:
1
32 5
−1
x y 3 · x 2 y2
SOLUTION:
1
23 5
1
32 32 5
−1
x y
3 ·x y
2 2
= x 3 y −1 · x 2 y2 distributivity
1 3 3 5
= x 3 · 2 y −1· 2 · x 2 y 2 power rule
1 3 5
= x 2 y−2 · x 2 y2
1 5 3
= x 2 + 2 y − 2 +2
1
= x3 y 2
447
7. RATIONAL EXPONENT CHAPTER 8. RADICALS
1
ANSWER: x3 y 2
EXAMPLE 7.1.4. Simplify and state the answer with positive exponents only:
13
a 2 b0 a−3
5 1
b− 7 a 4
SOLUTION: We can use the product and the quotient rules add and subtract corresponding
exponents:
13
a 2 b0 a−3
a 2 +(−3)− 4 b0−(− 7 )
13 1 5
13 13 6 13 − 6 7
= −3= − = =
− 57 1
2 2 2 2 2
b a 4
7 1 5
7 1 14 1 14 − 1 13
= a2−4 b7 − = − = =
2 4 4 4 4 4
13 5
= a 4 b7
13 5
ANSWER: a 4 b7
448
CHAPTER 8. RADICALS 7. RATIONAL EXPONENT
Homework 8.7.
Write the expression in radical form. Simplify and state the answer with positive
exponents only.
4
1. (2m) 5 1 3
3
17. yx3 · x y2
2. (10r) 4 5
3
18.4v 3 · v −1
3. (7x) 2 1 1 −1
4
19. a 2 b 2
4. (6b) 3
5
0
−2
20. x y 3
Write the expression in exponential form. ab3
21.
p
6
3ab0
5. v 1 1
p 2x 2 y 3
6. 5a 22. 4 7
Æ 2x 3 y − 4
7. (4x)5 1 3
23 2 5
p
3
23. x y 2 4 · x 3 y−2
8. c4
− 12
3 − 13 − 73
24. g h ·g
2 3
Evaluate the expression. 25. u2 · uv v 3
2
9. 83
26. (x · x y 2 )0
1
10. 16 4 1
32
0
5 27. x y 3 x0
11. 4 2
− 32
3 − 54 2 2
12. 9 2 28. u v · u 3
1
13. 25− 2 3
a 4 b−1 · b 4
7
3
29.
14. 16− 2 3b−1
5
15. (−8) − 43 2x −2 y 3
30. 5 5 1
x−4 y−3 · x y 2
− 13
16. (−27)
449
7. RATIONAL EXPONENT CHAPTER 8. RADICALS
1 5 47
a b 3 · 2b− 4 3
−2
31. m n
2
− 12 − 23 33.
4a b m−1 n− 3
4
5
3 y−4 1
−2
23
32. 1 y y
3
y −1 · 2 y − 3 34. 5 9
x−2 y−2
450
CHAPTER 8. RADICALS 7. RATIONAL EXPONENT
Homework 8.7 Answers.
b
p
5
1. 16m4 21.
p 3
3. 343x 3
x
1 23.
5. v 6
y2
5
7. (4x) 2
25. u3 v 3
9. 4
1
27. y2
11. 32
1 1 7 3
13. 29. b4 a4
5 3
1 3
15. a2
16 31. 1
4 5 2b 4
17. x y 3 2
35
1 m8
19. 1 1
33. 7
a2 b2 n6
451
7. PRACTICE TEST 8 CHAPTER 8. RADICALS
Practice Test 8
Simplify each expression, assuming that Solve each equation:
all variable bases and radicands are non- p p
negative: 15. 2x + 1 = x + 2
p p
1. 121 16. 5x + 1 − 6 = 0
p
17. x − 2 = 2x − 4
v
t4
2.
9 p
18. 2x − 5 + 4 = x
p
3
3. −8
p
4. x 2018
19. Find the length of the leg b of a
5. ( x − y)2
p
straight triangle, if the other leg a is 6 cm,
p and the hypotenuse c is 14 cm.
4
6. 81
20. Find the distance between points
(−5, −9) and 5, −3.
p
5
7. x 15
p
8. 72x 5
p p
9. 3 y 7 · 27 y 3
Evaluate the expression:
p
7x 7 3
10. p 21. 100 2
28x
2
(−64)− 3
v
t 54x 22.
11.
3y
p p p
12. 12(3 2 − 8)
Simplify the expression and state the an-
swer with positive exponents only:
Rationalize the denominator and simplify:
1
2
− 12 5
13.
6
p 23. x y 3 · x − 3 y −2
3+ 7
1 1
p p
2+ 3 a 2 b− 5
14. p p 24. 1 11
6− 5 a 3 b− 5
452
CHAPTER 8. RADICALS 7. PRACTICE TEST 8
Practice Test 8 Answers.
p
1. 11 13. 9 − 3 7
2 p p p p
2. 14. 10 + 2 3 + 15 + 3 2
3
3. −2 15. {1}
4. x 1009 16. {7}
5. x − y 17. {2, 4}
6. 3 18. {7}
p
7. x 3 19. 4 10
p p
8. 6x 2 2x 20. 2 34
9. 9 y 5 21. 1000
10. x 3 /2 22. 1/16
p
3 2x y 1
11. 23.
y x y3
1
a 6 b2
p
12. 2 6 24.
453
CHAPTER 9
Quadratic Equations
1. Complex Numbers
1.1. Imaginary Unit. Complex numbers were initially designed to assist in factoring poly-
nomials around 1545 by Gerolamo Cardano. It was observed that an equation such as
x 2 = −1
which is provably without a real solution, can be postulated to have a solution which is not a
real number, and nothing goes wrong as a result. In fact, one could say the situation improves
somewhat. It becomes possible to construct a set of complex numbers, which includes all real
numbers as well as all numbers which happen to be solutions of polynomial equations with
real or complex coefficients. All of the real number axioms hold true for the complex numbers,
so most theorems presented in this text hold for the complex numbers almost word for word.
DEFINITION 1.1.1 (Imaginary Unit). The solutions of the equation
x 2 = −1
are complex numbers i and its opposite −i:
i 2 = (−i)2 = −1
i is usually called the imaginary unit, and it happens to be the principal square root of the
negative unit: p
−1 = i
THEOREM 1.1.1. If x is a positive real number, then the principal square root of −x is
p p
−x = x · i
p
EXAMPLE 1.1.1. Simplify the expression −100
SOLUTION: p p
−100 = 100 · i = 10i
ANSWER: 10i
455
1. COMPLEX NUMBERS CHAPTER 9. QUADRATIC EQUATIONS
p
EXAMPLE 1.1.2. Simplify the expression −20
SOLUTION: p p p
−20 = 20 · i = 2 5 · i
p
ANSWER: 2 5 · i
1.2. The Set of Complex Numbers. Just like the set of real numbers can be visualized as
a number line, the set of complex numbers can be visualized as a number plane.
5i
Imaginary
4i
3i
3 + 2i
2i
i
Real
−5 −4 −3 −2 −1 1 2 3 4 5
−i
−2i
−3i
−4i
−5i
This plane works a lot like the Cartesian plane, but the coordinates are interpreted in a par-
ticular way. Every point on this plane represents a distinct complex number. The points on
the real axis are the familiar real numbers. The imaginary unit i is located one unit above the
origin, and the points on the imaginary axis are the real multiples of i. Every other point with
coordinates (a, b), where a and b are real numbers, is a complex number which can be written
as a + bi.
DEFINITION 1.2.1. A complex number is any number z which can be written in the form
a + bi
where a is a real number called the real part of z, b is a real number called the imaginary
part of z, and i is the imaginary unit.
456
CHAPTER 9. QUADRATIC EQUATIONS 1. COMPLEX NUMBERS
1.3. Adding and Multiplying Complex Numbers.
EXAMPLE 1.3.1. Simplify the expression (3 + 5i) − (2 + 7i)
SOLUTION: We treat i as if it was a variable, and combine the like terms:
(3 + 5i) − (2 + 7i) = 3 + 5i − 2 − 7i removed parentheses
= 1 − 2i combined like terms
ANSWER: 1 − 2i
EXAMPLE 1.3.2. Simplify the expression (1 + 6i)(2 − i)
SOLUTION: We treat i as if it was a variable, multiply the binomials, and combine the like
terms:
(1 + 6i)(2 − i) = 2 − 1i + 12i − 6i 2
= 2 + 11i − 6i 2 i 2 can be replaced by −1
= 2 + 11i − 6(−1)
= 2 + 11i + 6
= 8 + 11i combined like terms
ANSWER: 8 + 11i
p
8+ −16
EXAMPLE 1.3.3. Simplify the expression
4
SOLUTION: We simplify the radicand, and then cancel common factors:
p
8 + −16 8 + 4i
= rewrite in terms of i
4 4
4(2 + i)
= factored out GCF
4
= 2+i canceled common factor 4
ANSWER: 2 + i
457
1. COMPLEX NUMBERS CHAPTER 9. QUADRATIC EQUATIONS
p
20 − −50
EXAMPLE 1.3.4. Simplify the expression
10
SOLUTION:
p p
20 − −50 20 − 50 · i
= rewrite in terms of i
10 10
p
20 − 25 · 2 · i
= simplify the radicand
10
p p
20 − 25 2 · i p
= distributivity of over ·
10
p
20 − 5 2 · i p
= 25 = 5
10
p
5(4 − 2 · i)
= factored out GCF
10
p
4− 2·i
= canceled common factor 5
2
If desired, this expression could be stated in the form a + bi by rewriting the fraction as a
sum:
p p
4− 2·i 4 2
= − i
2 2 2
p
2
= 2− i
2
But we will be content with simplified fractions for answers:
p
4− 2·i
ANSWER:
2
458
CHAPTER 9. QUADRATIC EQUATIONS 1. COMPLEX NUMBERS
Homework 9.1.
Simplify the given expression and state the 16. (3 − i)(4 + i)
answer in the form bi, where b is real.
p 17. (10 − 3i)(2i)
1. −5
p 18. (−5i)(2 + 4i)
2. −6
p 19. 30(−2 + 0.5i)
3. −49
2
p 20. (24 − 30i)
4. −81 3
p 21. (7 + 5i)(7 − 5i)
5. −45
p 22. (3 − 6i)(3 + 6i)
6. −18
p 23. (4 + i)2
7. −120
p 24. (1 − 3i)2
8. −160
p
−6 + −4
25.
3
Simplify the given expression. p
15 − −9
26.
9. (7 + 3i) + (14 + 2i) 5
p
10. (12 − 4i) + (−1 + 6i) 18 + −8
27.
−2
11. (21 − 8i) − (5 + 5i) p
−10 + −25
28.
12. (−10 − i) − (2 − 4i) −10
p
13. (2 − 3i) − (9 − 3i) −24 − −27
29.
−6
14. (6 − 11i) + (17i − 6) p
−4 + −28
15. (1 + i)(1 − 2i) 30.
6
459
1. COMPLEX NUMBERS CHAPTER 9. QUADRATIC EQUATIONS
Homework 9.1 Answers.
p
1. 5·i 17. 6 + 20i
3. 7i 19. −60 + 15i
p
5. 3 5 · i 21. 24
p
7. 2 30 · i 23. 15 + 8i
9. 21 + 5i 2
25. −2 + i
3
11. 16 − 13i p
27. −9 − 2·i
13. −7 p
3
15. 3 − 2i 29. 4 + i
2
460
CHAPTER 9. QUADRATIC EQUATIONS 2. SQUARE ROOT PROPERTY
2. Square Root Property
2.1. Square Root Property.
THEOREM 2.1.1. For every non-zero real or complex number z there exist precisely two
distinct complex
p numbers w such that w2 = z. One ofpthe numbers is the principal square
root of z, z, and the other number is its opposite − z.
BASIC EXAMPLE 2.1.1. If w2 = 25, then w = 5 or w = −5. In the case when the square is
positive, both solutions are real.
If w2 = −4, then w = 2i or w = −2i:
(2i)2 = 22 i 2
= 4(−1)
= −4
−2i can be verified as a solution as well. In this case there are no real solutions, but there
are two distinct complex solutions.
THEOREM 2.1.2 (Square Root Property). Some equations can be solved by taking square
roots of both sides. If X is an algebraic expression and r is a real number, then solving
the equation
X2 = r
amounts to combining the solution sets of equations
p
X= r
and p
X =− r
EXAMPLE 2.1.1. Solve the equation
x2 = 9
SOLUTION: Take square roots of both sides. Either
p
x = 9=3
or p
x =− 9 = −3
ANSWER: {−3, 3}
461
2. SQUARE ROOT PROPERTY CHAPTER 9. QUADRATIC EQUATIONS
EXAMPLE 2.1.2. Solve the equation
y 2 = 12
SOLUTION: Take square roots of both sides. Either
p p
y = 12 = 2 3
or p p
y = − 12 = −2 3
p p
We can state this solution set using the roster notation −2 3, 2 3 , but it is more tradi-
tional to use the plus-minus notation when the quadratic solutions are irrational and/or
complex.
p
ANSWER: ±2 3
DEFINITION 2.1.1. Many quadratic equations have irrational solution sets of the form
¦ p p ©
a + b, a − b
and many others have complex solutions sets of the form
{a + bi, a − bi}
In cases like that we will use a plus-minus notation in order to be more concise:
¦ p p © p
a + b, a − b = a± b
{a + bi, a − bi} = a ± bi
BASIC EXAMPLE 2.1.2. Some solution sets expressed two ways:
roster notation plus-minus notation
p p p
2 − 3 5, 2 + 3 5 2±3 5
{−6i, 6i} ±6i
p p p
7 − 2 3 · i, 7 + 2 3 · i 7±2 3·i
8 − 9i 8 + 9i 8 ± 9i
§ ª
,
5 5 5
462
CHAPTER 9. QUADRATIC EQUATIONS 2. SQUARE ROOT PROPERTY
2.2. Solving Equations (nx + k)2 = r.
EXAMPLE 2.2.1. Solve the equation (x + 1)2 = 25.
SOLUTION: By the square root property, either
Æ p
(x + 1)2 = 25
p
x +1 = 25
x +1 = 5
x = 4
or
Æ p
(x + 1)2 = − 25
p
x +1 = − 25
x +1 = −5
x = −6
The solutions are rational, so we state the solution set in roster notation:
ANSWER: {−6, 4}
EXAMPLE 2.2.2. Solve the equation (t + 2)2 = 5
SOLUTION: By the square root property, either
p
t +2 = 5
p
t = −2 + 5
or
p
t +2 = − 5
p
t = −2 − 5
When the solution set for a quadratic equation is irrational or complex, it is traditional to
state the answer p p
−2 − 5, −2 + 5
p
in the plus-minus form −2 ± 5
p
ANSWER: −2 ± 5
463
2. SQUARE ROOT PROPERTY CHAPTER 9. QUADRATIC EQUATIONS
EXAMPLE 2.2.3. Solve the equation (2 y − 3)2 = 9
SOLUTION: By the square root property, either
Æ p
(2 y − 3)2 = 9
p
2y − 3 = 9
2y − 3 = 3
2y = 6 added 3 to both sides
y = 3 divided both sides by 2
or
Æ p
(2 y − 3)2 = − 9
p
2y − 3 = − 9
2y − 3 = −3
2y = 0
y = 0
The solutions are rational, so we state the solution set in roster notation:
ANSWER: {0, 3}
2.3. Complex Solutions.
EXAMPLE 2.3.1. Solve the equation z 2 = −16
SOLUTION: By the square root property, either
p
z = −16
p
= 16 · i
= 4i
or
p
z = − −16
= −4i
ANSWER: ±4i
464
CHAPTER 9. QUADRATIC EQUATIONS 2. SQUARE ROOT PROPERTY
EXAMPLE 2.3.2. Solve the equation (x − 5)2 = −100
SOLUTION: By the square root property, either
p
x −5 = −100
x −5 = 10i
x = 5 + 10i
or
p
x −5 = − −100
x −5 = −10i
x = 5 − 10i
ANSWER: 5 ± 10i
465
2. SQUARE ROOT PROPERTY CHAPTER 9. QUADRATIC EQUATIONS
Homework 9.2.
2
Solve the equation by finding all real solu- 3 1
tions. 12. x− =
4 16
1. x 2 = 36
2. y 2 = 144
3. (x − 6)2 = 4 Solve the equation by finding all complex
solutions.
4. (x + 7)2 = 16
13. z 2 = −2
5. (4n − 5) = 1 2
14. z 2 = −10
6. (5n − 1) = 49 2
15. ( y + 2)2 = −25
7. (x − 3) = 282
16. ( y − 4)2 = −36
8. (x + 9) = 402
17. (z − 4)2 = −9
9. (−3z + 4) = 0 2
18. (2z + 5)2 = −49
10. (5z + 8) = 0 2
2 19. (x + 4)2 = −75
1 9
11. x+ =
2 4 20. ( y − 5)2 = −32
466
CHAPTER 9. QUADRATIC EQUATIONS 2. SQUARE ROOT PROPERTY
Homework 9.2 Answers.
1. {−6, 6} 11. {−2, 1}
3. {4, 8} p
13. ± 2 · i
3
§ ª
5. ,1
2 15. −2 ± 5i
p
7. 3 ± 2 7 17. 4 ± 3i
4
§ ª
9. p
3 19. −3 ± 5 3 · i
467
3. COMPLETING THE SQUARE CHAPTER 9. QUADRATIC EQUATIONS
3. Completing the Square
5
y
(x + 2)2 4
y=
2
3
y = 3 − x2
2
1
x
−6 −5 −4 −3 −2 −1 1 2 3 4
−1
−2
3.1. Completing the Square for x 2 + bx. Our goal for this chapter is to be able to solve
every quadratic equation, and we will be able to do so because every quadratic equation can
be put in a form (x + v)2 = r. The crucial step in that transformation is rewriting an expression
of the type x 2 + bx as a sum of a binomial square and a constant:
x 2 + bx = (x + v)2 − t
As it happens, there is an easy way to figure out v and t from any given b.
THEOREM 3.1.1 (Completing the Square). For any real or complex number b, the polyno-
mial x 2 + bx has an equivalent form
x 2 + bx = (x + v)2 − t
2
b b
where v = and t =
2 2
PROOF.
2 2
b b
(x + v) − t
2
= x+ −
2 2
2 2
b b
= x 2 + bx + −
2 2
= x + bx
2
468
CHAPTER 9. QUADRATIC EQUATIONS 3. COMPLETING THE SQUARE
EXAMPLE 3.1.1. Complete the square for the expression x 2 + 6x
SOLUTION: Using the formula from the theorem 3.1.1, b = 6, so in order to complete the
square we need to add and subtract t, which we compute as follows:
2 2
b 6
t= = = 32 = 9
2 2
and so
x 2 + 6x = x 2 + 6x + 9 − 9
= (x + 3)2 − 9
ANSWER: (x + 3)2 − 9
EXAMPLE 3.1.2. Complete the square for the expression x 2 − 14x
SOLUTION: Using the formula from the theorem 3.1.1, b = −14, so in order to complete
the square we need to add and subtract t, which we compute as follows:
2
−14 2
b
t= = = (−7)2 = 49
2 2
and so
x 2 − 14x = x 2 − 14x + 49 − 49
= (x − 7)2 − 49
ANSWER: (x − 7)2 − 49
EXAMPLE 3.1.3. Complete the square for the expression x 2 + 5x
SOLUTION: Using the formula from the theorem 3.1.1, b = 5, so in order to complete the
square we need to add and subtract t, which we compute as follows:
2 2
b 5 25
t= = =
2 2 4
469
3. COMPLETING THE SQUARE CHAPTER 9. QUADRATIC EQUATIONS
and so
25 25
x 2 + 5x = x 2 + 5x + −
4 4
2
5 25
= x+ −
2 4
2
5 25
ANSWER: x+ −
2 4
3.2. Solving Quadratic Equations by Completing the Square.
EXAMPLE 3.2.1. Solve the equation x 2 − 8x + 2 = 0
SOLUTION: Complete the square for the binomial x 2 − 8 first. Since we are dealing with
an equation, we can add the constant required to complete the square to both sides. In
this case, the coefficient for the linear term is −8, and so the constant to complete the
square is
2
−8
= (−4)2 = 16
2
x 2 − 8x + 2 = 0 constant 2 belongs on the right
x 2 − 8x + 2 − 2 = 0−2
2
x − 8x = −2
x − 8x + 16
2
= −2 + 16 add 16 to both sides to complete the square
(x − 4)2 = 14 complete the square and simplify the right side
Now we apply the square root property to get the answers. Either
Æ p
(x − 4)2 = 14
p
x −4 = 14
p
x = 4 + 14
or
Æ p
(x − 4)2 = − 14
p
x −4 = − 14
p
x = 4 − 14
p
ANSWER: 4 ± 14
470
CHAPTER 9. QUADRATIC EQUATIONS 3. COMPLETING THE SQUARE
EXAMPLE 3.2.2. Solve the equation x 2 + 6x + 10 = 0
SOLUTION: Complete the square for the binomial x 2 + 6x first. Since we are dealing with
an equation, we will add the constant required to complete the square to both sides. In
our case, the coefficient for the linear term is 6, so the constant to complete the square is
2
−6
= (−3)2 = 9
2
x 2 + 6x + 10 = 0 the constant 10 belongs on the right
x 2 + 6x + 10 − 10 = 0 − 10
x 2 + 6x = −10
x + 6x + 9
2
= −10 + 9 add 9 to both sides to complete the square
(x + 3)2
= −1
Now we apply the square root property to get the answers. Either
p
x +3 = −1
x +3 = i
x = −3 + i
or
p
x +3 = − −1
x +3 = −i
x = −3 − i
As always in an equation, we can check each answer by comparing the values of the two
sides of our equation. If x = −3 + i, then
x 2 + 6x + 10 = (−3 + i)2 + 6(−3 + i) + 10
= 9 − 6i + i 2 − 18 + 6i + 10 distributed
= 1+i 2
combined like terms
= 0 i 2 = −1
If x = −3 − i, then
x 2 + 6x + 10 = (−3 − i)2 + 6(−3 − i) + 10
= 9 + 6i + i 2 − 18 − 6i + 10 distributed
= 1+i 2
combined like terms
= 0 i 2 = −1
ANSWER: −3 ± i
471
3. COMPLETING THE SQUARE CHAPTER 9. QUADRATIC EQUATIONS
EXAMPLE 3.2.3. Solve the equation 3x 2 − x − 6 = 0
SOLUTION: Before we can complete the square, we need to divide both sides by the
leading coefficient 3:
1
x2 − x − 2 = 0
3
Now we complete the square by adding
2 2
1 1 1
− ÷2 = − =
3 6 36
to both sides.
1
x2 − x = 2 constant belongs on the right
3
1 1 1
x2 − x+ = 2+
3 36 36
1 2 72 1
x− = + found LCD on the right
6 36 36
1 2 73
x− =
6 36
Now we can apply the square root property. This time we will introduce the plus-minus
notation right into the equation, saving ourselves some time:
v v
1 2
u t 73
= ±
t
x−
6 36
v
1 t 73
x− = ±
6 36
p
1 73
x− = ±
6 6
p
1 73
x = ±
6 6
p
1 ± 73
x =
6
p
1± 73
ANSWER:
6
472
CHAPTER 9. QUADRATIC EQUATIONS 3. COMPLETING THE SQUARE
Homework 9.3.
Complete the square for the given expres- 9. x 2 − 24x + 21 = 0
sion.
10. a2 − 10a + 20 = 0
1. x 2 + 8x
11. b2 − 8b + 16 = 0
2
2. x − 10x
12. c 2 + 10c + 25 = 0
3. x 2 − 20x
13. a2 − 3a − 10 = 0
4. x + 18x
2
14. x 2 − 3x − 4 = 0
5. x 2 + 13x
15. 5x 2 + 6x + 1 = 0
2
6. x − 3x
16. 2b2 − 7b − 10 = 0
Solve the given equation by completing the Find complex solutions for the equation by
square. completing the square.
7. x 2 + 2x − 4 = 0 17. x 2 − 6x + 25 = 0
8. x 2 + 4x − 5 = 0 18. x 2 + 2x + 5 = 0
473
3. COMPLETING THE SQUARE CHAPTER 9. QUADRATIC EQUATIONS
Homework 9.3 Answers.
1. (x + 4)2 − 16 11. {4}
3. (x − 10)2 − 100
13. {−2, 5}
5. (x + 6.5) − 42.25
2
p 15. {−0.2, −1}
7. −1 ± 5
p
9. 12 ± 123 17. 3 ± 4i
474
CHAPTER 9. QUADRATIC EQUATIONS 4. QUADRATIC FORMULA
4. Quadratic Formula
THEOREM 4.1.1 (Quadratic Formula). Let a, b, and c be real numbers, a 6= 0, and consider
the quadratic equation in one variable x in standard form:
ax 2 + bx + c = 0
where a is the coefficient of the quadratic term, b is the coefficient of the linear term, c
is the constant term. Then the solutions of the equation can be computed with
p p
−b + b2 − 4ac −b − b2 − 4ac
x1 = and x2 =
2a 2a
2
Moreover, the radicand b − 4ac is known as the discriminant. If the discriminant is
positive, then there are two distinct real solutions; if b2 − 4ac = 0, then there is a unique
real solution; and if b2 − 4ac < 0, then there are two distinct complex solutions.
PROOF. Deriving the formula amounts to solving this equation by completing the square:
a x 2 + bx + c = 0
a x 2 + bx = −c moved the constant to the right
bx −c
x2 + = divided both sides by the leading coefficient a
a a
bx b2 b2 c b2
x2 + + 2 = 2
− completed the square with
a 4a 4a a 4a2
b 2 b2 4ac
x+ = − found common denominator on the right
2a 4a2 4a2
b 2 b2 − 4ac
x+ = subtracted fractions on the right
2a 4a2
v
b t b2 − 4ac
x+ = ± applied the square root property
2a 4a2
p
b b2 − 4ac p
x+ = ± p distributed over the fraction
2a 4a2
p
b b2 − 4ac
x+ = ± simplified denominator
2a 2a
p
b b2 − 4ac
x = − ± isolated x on the left
2a 2a
p
−b ± b2 − 4ac
x = quadratic formula in plus-minus notation
2a
475
4. QUADRATIC FORMULA CHAPTER 9. QUADRATIC EQUATIONS
EXAMPLE 4.1.1. Solve the equation x 2 − 3x + 2 = 0.
SOLUTION: a = 1, b = −3, and c = 2, so the discriminant
b2 − 4ac = (−3)2 − 4 · 1 · 2 = 9 − 8 = 1,
and hence the solutions are
p
−(−3) + 1
x1 = =2
2·1
p
−(−3) − 1
x2 = =1
2·1
Just like for all equations, it is very easy to check the answer. If x 1 and x 2 are the solutions,
then substituting them for x in the original equation should produce identities:
22 − 3 · 2 + 2 = 0
4−6+2 = 0
0 = 0
12 − 3 · 1 + 2 = 0
1−3+2 = 0
0 = 0
ANSWER: {1, 2}
EXAMPLE 4.1.2. Solve the equation 4x 2 − 1 = 0.
SOLUTION: a = 4, b = 0, and c = −1, so the discriminant
b2 − 4ac = 02 − 4 · 4 · (−1) = 16
and hence the solutions are
p
−0 + 16 4 1
x1 = = =
2·4 8 2
p
−0 − 16 −4 1
x2 = = =−
2·4 8 2
ANSWER: {−1/2, 1/2}
476
CHAPTER 9. QUADRATIC EQUATIONS 4. QUADRATIC FORMULA
EXAMPLE 4.1.3. Solve the equation
x(x + 8) = 2x − 9
SOLUTION: First we have to bring the equation into the standard form:
x(x + 8) = 2x − 9
x + 8x
2
= 2x − 9
x + 8x − 2x + 9
2
= 0
x 2 + 6x + 9 = 0
a = 1, b = 6, and c = 9, and so the discriminant
b2 − 4ac = 36 − 4 · 1 · 9 = 36 − 36 = 0
and the only solution is p
−6 ± 0
x1 = x2 = = −3
2·1
ANSWER: {−3}
4.2. Complex Solutions.
EXAMPLE 4.2.1. Solve the equation
x 2 − 2x + 2 = 0
SOLUTION: a = 1, b = −2, and c = 2, so the discriminant
b2 − 4ac = (−2)2 − 4 · 1 · (2) = 4 − 8 = −4
Since −4 is negative, there are no real solutions, but we can still apply the quadratic
formula to obtain the complex solutions:
p
−(−2) ± −4 2 ± 2i
x1 = = =1±i
2·1 2
ANSWER: 1 ± i
477
4. QUADRATIC FORMULA CHAPTER 9. QUADRATIC EQUATIONS
Homework 9.4.
Use the quadratic formula to find all real 12. y 2 − 10 y = −25
solutions.
13. −9x 2 + 6x + 8 = 0
1. x 2 − 5x + 6 = 0
14. 2x(x + 3) = 5(x + 2)
2. x 2 + 3x − 40 = 0
15. x 2 − 4x + 7 = 0
3. x 2 − 2x − 3 = 0
16. x 2 + 6x − 2 = 0
4. y − 8 y + 20 = 0
2
5. x 2 − 7x = 0 Use the quadratic formula to find complex
solutions for each equation.
6. 6x 2 = −3x
17. x 2 + 1 = 0
7. x 2 − 9 = 0
1 18. −3x 2 = 75
8. 5 y =2
5
19. x 2 − 6x + 13 = 0
9. 3x 2 − 7x + 2 = 0
20. x 2 + 4x + 8 = 0
10. 3x 2 + 7x + 4 = 0
21. z 2 + 2z + 3 = 0
11. x 2 + 8x + 16 = 0
22. −x 2 + 7x − 12 = 0
478
CHAPTER 9. QUADRATIC EQUATIONS 4. QUADRATIC FORMULA
Homework 9.4 Answers.
2 4
§ ª
1. {2, 3}
13. − ,
3 3
3. {−1, 3}
p
5. {0, 7} 15. 2 ± 3·i
7. {−3, 3} 17. ±i
1
§ ª
9. ,2 19. −3 ± 2i
3
p
11. {4} 21. −1 ± 2·i
479
4. PRACTICE FINAL CHAPTER 9. QUADRATIC EQUATIONS
Practice Final
1. Solve the formula for y: 9. Solve the system:
17x + 20 y = 26
4 y + 3x = 5 − y
−17x + 10 y = 64
2. Solve the inequality
10. Solve the system:
2(x − 3) ≥ 5x + 12 2(x − 1) = 6 y
3y − x + 2 = 0
3. Frodo’s journey to Mount Doom was
three times longer than his journey back
home, and the whole trip, there and back
again, took 240 days. Find the duration of Simplify the expression and state the an-
the journey to the Mount Doom and the du- swer in scientific notation:
ration of the homeward journey.
42 × 10−2
11.
4. One week Alice worked 10 hours wait- 5 × 10−4
ing tables and 3 hours editing a book, and
12. (5.5 × 103 )(3.4 × 10−5 )
she made $195. Next week she waited ta-
bles for 5 hours and spent 5 hours editing a
book, and she made $185. Find the hourly
wage for each job. Simplify the expression and state the an-
swer with positive exponents only:
5. Find an equation of the line with slope −2
1 x 3 y −2
− and y-intercept (0, −2). State the an- 13.
3 x −1 y −4
swer in the slope-intercept form.
14. (−3x −5 y 7 )(−5x 0 y −4 )
6. Find an equation of the line with slope
−2 that passes through the point (−1, 3).
State the answer in the slope-intercept
form. Factor the expression completely:
7. Find the slope and an equation of the 15. z 4 − 81 y 4
line that passes through the points (1, −5)
and (3, −1). State the answer in the point- 16. 8x − 4 + 6x 3 − 3x 2
slope form.
8. Solve the system of linear equations by
graphing:
y+x = 3
2 y − x = −6
480
CHAPTER 9. QUADRATIC EQUATIONS 4. PRACTICE FINAL
Solve the equation by factoring: 26.
p
x −3+6=4
17.
6x 2 = −10x 27.
p p
5 x = x +2
18.
100 − 4x 2 = 0 28.
p
19. 2x + 5 − 1 = x
x 3 − 2x 2 = 24x
Simplify the expression: 29. Solve the equation by taking square
roots of both sides:
p
20. 100x 2 (x + 3)2 = 20
p
3
21. x 24 30. Solve the equation by completing the
p square:
22. 45x 2
x 2 + 18x + 1 = 0
Æ
23. 242x 6 y 7
p
8x 7
24. p Solve the equation by using the quadratic
2x y 2 formula:
31.
Solve the equation: x 2 − 4x + 1 = 0
25. 32.
7 x
= 4x 2 − 4x − 35 = 0
x +3 4
481
4. PRACTICE FINAL CHAPTER 9. QUADRATIC EQUATIONS
Practice Final Answers.
5 − 3x 11. 8.4 × 102
1. y =
5
12. 1.87 × 10−1
2. (−∞, −6]
1
13.
−6 y4 x8
3. 15 y 3
x and y are the durations for going there 14.
x5
and back again respectively, in days
x = 3y 15. (z + 3 y)(z − 3 y)(z 2 + 9 y 2 )
x + y = 240 16. (4 + 3x 2 )(2x − 1)
solution: x = 180, y = 60
17. {0, −5/3}
4.
x and y are the wages for waiting and edit- 18. {−5, 5}
ing respectively, in dollars per day
19. {0, −4, 6}
10x + 3 y = 195
5x + 5 y = 185 20. 10x
solution: x = 12, y = 25
21. x 8
1 p
5. y = − x − 2
3 22. 3x 5
6. y = −2x + 1 p
23. 11x 3 y 3 2y
7. y + 5 = 2(x − 1)
2x 3
8. (4, −1) 24.
y
5
y 25. {−7, 4}
4
3 26. ∅
2
1 27. {1/4}
x
−5 −4 −3 −2 −1 1 2 3 4 5 28. {2}
−1
−2 p
29. −3 ± 2 5
−3
p
−4 30. −9 ± 4 5
−5 p
31. 2 ± 3
9. (−2, 3)
10. no solution 32. {−2.5, 3.5}
482
APPENDIX A
To Instructors
1. Lecture Notes
1.1. Concepts.
1. Expressions
Define: expressions, equivalent expressions, equations, relations, substitution, sum of terms,
product of factors, coefficient. Show how to split sums into terms, and products into factors.
2. Axioms
Stop before identity to show that terms of sums can be added in any order, and factors of
a product can be multiplied in any order. Stop before distributivity to discuss opposites and
reciprocals. Show distributivity both ways.
EXAMPLE 1.1.1. Rewrite the expression without parentheses and then simplify by appeal-
ing to the axioms: x + 3(2 + x)
SOLUTION:
equivalent expressions axiom used for substitution
x + 3(2 + x)
x + (3 · 2 + 3 · x) distributivity
x + (3 · x + 3 · 2) commutativity of +
x + (3 · x + 6) closure
(x + 3 · x) + 6 associativity of +
(1 · x + 3 · x) + 6 identity of ·
(1 + 3)x + 6 distributivity
4x + 6 closure
Define subtraction and division, show sums with negative terms and products with reciprocal
factors.
483
1. LECTURE NOTES CHAPTER A. TO INSTRUCTORS
3. Sets
Define sets, roster, integers: positive and non-negative.
Show the number line and the addition/subtraction of negative numbers.
Define set builder, show odd numbers or something like that:
2k + 1 k is an integer
Show how to list a few members of a set
2
m m is a positive integer
Define rationals and reals, mention irrational numbers.
Show absolute value and inequality relations.
4. Reals
Show products of negative numbers, opposite of a sum, positive integer exponent, order of
operations, like terms.
5. Primes
Define primes, show prime factorizations
Show cancellation of common factors in fractions, define fractions in lowest terms.
6. Fractions
Show reciprocal of a product, rewriting a fraction as a product, multiplying fractions with
invisible parentheses, applications to areas, dividing fractions, and changing the denominator.
7. LCD
Show distributivity with common denominator, define LCM and LCD.
Show rational addition and mixed numbers.
8. Translation
Show the four arithmetic patterns, mention combinations like “X is 10 units greater than 4
times y”.
484
CHAPTER A. TO INSTRUCTORS 1. LECTURE NOTES
1.2. Linear Equations.
1. Properties
Define solution, equivalent equations, show addition and multiplication properties.
EXAMPLE 1.2.1. Ivan’s math class has 1.2 times more students than Ivan’s statistics class.
Find how many people are enrolled in the statistics class has if the math class has 30
students.
SOLUTION: Let m and s be the enrollment figures for math and stat class respectively
(units are students). The statement
math class has 1.2 times more students than statistics class
can be translated as
m = 1.2s
Since we are given that the math class has 30 students, we can substitute 30 for m in the
equation and solve for s:
30 = 1.2s
25 = s
So the stats class has 25 students enrolled.
ANSWER:
m and s count students enrolled in math and stats class respectively,
equation: m = 1.2s
solution: s = 25
2. Solving
Define linear equations. Show how to solve by isolating the variable: a simple case, with
parentheses, and with fractions.
EXAMPLE 1.2.2. Bart, Millhouse, and Martin Prince saved up 149 dollars for a collectible
comic book. Find how much money each one saved up if Bart contributed 3 times less
than Millhouse, and Martin contributed 9 dollars more than Millhouse.
SOLUTION: Let b, m, and p be the amounts contributed by Bart, Millhouse, and Martin
Prince, in dollars. We can translate the statements as
b+m+p = 149 together they saved up $149
b = m÷3
p = m+9
485
1. LECTURE NOTES CHAPTER A. TO INSTRUCTORS
We can substitute expressions for b and p into the first equation and then solve for m:
b+m+p = 149
(m ÷ 3) + m + (m + 9) = 149
m + 3m + 3m + 27 = 447
7m = 420
m = 60
We can find b and p now using the other two equations which are solved for these vari-
ables:
b = m ÷ 3 = 60 ÷ 3 = 20
p = m + 9 = 60 + 9 = 69
ANSWER:
b, m, and p be the amounts contributed by Bart, Millhouse, and Martin Prince, in dollars,
b + m + p = 149
b = m÷3
p = m+9
solution: b = 20, m = 60, p = 69
3. Formulas
Show how to solve linear formulas for a variable. Examples to be solved for y:
3x − 4 y = x −4
2(x − y) = 4 y + 10x − 1
7x y 4
=
3 b
8y + 2 = 4(x − x y)
4. Percent
Define percent, show decimal representation, show three easy translation patterns. Define
percent increase and decrease, show basic examples and applications.
EXAMPLE 1.2.3. A quantity increased by 14% and is now 171. Find the quantity before
the increase.
486
CHAPTER A. TO INSTRUCTORS 1. LECTURE NOTES
SOLUTION:
171 = P(1 + 0.14)
150 = P
EXAMPLE 1.2.4. A quantity decreased from 140 to 133. Find the percent decrease.
SOLUTION:
k
133 = 140 − · 140
100
k
−7 = − · 140
100
7 · 100
= k
140
5 = k
ANSWER: 5%
EXAMPLE 1.2.5. John paid 24% income tax on the total income of $28, 000. Find how
much money John has made after tax, and the amount he paid to the government.
ANSWER: John has made $21, 280 after paying $6, 720 tax.
EXAMPLE 1.2.6. A contractor buys a residential property, renovates it, and then sells (flips)
it with 25% markup. If the the sale price is $160, 000, find how much the contractor has
paid for the property, as well as his profit from flipping it.
ANSWER: The original price was $128, 000, and the profit was $32, 000.
5. Applications
Show the work equation and give few basic examples of assigning/comparing units:
• distance = speed · duration
• paycheck = hourly rate · duration
• volume of water = pumping rate · duration
• total price = unit price · quantity
487
1. LECTURE NOTES CHAPTER A. TO INSTRUCTORS
EXAMPLE 1.2.7. Ivan presented 9 examples over 2 hours and 15 minutes. Describe vari-
ables and units in
w = rt
and find the rate.
SOLUTION: w is the number of examples presented, t is duration in hours, r is the rate in
examples per hour. Note that 15 minutes is 0.25 of an hour.
9 = r · 2.25
4 = r
So the rate is 4 examples per hour.
EXAMPLE 1.2.8. Shannon’s business serves 14 clients per day on average. Describe vari-
ables and units in
w = rt
and find the time needed to serve 518 clients?
SOLUTION: The rate is in clients per day, so the measure of “work” is just the number of
clients, and the units of time are days.
518 = 14t
37 = t days
EXAMPLE 1.2.9. Two cars, a Lotus and a Renault, are 635 kilometers apart. They start at
the same time and drive toward each other. The Lotus travels at a rate of 70 kilometers
per hour and the Renault travels at 57 kilometers per hour. In how many hours will the
two cars meet?
SOLUTION:
70t + 57t = 635 distance equation
127t = 635
t = 5 hours
EXAMPLE 1.2.10. Jane’s monthly car loan payment is one quarter of her monthly house
loan payment. Find each monthly payment rate if Jane pays 19200 dollars in total over
the course of the year.
488
CHAPTER A. TO INSTRUCTORS 1. LECTURE NOTES
SOLUTION: c and h are payment rates for the car and the house respectively, in dollars
per month.
c = h÷4 car payment is a quarter of house payment
c · 12 + h · 12 = 19200 “work” equation
Solve by substitution:
12(h ÷ 4) + 12h = 19200 substitute h/4 for c
3h + 12h = 19200
15h = 19200
h = 1280 house payment
Back to the first equation to find c:
c = 1280/4
c = 320 car payment
6. Inequalities
Define linear inequalities and solutions. Show graphs and interval notation. Show addition
and multiplication properties, examples.
1.3. Graphing.
1. Reading
Define ordered pair, coordinate plane, graph of a set of points. Plot/read some points, show
axis scaling, show how to find intercepts and extrema.
2. Linear Equations
Define solution set. Show some fun solution sets, like y = x 2 or y = x 3 . Show vertical and
horizontal lines. Show how to graph linear equations which can be solved for a variable by
finding 2 points.
3. Intercepts
Define intercepts, show how to find them from equations, and how to use them for plotting.
Mention vertical and horizontal lines.
4. Slope
Define grade, slope of a segment, slope of a line. Mention positive and negative slopes. Show
how to plot using a rational slope.
489
1. LECTURE NOTES CHAPTER A. TO INSTRUCTORS
5. Slope-intercept
Define slope-intercept form. Show how to read, write, and to solve for y to get it. Define
parallel and perpendicular lines, show how to compare given lines using their slopes, mention
vertical and horizontal lines.
6. Point-slope
Define point-slope form, show how to read. Show how to construct an equation of a line given
slope and point, and also given two points. Show how to use point-slope form for plotting on
the grid.
7. Function Notation
This section is optional in the sense that nothing else in the text depends on it.
Define functions and function notation, show examples of use.
EXAMPLE 1.3.1. Acme CO’s profit in 2000 is 4.7 billion dollars, and 5.3 billion dollars in
2015. Let x be the year since 2000, and let y(x) be the profit in the corresponding year,
in billions of dollars. Find a linear model y(x), use the model to predict the profit in
2018, and to predict the year when the profit is 6 billion dollars.
SOLUTION:
m = 0.6/15 = 0.04 slope
y(x) = 0.04x + 4.7 model
y(18) = 5.42 billion dollars
6 = 0.04x + 4.7
1.3 = 0.04x
32.5 = x approximately year 2032
1.4. Systems.
1. Graphing
Define the solution for a system with 2 equations and 2 variables. Show the 3 cases for the
shape of the solution set. Show how to solve by graphing:
2x + 4 y = 4
(4, −1)
2x − y = 9
y+x = 3
no solution
2 y = 7 − 2x
490
CHAPTER A. TO INSTRUCTORS 1. LECTURE NOTES
2. Substitution
Solve by substitution:
y = 5x + 13
(−2, 3)
4x − 3 y = −17
3 y − 2x = −4
(−4, −4)
4x + 5 y = −36
3. Elimination
Show multiplication and addition properties for systems. Solve by elimination:
5x − 6 y = 11
(7, 4)
5x + 6 y = 59
3x − 7 y = 11
(−1, −2)
9x + 5 y = −19
4. Applications
EXAMPLE 1.4.1. Every Monday Helen shares coffee and donuts with the office where she
works. One week Helen buys 4 cups of coffee and 6 donuts, and pays 18 dollars. The
next week Helen buys 6 cups of coffee and 12 donuts, and pays 30 dollars. Find the price
of one cup of coffee and the price of one donut.
SOLUTION: Let c and d be the price of one cup of coffee and the price of one donut
respectively, in dollars.
4c + 6d = 18
6c + 12d = 30
ANSWER: (3, 1)
EXAMPLE 1.4.2 (Volume Mixing). A chemist needs to mix an 18% acid solution with a
45% acid solution to obtain 12 liters of a 36% solution. How many liters of each of the
acid solutions must be used?
491
1. LECTURE NOTES CHAPTER A. TO INSTRUCTORS
SOLUTION: Let x and y be amounts of solutions in liters.
x + y = 12
0.18x + 0.45 y = 0.36(12)
EXAMPLE 1.4.3 (Value Mixing). How many pounds of chamomile tea that cost $9 per
pound must be mixed with 8 pounds of orange tea that cost $6 per pound to make a
mixture that costs $7.80 per pound?
SOLUTION: Let c be the weight of chamomile tea in pounds, and then
9c + 6(8) = 7.8(c + 8)
Alternatively, we can also let t be the total weight of the mix, and write a system of two
equations:
c+8= t
9c + 6(8) = 7.8t
5. Inequalities
Define the solution for an inequality in 2 variables, describe solution sets, show examples,
mention vertical and horizontal lines.
1.5. Polynomials.
1. Exponent
Define integer exponent, mention base 1. Show product rule and quotient rule. Mention sums
as bases: (x + y)n
2. Properties
Show power rule and distributivity over products and fractions.
3. Polynomials
Define monomial and its simplified form, monomial degree, monomial degree names, polyno-
mial and its standard form, leading coefficient, polynomial names, polynomial degree. Show
how to combine like terms and how to evaluate polynomials.
4. Sums
Show how to add and subtract polynomials by combining like terms.
492
CHAPTER A. TO INSTRUCTORS 1. LECTURE NOTES
5. Products
Show how to multiply monomials, monomial times polynomial, and polynomial times polyno-
mial (show the box).
6. Special Products
Show the difference of squares both ways, and the square of a binomial.
7. Quotients
Show how to divide by a monomial, and the polynomial long division.
8. Negative Exponent
Mention definition again, show how to get rid of negative exponents in products and fractions.
Define scientific notation.
1.6. Factoring.
1. GCF
Define irreducible and prime polynomials. Show polynomial factorizations. Mention that lin-
ear polynomials are irreducible.
Define GCF for integers and GCF for monomials.
2. Grouping
Show factoring by grouping.
x y + 2 y + 3x + 6 4x 3 + 2x 2 + 2x + 1
24x 3 − 6x 2 + 8x − 2 3x 3 − 6x 2 − 15x + 30
3. Diamond
Show the diamond pattern. Show the basic factoring strategy combining monomial GCF, group-
ing, and diamond.
4. Trinomials
Show how to factor a trinomial.
6z 2 + 11z + 4 −4h2 + 11h + 3
2n2 − 13n − 7 3x 2 + x y − 14 y 2
4r 2 + 11rs − 3s2 5x 2 + 14x − 3
493
1. LECTURE NOTES CHAPTER A. TO INSTRUCTORS
5. Special Products
Show how to factor special products.
6. Strategy
List the types of irreducible polynomials: linear, prime trinomials, and sums of squares. Present
the factoring strategy, combining GCF, grouping, difference of squares, diamond, and trinomial
pattern.
6x 4 − 13x 3 + 5x 2 5x 5 − 20x
4x 2 y − 9 y + 8x 2 − 18 36x 4 − 15x 3 − 9x 2
7. Equations
Show the zero product property and how to solve non-linear polynomial equations.
3x 2 − 12x = 0
x + 5x − 9x − 45
3 2
= 0
3x 5 − 6x 4 + 3x 3 = 0
6x 3 − 4x 2 − 16x = 0
2x 3 + 10x − 3x 2 − 15 = 0
8. Applications
Show applications to areas of rectangles and triangles.
1.7. Rational Expressions.
1. Simplifying
Define rational expressions. Show how to find which values of a variable make the expressions
undefined. Show how to simplify by canceling common polynomial factors.
6x 3 − 12x 2 − 18x
3x 4 + 6x 3 + 3x 2
2. Products
Show how to multiply and divide rational expressions.
30x 2 − 15x x 2 − 49 x −4 x 2 − x − 12
· ÷
6x − 42 20x 2 − 20x + 5 x 2 − 9 x 2 + 6x + 9
3. LCD
Define common multiple, LCM, and LCD. Show how to find LCMs and LCDs.
494
CHAPTER A. TO INSTRUCTORS 1. LECTURE NOTES
4. Sums
Show how to simplify sums with the same denominator, and sums with different denominators.
x2 25 2 t −2 2x 2 x 1
− − 2 + −
x −5 x −5 t +1 t − t −2 x −4 x −2 x +2
2
5. Complex Fractions
Define complex fractions, show how to simplify them.
9 y x 1
3+ − +4
x −3 x y x
12 1 1 2
4+ + 3+ 2
x −3 y x x
6. Equations
Define extraneous solutions. Show how to detect extraneous solutions for equations involving
rational expressions. Show how to solve the equations.
6 −3
+9 =
x −3 x −3
3 1 2
= − 2
x2 − 2x x −2 x
x 2 5
+ 2 =
x + 2 x + 5x + 6 x +3
x 7x 3
= −
x −5 x 2 − 3x − 10 x + 2
5 4 1
= + 2
n3 + 5n2 n+5 n
1.8. Radicals.
1. Intro
Define principal square root. Mention that the square root of a negative number is not real,
but complex. Show how the square root cancels the even exponent. Define principal nth root
and show how it cancels the exponent which is multiple of n. Mention sums as radicands.
2. Products
495
1. LECTURE NOTES CHAPTER A. TO INSTRUCTORS
Show distributivity over multiplication. Use it to simplify the radicand involving numbers as
well as variables: p p
24 a7
p p p p p
5x · 5x 3 6(2 3 + 3 6)
3. Quotients
Show distributivity over fractions. Define rationalized denominator. Show how to simplify
fractions and rewrite them with rationalized denominator:
p p
1 8 40x
p p p
5 2y 2x 2
4. Sums
Define radical like terms. Show how to simplify sums by simplifying radicals and combining
like terms: p p p p
3 + 27 (3 − 2)(5 − 2 2)
p p
Define radical conjugates A B ± C D. Show how to rationalize a denominator with a sum of
two terms.
5. Equations
Show that squaring both sides produces an almost equivalent equation. Show how to check
for extraneous solutions:
p
2x + 1 = x − 1, check x = 4 and x = 0
Show how to solve equations:
p
7− 3x + 7 = 3
p
2 + 6 2x 2 − 3x + 5 = 0
p
4 + x 2 − 4x = x
Æ p
x + x −1+2 = 3
p p
3x − 5 = 2− x −1
6. Applications
Show the distance formula and applications to right triangles.
7. Rational Exponent
Define rational exponent and show why it has to be in lowest terms before it can be evaluated.
Show how to change expressions from radical to rational exponent form and back. Show
how to evaluate expressions. Show how to simplify expressions involving rational exponents.
Optionally, define the imaginary unit and show how the distributivity and the exponent of
exponent property fail for negative bases.
496
CHAPTER A. TO INSTRUCTORS 1. LECTURE NOTES
1.9. Quadratics. It is possible to skip introducing complex numbers, if so desired, by skip-
ping the first section. The homework in the sections that follow is divided into two parts: first
part with real solutions, and the second part with complex solutions, so one can easily assign
one or the other or both topics to practice.
1. Complex Numbers
Simplify square roots of negative
p numbers, sums and products of complex numbers, and ex-
−b ± D
pressions of the form
2a
2. Square Root Property
State that every w2 = z has two solutions for non-zero z. Show the square root property and
the plus-minus notation. Show how to solve equations (nx + k)2 = r.
3. Completing The Square
Show how to complete the square x 2 + bx as well as ax 2 + bx. Show how to solve equations
using this method.
4. Quadratic Formula
Derive the quadratic formula by completing the square in
ax 2 + bx + c = 0
Show a few examples.
497
APPENDIX B
To Developers
1. Contributing
1.1. Submitting Contributions. If you would like to suggest an edit, please do not hesitate
to contact the maintainer:
"Ivan G. Zaigralin" <melikamp@melikamp.com>
Both this text and its source are licensed under free and libre licenses of the copyleft type,
meaning that anyone can use them in any way whatever, up to and including distributing
modified copies, as long as they use the same licenses for derivative works. All your contri-
butions automatically assume the same project license, but nontrivial LATEX source copy may
qualify you for becoming a co-author and a consequent copyright assignment.
Awesome ideas about content, structure, and appearance are welcome, and so is constructive
criticism. If you want more exercises or examples of a certain type, we urge you strongly to
do as much work as possible, and describe the statements, the solutions, and the answers in
detail, so that we can typeset your idea with minimal research. And if you happen to know
LATEX and opt to submit code contributions which are formatted as described below, you should
be prepared to be listed as a co-author.
1.2. Document Hierarchy. Standard book template hierarchy is used, with chapters, sec-
tions, and subsections. The chapter source files are included from basic-algebra.tex. Each
chapter file contains the title and the list of included sections. Each section source lives in its
own file. When working on a specific section, it is possible to greatly speed up the build time
by commenting out other chapters in basic-algebra.tex, and other sections in the corre-
sponding chapter source file. The chapter title is printed on the same page as the first section
in that chapter, so do not forget to comment out the title as well when working on sections
other than the first, or else the document will re-flow incorrectly.
499
2. DOCUMENT FORMATTING GUIDELINES CHAPTER B. TO DEVELOPERS
2. Document Formatting Guidelines
2.1. Mathematical Notions. We use fairly vanilla LATEX throughout the text, only defining
special commands when we really need to enforce the way a particular concept is typeset.
• Interval graphs
\InfOpen{17}
17
\OpenInf{17}
17
\InfClosed{17}
17
\ClosedInf{17}
17
• Interval notation
\opop{1,2}
(1, 2)
\opcl{1,2}
(1, 2]
\clop{1,2}
[1, 2)
\clcl{1,2}
[1, 2]
• Large delimiters
\pipes{x+\sqrt x}
p
x+ x
\brackets{x+y}
[x + y]
\parens{\frac1x + 3}
1
+3
x
500
CHAPTER B. TO DEVELOPERS 2. DOCUMENT FORMATTING GUIDELINES
• Mixed number
\mixed{3}{4}{5}
3 45
• Set builder
\setb{2k}{k\in A}
2k k ∈ A
• Set roster
\aggr{1,2,3}
{1, 2, 3}
• Complex fractions set in normal font size
\compfrac{\frac 12 + \frac 34}{\frac1x - \frac2x}
1 3
+
2 4
1 2
−
x x
• System of two equations
\twoEqSystem{x + y}{2x + 2}{y}{4x - 5}
x + y = 2x + 2
y = 4x − 5
2.2. Sets of Equations and Expressions. Equivalent expressions and relations are typeset
using align, with commands defined to make the relation appear nice:
\begin{align*}
x - 1 &\ee 5 &\ecomment{ecomment is for annotations in equations}\\
x &\ee 6 &\ecomment{ee command makes equal sign with space around it}
\end{align*}
x −1 = 5 ecomment is for annotations in equations
x = 6 ee command makes equal sign with space around it
\begin{align*}
x - 1 &\qq\geq 5\\
x &\qq\geq 6 &\ecomment{qq command inserts space around the argument}
\end{align*}
x −1 ≥ 5
x ≥ 6 qq command inserts space around the argument
501
2. DOCUMENT FORMATTING GUIDELINES CHAPTER B. TO DEVELOPERS
2.3. Frames. There are several mdframed environments defined, and should be used con-
sistently throughout.
\begin{axiom}[This is an axiom]
This environment is used for the real and complex number axioms only.
\end{axiom}
AXIOM 2.3.1 (This is an axiom). This environment is used for the real and complex number
axioms only.
\begin{definition}[This is a definition]
This environment is used for definitions.
\end{definition}
DEFINITION 2.3.1 (This is a definition). This environment is used for definitions.
\begin{theorem}[This is a true fact]
This environment is used for all general enough true statements
which can be proven from the axioms and the logical principles.
\end{definition}
THEOREM 2.3.1 (This is a true fact). This environment is used for all general enough true
statements which can be proven from the axioms and the logical principles.
\begin{trivia}[This is an easy example]
This environment is used for examples which are trivial enough.
\end{trivia}
BASIC EXAMPLE 2.3.1 (This is an easy example). This environment is used for examples
which are trivial enough.
502
CHAPTER B. TO DEVELOPERS 2. DOCUMENT FORMATTING GUIDELINES
\begin{example}[This is a nontrivial example]
This environment is used for stating examples which are similar to
the homework. It is almost always followed by the other two
environments: the solution and the answer.
\end{example}
\begin{exsol}
This environment provides a solution for the example stated above.
\end{exsol}
\begin{exans}
This environment states the answer to the question above in an
optimal way which the readers are expected to follow as they are
working through the homework.
\end{exans}
EXAMPLE 2.3.1 (This is a nontrivial example). This environment is used for stating exam-
ples which are similar to the homework. It is almost always followed by the other two
environments: the solution and the answer.
SOLUTION: This environment provides a solution for the example stated above.
ANSWER: This environment states the answer to the question above in an optimal way
which the readers are expected to follow as they are working through the homework.
503
2. DOCUMENT FORMATTING GUIDELINES CHAPTER B. TO DEVELOPERS
2.4. Homework and Answers.
\begin{hw-and-answers}
\begin{exercise}
These three environments allow to typeset exercise sections with
or without answers. They are numbered as subsections and they insert
page breaks. Does this exercise have an answer?
\begin{answer}
Yes.
\end{answer}
\end{exercise}
\begin{exercise}
This exercise does not have an answer. Why should it?
\end{exercise}
\end{hw-and-answers}
504
CHAPTER B. TO DEVELOPERS 2. DOCUMENT FORMATTING GUIDELINES
Homework B.2.
1. These three environments allow to type- and they insert page breaks. Does this ex-
set exercise sections with or without an- ercise have an answer?
swers. They are numbered as subsections
2. This exercise does not have an answer.
Why should it?
505
2. DOCUMENT FORMATTING GUIDELINES CHAPTER B. TO DEVELOPERS
Homework B.2 Answers.
1. Yes.
506
CHAPTER B. TO DEVELOPERS 4. WORKING WITH THE SOURCE
3. To Do
3.1. To Do Items.
• More notes for applications of w = r t.
• More applications of difference in chapter 1.
• More applications of slope as rate of change.
• More mixture notes.
• Add 1-variable mixture problems in chapter 2.
• Add applications of equations with rational expressions?
• Add practice test 9?
4. Working with the Source
4.1. Getting the Source. The easiest way to get the source is to git it. You can clone the
repository with
git clone https://git.albertleadata.org/melikamp/basic-algebra
4.2. Building the Source with LATEX. This text is typeset in LATEX, texlive to be specific,
and a GNU+Linux environment which provides a minimal userland such as ls and a bash-
like shell, a pdflatex command and GNU make. If you use any kind of vanilla GNU+Linux
OS such as Freenix or Debian, then all you need to do is fire up the console, change into the
directory with the source, and run
make
to build print/basic-algebra.pdf, which is the version intended as the final result.
make clean
will remove temporary build files, staged in build/.
If you happen to use a bizarro OS without a GNU-style make and/or texlive, you may still
be able to build the PDF. At the very least you need some kind of pdflatex and makeindex
commands, and all the LATEX packages used. Then you just need to run something like
pdflatex basic-algebra
pdflatex basic-algebra
makeindex basic-algebra
pdflatex basic-algebra
pdflatex basic-algebra
to build the final version of the PDF in the current directory. Good luck.
507
5. ACKNOWLEDGMENTS CHAPTER B. TO DEVELOPERS
4.3. Branches. The development happens in the branch numbered after the next edition
(first, second, and so on). Once the new edition is ready for use, the changes are merged into
the master branch, and a corresponding tag is created. The top of the master branch is then
merged into the color and mono branches. The master branch, while printable, is intended
to be viewed on a screen; color is a print-ready branch with decolored web-links, and mono
is a print-ready branch done in shades of gray.
5. Acknowledgments
Many homework exercises were shamelessly borrowed from Beginning and Intermediate Al-
gebra, a free and libre (CC-BY) textbook by Tyler Wallace, available for free download at
http://wallace.ccfaculty.org/book/book.html
Many thanks to the Elementary Algebra class of Spring 2018 at Cosumnes River College,
where students graciously reviewed a very early draft of the Concepts chapter.
Even more thanks to the Elementary Algebra class of Summer 2018 at CRC, and especially
Michelle Cao, Margaret Nyambura, and Brittany Pace for proofing the early draft.
Immense amount of thanks to Carlos Campa Navarro for proofing most of the text methodically
and thoroughly, as well as Duy Ly, Gloria Williams, Samreen Javed, and the rest of the Fall 2018
Elementary Algebra class at CRC.
Muito obrigado to Gilberta Iorg and the rest of the Spring 2020 Elementary Algebra class at
CRC.
Yet more thanks to the Fall 2020 classat CRC, in particular Kalsum Mirzai, Katharine Owens-
Byrd, Nguyen Pham, Thuy Vuong, and especially Alla Tkhay.
508
Index
∞, 139 dividing fractions, 73
∅, 34 division, 29
absolute value, 40 element of a set, 34
addition property for equations, 97 elimination for linear systems, 226
addition property for inequalities, 141 empty set, 34
addition property for systems, 226 equation, 15
additive identity, 24 equations with rational expressions, 399
additive inverse, 24 equivalent equations, 96
algebraic expression, 15 equivalent expressions, 15
algebraic substitution, 17 equivalent inequalities, 141
associativity of addition, 23 equivalent systems, 226
associativity of multiplication, 24 Euclidean distance, 441
axioms, 23 expression, 15
extraneous solution, 399
binary relation, 16
binomial, 267 factor (of a product), 19
bivariate polynomial, 266 factoring ax 2 + bx + c by guessing, 332
factoring x 2 + bx + c by guessing, 325
Cartesian plane, 149 factoring by grouping, 321
closure of arithmetic operations, 23 formula, 117
coefficient, 20 fraction in lowest terms, 64
common multiple for polynomials, 377 fraction notation, 30
commutativity of addition, 23 function defined by an expression, 202
commutativity of multiplication, 23 function notation, 202
completing the square, 468 fundamental theorem of arithmetic, 62
complex numbers, 456
composite numbers, 62 Gaussian elimination, 226
conjugate of a sum with radicals, 429 GCF, 314
constant monomial, 266 GCF for monomials, 316
constant rate applications of systems, 232 grade of the road, 174
coordinate, 149 graph, 150
coordinate plane, 149 graph of a relation, 160
cubic monomial, 266 greatest common factor for monomials, 316
greatest common factor, 314
difference of squares, 284 grouping, 321
dimensional analysis, 130
discriminant, 475 i, 455
distance, 441 imaginary unit, 455
distributivity of exponent over division, 260 improper fraction, 82
distributivity of exponent over multiplication, 258 inequality relations, 42
distributivity of multiplication over addition, 25 infinity, 139
509
CHAPTER B. INDEX
integer divisor, 62 number line, 35
integer exponent, 249 numerical coefficient, 20
integer reciprocal exponent, 445
integers, 35 one, 24
intercepts, 168 opposite, 24
interval notation, 139 opposite of a sum, 49
invisible parentheses, 30 opposite of opposite, 28
irrational numbers, 39 order of arithmetic operations, 54
irreducible over integers, 313 order of operations in fractions, 30
irreducible over reals, 313 ordered pair, 149
irreducible polynomial, 313 ordered pair solution, 159
isolating a variable, 105 orthogonal lines, 188
parallel lines, 186
leading coefficient, 266
percent, 123
least common multiple, 80
percent increase and decrease, 125
least common multiple for polynomials, 377
perimeter of a rectangle, 117
like terms, 57
perpendicular lines, 188
like terms with radicals, 427
plus-minus notation, 462
linear equation, 104
point on a plane, 149
linear equation graph, 162
point-slope form, 195
linear function, 204
polynomial factorization, 314
linear inequality, 138
polynomial common multiple, 377
linear inequality in two variables, 238
polynomial degree, 267
linear model, 204
polynomial division, 290
linear monomial, 266
polynomial expression, 266
linear system, 213
polynomial in standard form, 266
linear systems and substitution, 222
polynomial least common multiple, 377
lowest common denominator, 81
polynomial long division, 292
lowest common denominator for rational
positive integer exponent, 50
expressions, 377
positive integers, 34
lowest terms fraction, 64
power rule for exponents, 257
prime numbers, 62
member of a set, 34
prime polynomial, 313
mixed numbers, 82
principal nth root, 411
mixing applications of systems, 233
principal square root, 409
monomial, 267
product (expression), 19
monomial coefficient, 265
product of fractions, 71
monomial degree, 266
product rule for exponents, 250
monomial expression, 265
proper fraction, 82
monomial GCF, 316
movement at constant rate, 130 Q, 37
multiplication property for equations, 98 quadratic formula, 475
multiplication property for inequalities, 142 quadratic monomial, 266
multiplication property for systems, 226 quartic monomial, 266
multiplicative identity, 24 quintic monomial, 266
multiplicative inverse, 25 quotient of fractions, 73
multiplying fractions, 71 quotient rule for exponents, 252
natural numbers, 34 R, 38
negative reciprocal, 188 radical, 409
negative integer exponent, 300 radical conjugate, 429
non-negative integers, 34 radix, 409
non-strict inequality, 42 rational exponent, 446
510
CHAPTER B. INDEX
rational expression, 363 zero, 24
rational numbers, 37 zero product in equations, 348
rationalizing the denominator, 429 zero product property, 348
real number axioms, 23
real numbers, 38
reciprocal, 25
reciprocal of a product, 70
reciprocal of reciprocal, 29
relation, 16
relatively prime, 314
relatively prime monomials, 316
rise, 174
rise over run, 174
roster notation, 34
run, 174
scientific notation, 305
set, 34
set roster notation, 34
set-builder notation, 36
similar terms, 57
similar terms with radicals, 427
simplifying fractions, 64
slope formula, 174
slope of a line, 175
slope of a segment, 174
slope-intercept form, 182
solution, 95
solution set, 95
solution set for bivariate inequality, 238
solving a formula for a variable, 117
square of a binomial, 286
square root, 409
square root property, 461
strict inequality, 42
substitution, 17
substitution for linear systems, 222
subtraction, 29
sum (expression), 18
systems of linear equations, 213
term (of a sum), 18
trinomial, 267
unique factorization domains, 314
unit, 24
univariate polynomial, 266
value of an expression, 15
variable restrictions in expressions, 364
volume of a box, 117
whole numbers, 34
work at constant rate, 130
511