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Linear Equations and Inequalities in One Variable Some ancient peoples chewed on leaves to cure their headaches. Thousands of

years ago, the Egyptians used honey, salt, cedar oil, and sycamore bark to cure

illnesses. Currently, some of the indigenous people of North America use black

birch as a pain reliever.

Today, we are grateful for modern medicine and the seemingly simple cures

for illnesses. From our own experiences we know that just the right amount of a

drug can work wonders but too much of a drug can do great harm. Even though

physicians often prescribe the same drug for children and adults, the amount

given must be tailored to the

individual. The portion of a drug

given to children is usually reduced

on the basis of factors such as the

weight and height of the child.

Likewise, older adults frequently

need a lower dosage of medication

than what would be prescribed for a

younger, more active person.

Various algebraic formulas have

been developed for determining the

proper dosage for a child and an

older adult.

2.1 The Addition and Multiplication Properties of Equality

2.2 Solving General Linear Equations

2.3 More Equations

2.4 Formulas and Functions

2.5 Translating Verbal Expressions into Algebraic Expressions

2.6 Number, Geometric, and Uniform Motion Applications

2.7 Discount, Investment, and Mixture Applications

2.8 Inequalities

2.9 Solving Inequalities and Applications

2C h a p te r

In Exercises 91 and 92 of Section 2.4 you will see two

formulas that are used to determine a child’s dosage by

using the adult dosage and the child’s age.

1 0

500

1000

2 3 4 5 6 7 8 9 10 11 12

C hi

ld ’s

d os

ag e

(m g)

Age of child (yr)

Adult dose

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86 Chapter 2 Linear Equations and Inequalities in One Variable 2-2

2.1 The Addition and Multiplication Properties of Equality In Section 1.6, an equation was defined as a statement that two expressions are equal. A solution to an equation is a number that can be used in place of the variable to make the equation a true statement. The solution set is the set of all solutions to an equation. Equations with the same solution set are equivalent equations. To solve an equation means to find all solutions to the equation. In this section you will learn systematic procedures for solving equations.

U1V The Addition Property of Equality If two workers have equal salaries and each gets a $1000 raise, then they will have equal salaries after the raise. If two people are the same age now, then in 5 years they will still be the same age. If you add the same number to two equal quantities, the results will be equal. This idea is called the addition property of equality:

In This Section

U1V The Addition Property of Equality

U2V The Multiplication Property of Equality

U3V Variables on Both Sides

U4V Applications

The Addition Property of Equality

Adding the same number to both sides of an equation does not change the solution to the equation. In symbols, a � b and

a � c � b � c

are equivalent equations.

E X A M P L E 1 Adding the same number to both sides Solve x � 3 � �7.

Solution Because 3 is subtracted from x in x � 3 � �7, adding 3 to each side of the equation will isolate x:

x � 3 � �7

x � 3 � 3 � �7 � 3 Add 3 to each side.

x � 0 � �4 Simplify each side.

x � �4 Zero is the additive identity.

Since �4 satisfies the last equation, it should also satisfy the original equation because all of the previous equations are equivalent. Check that �4 satisfies the original equation by replacing x by �4:

x � 3 � �7 Original equation

�4 � 3 � �7 Replace x by �4.

�7 � �7 Simplify.

Since �4 � 3 � �7 is correct, ��4 � is the solution set to the equation. Now do Exercises 1–8

Consider the equation x � 5. The only possible number that could be used in place of x to get a true statement is 5, because 5 � 5 is true. So the solution set is {5}. We say that x in x � 5 is isolated because it occurs only once in the equation and it is by itself. The variable in x � 3 � �7 is not isolated. In Example 1, we solve x � 3 � �7 by using the addition property of equality to isolate the variable.

U Helpful Hint V

Think of an equation like a balance scale. To keep the scale in balance, what you add to one side you must also add to the other side.

3 3

x � 3 �7

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E X A M P L E 2 Subtracting the same number from both sides Solve 9 � x � �2.

Solution We can remove the 9 from the left side by adding �9 to each side or by subtracting 9 from each side of the equation:

9 � x � �2

9 � x � 9 � �2 � 9 Subtract 9 from each side.

x � �11 Simplify each side.

Check that �11 satisfies the original equation by replacing x by �11:

9 � x � �2 Original equation

9 � (�11) � �2 Replace x by �11.

Since 9 � (�11) � �2 is correct, ��11� is the solution set to the equation. Now do Exercises 9–18

Note that enclosing the solutions to an equation in braces is not absolutely neces- sary. It is simply a formal way of stating the answer. At times we may simply state that the solution to the equation is �4.

The equations that we work with in this section and Sections 2.2 and 2.3 are called linear equations. The name comes from the fact that similar equations in two variables that we will study in Chapter 3 have graphs that are straight lines.

2-3 2.1 The Addition and Multiplication Properties of Equality 87

Linear Equation

A linear equation in one variable x is an equation that can be written in the form

ax � b

where a and b are real numbers and a � 0.

An equation such as 2x � 3 is a linear equation. We also refer to equations such as

x � 8 � 0, 2x � 5 � 9 � 5x, and 3 � 5(x � 1) � �7 � x

as linear equations, because these equations could be written in the form ax � b using the properties of equality.

In Example 1, we used addition to isolate the variable on the left-hand side of the equation. Once the variable is isolated, we can determine the solution to the equation. Because subtraction is defined in terms of addition, we can also use subtraction to isolate the variable.

Our goal in solving equations is to isolate the variable. In Examples 1 and 2, the variable was isolated on the left side of the equation. In Example 3, we isolate the vari- able on the right side of the equation.

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U2V The Multiplication Property of Equality To isolate a variable that is involved in a product or a quotient, we need the multipli- cation property of equality.

88 Chapter 2 Linear Equations and Inequalities in One Variable 2-4

E X A M P L E 3 Isolating the variable on the right side Solve �1

2 � � ��

1 4

� � y.

Solution We can remove ��1

4 � from the right side by adding �1

4 � to both sides of the equation:

� 1 2

� � �� 1 4

� � y

� 1 2

� � � 1 4

� � �� 1 4

� � y � � 1 4

� Add �1 4

� to each side.

� 3 4

� � y � 1 2

� � � 1 4

� � � 2 4

� � � 1 4

� � � 3 4

�

Check that �3 4

� satisfies the original equation by replacing y by �3 4

�:

� 1 2

� � �� 1 4

� � y Original equation

� 1 2

� � �� 1 4

� � � 3 4

� Replace y by �3 4

�.

� 1 2

� � � 2 4

� Simplify.

Since �1 2

� � � 2 4

� is correct, ��34�� is the solution set to the equation. Now do Exercises 19–26

The Multiplication Property of Equality

Multiplying both sides of an equation by the same nonzero number does not change the solution to the equation. In symbols, for c � 0, a � b and

ac � bc

are equivalent equations.

We specified that c � 0 in the multiplication property of equality because multiplying by 0 can change the solution to an equation. For example, x � 4 is satisfied only by 4, but 0 � x � 0 � 4 is true for any real number x.

In Example 4, we use the multiplication property of equality to solve an equation.

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E X A M P L E 5 Dividing both sides by the same number Solve �5w � 30.

Solution Since w is multiplied by �5, we can isolate w by dividing by �5:

�5w � 30 Original equation

� �

�

5 5 w

� � � �

30 5 � Divide each side by �5.

1 � w � �6 Because ��

5 5 � � 1

w � �6 Multiplicative identity

We could also solve this equation by multiplying each side by �� 1 5

�:

�� 1 5� � �5w � ��

1 5� � 30

1 � w � �6

w � �6

Because �5(�6) � 30, ��6� is the solution set to the equation. Now do Exercises 35–44

Because dividing by a number is the same as multiplying by its reciprocal, the multiplication property of equality allows us to divide each side of the equation by any nonzero number.

2-5 2.1 The Addition and Multiplication Properties of Equality 89

E X A M P L E 4 Multiplying both sides by the same number Solve �

z � � 6.

Solution We isolate the variable z by multiplying each side of the equation by 2.

� 2 z

� � 6 Original equation

2 � � 2 z

� � 2 � 6 Multiply each side by 2.

1 � z � 12 Because 2 � � 2 z

� � 2 � �1 2

�z � 1z

z � 12 Multiplicative identity

Because �1 2 2 � � 6, �12� is the solution set to the equation.

Now do Exercises 27–34

In Example 6, the coefficient of the variable is a fraction. We could divide each side by the coefficient as we did in Example 5, but it is easier to multiply each side by the reciprocal of the coefficient.

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If the coefficient of the variable is an integer, we usually divide each side by that integer, as we did in solving �5w � 30 in Example 5. Of course, we could also solve that equation by multiplying each side by ��

1 5

�. If the coefficient of the variable is a fraction, we usually multiply each side by the reciprocal of the fraction as we did in solving �4

5 � p � 40 in Example 6. Of course, we could also solve that equation by

dividing each side by �4 5

�. If �x appears in an equation, we can multiply by �1 to get x or divide by �1 to get x, because �1(�x) � x and �

�

1 x

� � x.

90 Chapter 2 Linear Equations and Inequalities in One Variable 2-6

E X A M P L E 6 Multiplying by the reciprocal Solve �4

5 � p � 40.

Solution Multiply each side by �5

4 �, the reciprocal of �4

5 �, to isolate p on the left side.

� 4 5

� p � 40

� 5 4

� � � 4 5

� p � � 5 4

� � 40 Multiply each side by �5 4

�.

1 � p � 50 Multiplicative inverses

p � 50 Multiplicative identity

Because �4 5

� � 50 � 40, we can be sure that the solution set is �50�.

Now do Exercises 45–52

U Helpful Hint V

You could solve this equation by multiplying each side by 5 to get 4p � 200, and then dividing each side by 4 to get p � 50.

E X A M P L E 7 Multiplying or dividing by �1 Solve �h � 12.

Solution This equation can be solved by multiplying each side by �1 or dividing each side by �1. We show both methods here. First replace �h with �1 � h:

Multiplying by �1 Dividing by �1

�h � 12 �h � 12

�1(�1 � h) � �1 � 12 � �

�

1 1 � h � � �

�

12 1 �

h � �12 h � �12

Since �(�12) � 12, the solution set is ��12�. Now do Exercises 53–60

U3V Variables on Both Sides In Example 8, the variable occurs on both sides of the equation. Because the variable represents a real number, we can still isolate the variable by using the addition property

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U4V Applications In Example 9, we use the multiplication property of equality in an applied situation.

2-7 2.1 The Addition and Multiplication Properties of Equality 91

E X A M P L E 8 Subtracting an algebraic expression from both sides Solve �9 � 6y � 7y.

Solution The expression 6y can be removed from the left side of the equation by subtracting 6y from both sides.

�9 � 6y � 7y

�9 � 6y � 6y � 7y � 6y Subtract 6y from each side.

�9 � y Simplify each side.

Check by replacing y by �9 in the original equation:

�9 � 6(�9) � 7(�9)

�63 � �63

The solution set to the equation is ��9�. Now do Exercises 61–68

E X A M P L E 9 Comparing populations In the 2000 census, Georgia had �2

3 � as many people as Illinois (U.S. Bureau of Census,

www.census.gov). If the population of Georgia was 8 million, then what was the popula- tion of Illinois?

Solution If p represents the population of Illinois, then �2

3 � p represents the population of Georgia.

Since the population of Georgia was 8 million, we can write the equation �2 3

� p � 8. To find

p, solve the equation:

� 2 3

�p � 8

� 3 2

� � � 2 3

�p � � 3 2

� � 8 Multiply each side by �3 2

�.

p � 12 Simplify.

So the population of Illinois was 12 million in 2000.

Now do Exercises 89–94

U Helpful Hint V

It does not matter whether the variable ends up on the left or right side of the equation. Whether we get y � �9 or �9 � y we can still con- clude that the solution is �9.

of equality. Note that it does not matter whether the variable ends up on the right side or the left side.

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92 Chapter 2 Linear Equations and Inequalities in One Variable 2-8

Warm-Ups ▼

Fill in the blank. 1. An is a sentence that expresses the equality of

two algebraic expressions.

2. The is the set of all solutions to an equation.

3. A number an equation if the equation is true when the variable is replaced by the number.

4. Equations that have the same solution set are .

5. A equation in one variable has the form ax � b,

with a � 0.

6. According to the , adding the same number to both sides of an equation does not change the solution set.

True or false? 7. The solution to x – 5 � 5 is 10.

8. The equation � 2 x

� � 4 is equivalent to x � 8.

9. To solve �3 4

� y � 12, we should multiply each side by �3 4

�.

10. The equation � 7 x

� � 4 is equivalent to �1 7

� x � 4.

11. The equations 5x � 0 and 4x � 0 are equivalent.

12. To isolate t in 2t � 7 � t, we subtract t from each side.

13. The solution set to 2x – 3 � x – 1 is {4}.

U1V The Addition Property of Equality

Solve each equation. Show your work and check your answer. See Example 1.

1. x � 6 � �5 2. x � 7 � �2

3. �13 � x � �4 4. �8 � x � �12

5. y � � 1 2

� � � 1 2

� 6. y � � 1 4

� � � 1 2

�

7. w � � 1 3

� � � 1 3

� 8. w � � 1 3

� � � 1 2

�

Solve each equation. Show your work and check your answer. See Example 2.

9. x � 3 � �6 10. x � 4 � �3

11. 12 � x � �7 12. 19 � x � �11

13. t � � 1

2 � � �

4 � 14. t � �

3 � � 1

15. � 1 1 9 � � m � �

1 1 9 � 16. �

1 3

� � n � � 1 2

�

17. a � 0.05 � 6 18. b � 4 � �0.7

Solve each equation. Show your work and check your answer. See Example 3.

19. 2 � x � 7 20. 3 � x � 5

21. �13 � y � 9 22. �14 � z � 12

23. 0.5 � �2.5 � x 24. 0.6 � �1.2 � x

25. � 1 8

� � �� 1 8

� � r 26. � 1 6

� � �� 1 6

� � h

Exercises

U Study Tips V • Get to know your classmates whether you are an online student or in a classroom. • Talk about what you are learning. Verbalizing ideas helps you get them straight in your mind.

2 .1

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2-9 2.1 The Addition and Multiplication Properties of Equality 93

U2V The Multiplication Property of Equality

Solve each equation. Show your work and check your answer. See Example 4.

27. � 2 x

� � �4 28. � 3 x

� � �6

29. 0.03 � � 6 y

0 � 30. 0.05 � �

8 y

0 �

31. � a

2 � � �

1 3

� 32. � b

2 � � �

1 5

�

33. � 1 6

� � � 3 c

� 34. � 1 1 2 � � �

3 �

Solve each equation. Show your work and check your answer. See Example 5.

35. �3x � 15 36. �5x � �20

37. 20 � 4y 38. 18 � �3a

39. 2w � 2.5 40. �2x � �5.6

41. 5 � 20x 42. �3 � 27d

43. 5x � � 3 4

� 44. 3x � �� 2 3

�

Solve each equation. Show your work and check your answer. See Example 6.

45. � 3 2

� x � �3 46. � 2 3

� x � �8

47. 90 � � 3 4 y � 48. 14 � �

7 8 y �

49. �� 3 5

� w � �� 1 3

� 50. �� 5 2

� t � �� 3 5

�

51. � 2 3

� � �� 4 3 x � 52. �

4 � � ��

p �

Solve each equation. Show your work and check your answer. See Example 7.

53. �x � 8 54. �x � 4

55. �y � �� 1 3

� 56. �y � �� 7 8

�

57. 3.4 � �z 58. 4.9 � �t

59. �k � �99 60. �m � �17

U3V Variables on Both Sides

Solve each equation. Show your work and check your answer. See Example 8.

61. 4x � 3x � 7 62. 3x � 2x � 9

63. 9 � 6y � �5y 64. 12 � 18w � �17w

65. �6x � 8 � 7x 66. �3x � �6 � 4x

67. � 1 2

� c � 5 � � 1 2

� c 68. �� 1 2

� h � 13 � � 3 2

� h

Miscellaneous

Use the appropriate property of equality to solve each equation.

69. 12 � x � 17 70. �3 � x � 6

71. � 3

4 � y � �6 72. �

9 � z � �10

73. �3.2 � x � �1.2 74. t � 3.8 � �2.9

75. 2a � � 1

3 � 76. �3w � �

2 �

77. �9m � 3 78. �4h � �2

79. �b � �44 80. �r � 55

81. � 2

3 � x � �

2 � 82. �

4 � x � �

3 �

83. �5x � 7 � 6x 84. �� 1

2 � � 3y � 4y

85. � 5

a � � �10 86. �

r � � �14

87. � 1

2 � v � ��

1 2

� v � � 3 8

� 88. � 1

3 � s � �

9 � � �

3 � s

U4V Applications

Solve each problem by writing and solving an equation. See Example 9.

89. Births to teenagers. In 2006 there were 41.8 births per 1000 females 15 to 19 years of age (National Center for Health Statistics, www.cdc.gov/nchs). This birth rate is

� 2 3

� of the birth rate for teenagers in 1991.

a) Write an equation and solve it to find the birth rate for teenagers in 1991.

b) Use the accompanying graph to estimate the birth rate to teenagers in 2000.

Figure for Exercise 89

42 6 108 1412 16 Years since 1990

B ir

th s

pe r 1

00 0

fe m

al es

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94 Chapter 2 Linear Equations and Inequalities in One Variable 2-10

90. World grain demand. Freeport McMoRan projects that in 2015 world grain supply will be 2.1 trillion metric tons and the supply will be only �34� of world grain demand. What will world grain demand be in 2015?

91. Advancers and decliners. On Thursday, �2 3

� of the stocks traded on the New York Stock Exchange advanced in price. If 1918 stocks advanced, then how many stocks were traded on that day?

92. Births in the United States. In 2009, two-fifths of all births in the United States were to unmarried women (National Center for Health Statistics, www.cdc.gov/nchs). If there were 1,707,600 births to unmarried women, then how many births were there in 2009?

93. College students. At Springfield College 40% of the students are male. If there are 1200 males, then how many students are there at the college?

94. Credit card revenue. Seventy percent of the annual revenue for a credit card company comes from interest and penalties. If the amount for interest and penalties was $210 million, then what was the annual revenue?

Photo for Exercise 90

2.2 Solving General Linear Equations

All of the equations that we solved in Section 2.1 required only a single application of a property of equality. In this section you will solve equations that require more than one application of a property of equality.

U1V Equations of the Form ax � b � 0 To solve an equation of the form ax � b � 0 we might need to apply both the addition property of equality and the multiplication property of equality.

In This Section

U1V Equations of the Form ax � b � 0

U2V Equations of the Form ax � b � cx � d

U3V Equations with Parentheses

U4V Applications

E X A M P L E 1 Using the addition and multiplication properties of equality Solve 3r � 5 � 0.

Solution To isolate r, first add 5 to each side, and then divide each side by 3.

3r � 5 � 0 Original equation

3r � 5 � 5 � 0 � 5 Add 5 to each side.

3r � 5 Combine like terms.

� 3 3 r � � �

5 3

� Divide each side by 3.

r � � 5 3

� Simplify.

U Helpful Hint V

If we divide each side by 3 first, we must divide each term on the left side by 3 to get r � �5

3 � � 0. Then add �5

3 � to

each side to get r � �5 3

�. Although we get the correct answer, we usually save division to the last step so that fractions do not appear until necessary.

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In solving ax � b � 0, we usually use the addition property of equality first and the multiplication property last. Note that this is the reverse of the order of operations (multiplication before addition), because we are undoing the operations that are done in the expression ax � b.

CAUTION

2-11 2.2 Solving General Linear Equations 95

E X A M P L E 2 Using the addition and multiplication properties of equality Solve ��

2 3

� x � 8 � 0.

Solution To isolate x, first subtract 8 from each side, and then multiply each side by ��3

2 �.

�� 2 3

�x � 8 � 0 Original equation

�� 2 3

� x � 8 � 8 � 0 � 8 Subtract 8 from each side.

�� 2 3

� x � �8 Combine like terms.

�� 3 2

��23� x� � ��32�(�8) Multiply each side by ��32�. x � 12 Simplify.

Checking 12 in the original equation gives

�� 2 3

�(12) � 8 � �8 � 8 � 0.

So �12� is the solution set to the equation. Now do Exercises 7–14

Checking �5 3

� in the original equation gives

3 � � 5 3

� � 5 � 5 � 5 � 0.

So ��53�� is the solution set to the equation. Now do Exercises 1–6

U2V Equations of the Form ax � b � cx � d In solving equations, our goal is to isolate the variable. We use the addition property of equality to eliminate unwanted terms. Note that it does not matter whether the vari- able ends up on the right or left side. For some equations, we will perform fewer steps if we isolate the variable on the right side.

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96 Chapter 2 Linear Equations and Inequalities in One Variable 2-12

E X A M P L E 3 Isolating the variable on the right side Solve 3w � 8 � 7w.

Solution To eliminate the 3w from the left side, we can subtract 3w from both sides.

3w � 8 � 7w Original equation

3w � 8 � 3w � 7w � 3w Subtract 3w from each side.

�8 � 4w Simplify each side.

�� 8 4

� � � 4 4 w � Divide each side by 4.

�2 � w Simplify.

To check, replace w with �2 in the original equation:

3w � 8 � 7w Original equation

3(�2) � 8 � 7(�2)

�14 � �14

Since �2 satisfies the original equation, the solution set is ��2�. Now do Exercises 15–22

You should solve the equation in Example 3 by isolating the variable on the left side to see that it takes more steps. In Example 4, it is simplest to isolate the variable on the left side.

E X A M P L E 4 Isolating the variable on the left side Solve �1

2 � b � 8 � 12.

Solution To eliminate the 8 from the left side, we add 8 to each side.

� 2 1

�b � 8 � 12 Original equation

� 2 1

�b � 8 � 8 � 12 � 8 Add 8 to each side.

� 2 1

�b � 20 Simplify each side.

2 � � 1 2

�b � 2 � 20 Multiply each side by 2.

b � 40 Simplify.

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In Example 5, both sides of the equation contain two terms.

2-13 2.2 Solving General Linear Equations 97

To check, replace b with 40 in the original equation:

� 2 1

�b � 8 � 12 Original equation

� 1 2

�(40) � 8 � 12

12 � 12

Since 40 satisfies the original equation, the solution set is �40�. Now do Exercises 23–30

U3V Equations with Parentheses Equations that contain parentheses or like terms on the same side should be simplified as much as possible before applying any properties of equality.

E X A M P L E 5 Solving ax � b � cx � d Solve 2m � 4 � 4m � 10.

Solution First, we decide to isolate the variable on the left side. So we must eliminate the 4 from the left side and eliminate 4m from the right side:

2m � 4 � 4m � 10

2m � 4 � 4 � 4m � 10 � 4 Add 4 to each side.

2m � 4m � 6 Simplify each side.

2m � 4m � 4m � 6 � 4m Subtract 4m from each side.

�2m � �6 Simplify each side.

� �

�

2 2 m

� � � �

�2 6 � Divide each side by �2.

m � 3 Simplify.

To check, replace m by 3 in the original equation:

2m � 4 � 4m � 10 Original equation

2 � 3 � 4 � 4 � 3 � 10

2 � 2

Since 3 satisfies the original equation, the solution set is �3�. Now do Exercises 31–38

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Linear equations can vary greatly in appearance, but there is a strategy that you can use for solving any of them. The following strategy summarizes the techniques that we have been using in the examples. Keep it in mind when you are solving linear equations.

98 Chapter 2 Linear Equations and Inequalities in One Variable 2-14

E X A M P L E 6 Simplifying before using properties of equality Solve 2(q � 3) � 5q � 8(q � 1).

Solution First remove parentheses and combine like terms on each side of the equation.

2(q � 3) � 5q � 8(q � 1) Original equation

2q � 6 � 5q � 8q � 8 Distributive property

7q � 6 � 8q � 8 Combine like terms.

7q � 6 � 6 � 8q � 8 � 6 Add 6 to each side.

7q � 8q � 2 Combine like terms.

7q � 8q � 8q � 2 � 8q Subtract 8q from each side.

�q � �2

�1(�q) � �1(�2) Multiply each side by �1.

q � 2 Simplify.

To check, we replace q by 2 in the original equation and simplify:

2(q � 3) � 5q � 8(q � 1) Original equation

2(2 � 3) � 5(2) � 8(2 � 1) Replace q by 2.

2(�1) � 10 � 8(1)

8 � 8

Because both sides have the same value, the solution set is �2�. Now do Exercises 39–46

Strategy for Solving Equations

1. Remove parentheses by using the distributive property and then combine like terms to simplify each side as much as possible.

2. Use the addition property of equality to get like terms from opposite sides onto the same side so that they can be combined.

3. The multiplication property of equality is generally used last.

4. Check that the solution satisfies the original equation.

U Calculator Close-Up V

You can check an equation by enter- ing the equation on the home screen as shown here. The equal sign is in the TEST menu.

When you press ENTER, the calcu- lator returns the number 1 if the equation is true or 0 if the equation is false. Since the calculator shows a 1, we can be sure that 2 is the solution.

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U4V Applications Linear equations occur in business situations where there is a fixed cost and a per item cost. A mail-order company might charge $3 plus $2 per CD for shipping and han- dling. A lawyer might charge $300 plus $65 per hour for handling your lawsuit. AT&T might charge 5 cents per minute plus $2.95 for long distance calls. Example 7 illus- trates the kind of problem that can be solved in this situation.

2-15 2.2 Solving General Linear Equations 99

E X A M P L E 7 Long-distance charges With AT&T’s One Rate plan you are charged 5 cents per minute plus $2.95 for long- distance service for one month. If a long-distance bill is $4.80, then what is the number of minutes used?

Solution Let x represent the number of minutes of calls in the month. At $0.05 per minute, the cost for x minutes is the product 0.05x dollars. Since there is a fixed cost of $2.95, an expression for the total cost is 0.05x � 2.95 dollars. Since the total cost is $4.80, we have 0.05x � 2.95 � 4.80. Solve this equation to find x.

0.05x � 2.95 � 4.80

0.05x � 2.95 � 2.95 � 4.80 � 2.95 Subtract 2.95 from each side.

0.05x � 1.85 Simplify.

� 0 0 . . 0 0 5 5 x

� � � 1 0 . . 8 0 5 5

� Divide each side by 0.05.

x � 37 Simplify.

So the bill is for 37 minutes.

Now do Exercises 87–94

Warm-Ups ▼

Fill in the blank. 1. To solve �x � 8 we use the property of

equality.

2. To solve x � 5 � 9 we use the property of equality.

3. To solve 3x � 7 � 11 we apply the property of equality and then the property of equality.

True or false? 4. The solution set to 4x � 3 � 3x is {3}.

5. The equation 2x � 7 � 8 is equivalent to 2x � 1.

6. To solve 3x � 5 � 8x � 7, you could add 5 to each side and then subtract 8x from each side.

7. To solve 5 � 4x � 9 � 7x, you could subtract 9 from each side and then subtract 7x from each side.

8. The equation �n � 9 is equivalent to n � �9.

9. The equation �y � �7 is equivalent to y � 7.

10. The solution to 7x � 5x is 0.

11. To isolate y in 3y � 7 � 6, you could divide each side by 3 and then add 7 to each side.

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U1V Equations of the Form ax � b � 0

Solve each equation. Show your work and check your answer. See Examples 1 and 2.

1. 5a � 10 � 0 2. 8y � 24 � 0

3. �3y � 6 � 0 4. �9w � 54 � 0

5. 3x � 2 � 0 6. 5y � 1 � 0

7. � 1 2

� w � 3 � 0 8. � 3 8

� t � 6 � 0

9. �� 2 3

� x � 8 � 0 10. �� 1 7

� z � 5 � 0

11. �m � � 1 2

� � 0 12. �y � � 3 4

� � 0

13. 3p � � 1 2

� � 0 14. 9z � � 1 4

� � 0

U2V Equations of the Form ax � b � cx � d

Solve each equation. See Examples 3 and 4.

15. 6x � 8 � 4x 16. 9y � 14 � 2y

17. 4z � 5 � 2z 18. 3t � t � 3

19. 4a � 9 � 7 20. 7r � 5 � 47

21. 9 � �6 � 3b 22. 13 � 3 � 10s

23. � 1 2

� w � 4 � 13 24. � 1 3

� q � 13 � �5

25. 6 � � 1 3

� d � � 1 3

� d 26. 9 � � 1 2

� a � � 1 4

� a

27. 2w � 0.4 � 2 28. 10h � 1.3 � 6

29. x � 3.3 � 0.1x 30. y � 2.4 � 0.2y

Solve each equation. See Example 5.

31. 3x � 3 � x � 5 32. 9y � 1 � 6y � 5

33. 4 � 7d � 13 � 4d 34. y � 9 � 12 � 6y

35. c � � 1 2

� � 3c � � 1 2

� 36. x � � 1 4

� � � 1 2

� � x

37. � 2 3

� a � 5 � � 1 3

� a � 5 38. � 1 2

� t � 3 � � 1 4

� t � 9

U3V Equations with Parentheses

Solve each equation. See Example 6.

39. 5(a � 1) � 3 � 28

40. 2(w � 4) � 1 � 1

41. 2 � 3(q � 1) � 10 � (q � 1)

42. �2(y � 6) � 3(7 � y) � 5

43. 2(x � 1) � 3x � 6x � 20

44. 3 � (r � 1) � 2(r � 1) � r

45. 2�y � �12�� 4�y � � 1 4

�� y 46. �

1 2

� (4m � 6) � � 2 3

� (6m � 9) � 3

Miscellaneous

Solve each equation. Show your work and check your answer. See the Strategy for Solving Equations box on page 98.

47. 2x � � 1 3

� 48. 3x � � 1 6 1 �

49. 5t � �2 � 4t 50. 8y � 6 � 7y

51. 3x � 7 � 0 52. 5x � 4 � 0

53. �x � 6 � 5 54. �x � 2 � 9 55. �9 � a � �3 56. 4 � r � 6

57. 2q � 5 � q � 7 58. 3z � 6 � 2z � 7

59. �3x � 1 � 5 � 2x 60. 5 � 2x � 6 � x

61. �12 � 5x � �4x � 1 62. �3x � 4 � �2x � 8

63. 3x � 0.3 � 2 � 2x 64. 2y � 0.05 � y � 1

65. k � 0.6 � 0.2k � 1 66. 2.3h � 6 � 1.8h � 1

67. 0.2x � 4 � 0.6 � 0.8x 68. 0.3x � 1 � 0.7x

69. �3(k � 6) � 2 � k 70. �2(h � 5) � 3 � h

71. 2(p � 1) � p � 36 72. 3(q � 1) � q � 23

73. 7 � 3(5 � u) � 5(u � 4)

74. v � 4(4 � v) � �2(2v � 1)

75. 4(x � 3) � 12 76. 5(x � 3) � �15

Exercises

U Study Tips V • Don’t simply work exercises to get answers. Keep reminding yourself of what you are actually doing. • Look for the big picture.Where have we come from? Where are we going next? When will the picture be complete?

2 .2

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2-17 2.2 Solving General Linear Equations 101

77. � w 5

� � 4 � �6 78. � q

2 � � 13 � �22

79. � 2 3

� y � 5 � 7 80. � 3 4

� u � 9 � �6

81. 4 � � 2 5 n � � 12 82. 9 � �

2 7 m � � 19

83. �� 1 3

� p � � 1 2

� � � 1 2

� 84. �� 3 4

� z � � 2 3

� � � 1 3

�

85. 3.5x � 23.7 � �38.75

86. 3(x � 0.87) � 2x � 4.98

U4V Applications

Solve each problem. See Example 7.

87. The practice. A lawyer charges $300 plus $65 per hour for a divorce. If the total charge for Bill’s divorce was $1405, then for what number of hours did the lawyer work on the case?

88. The plumber. Tamika paid $165 to her plumber for a service call. If her plumber charges $45 plus $40 per hour for a service call, then for how many hours did the plumber work?

89. Celsius temperature. If the air temperature in Quebec is 68° Fahrenheit, then the solution to the equation �9

5 � C �

32 � 68 gives the Celsius temperature of the air. Find the Celsius temperature.

90. Fahrenheit temperature. Water boils at 212°F. a) Use the accompanying graph to determine the Celsius

temperature at which water boils. b) Find the Fahrenheit temperature of hot tap water at

70°C by solving the equation

70 � � 5 9

� (F � 32).

91. Rectangular patio. If the rectangular patio in the accom- panying figure has a length that is 3 feet longer than its width and a perimeter of 42 feet, then the width can be found by solving the equation 2x � 2(x � 3) � 42. What is the width?

92. Perimeter of a triangle. The perimeter of the triangle shown in the accompanying figure is 12 meters. Determine the values of x, x � 1, and x � 2 by solving the equation

x � (x � 1) � (x � 2) � 12.

93. Cost of a car. Jane paid 9% sales tax and a $150 title and license fee when she bought her new Saturn for a total of $16,009.50. If x represents the price of the car, then x satisfies x � 0.09x � 150 � 16,009.50. Find the price of the car by solving the equation.

94. Cost of labor. An electrician charged Eunice $29.96 for a service call plus $39.96 per hour for a total of $169.82 for installing her electric dryer. If n represents the number of hours for labor, then n satisfies

39.96n � 29.96 � 169.82.

Find n by solving this equation.

Figure for Exercise 92

x � 2 m

x � 1 m

x m

Figure for Exercise 91

x ft

x � 3 ft

Figure for Exercise 90

100

C el

si us

te m

pe ra

tu re

Fahrenheit temperature

0 100 212

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102 Chapter 2 Linear Equations and Inequalities in One Variable 2-18

2.3 More Equations

In this section we will solve more equations of the type that we solved in Sections 2.1 and 2.2. However, some equations in this section will contain fractions or decimal numbers. Some equations will have infinitely many solutions, and some will have no solution.

U1V Equations Involving Fractions We solved some equations involving fractions in Sections 2.1 and 2.2. Here, we will solve equations with fractions by eliminating all fractions in the first step. All of the fractions will be eliminated if we multiply each side by the least common denominator.

In This Section

U1V Equations Involving Fractions

U2V Equations Involving Decimals

U3V Simplifying the Process

U4V Identities, Conditional Equations, and Inconsistent Equations

U5V Applications

E X A M P L E 1 Multiplying by the least common denominator Solve �

y � � 1 � �

y � � 1.

Solution The least common denominator (LCD) for the denominators 2 and 3 is 6. Since both 2 and 3 divide into 6 evenly, multiplying each side by 6 will eliminate the fractions:

6��2y� � 1� � 6��3 y

� � 1� Multiply each side by 6. 6 � �

2 y

� � 6 � 1 � 6 � � 3 y

� � 6 � 1 Distributive property

3y � 6 � 2y � 6 Simplify: 6 � � 2 y

� � 3y

3y � 2y � 12 Add 6 to each side.

y � 12 Subtract 2y from each side.

Check 12 in the original equation:

� 1 2 2 � � 1 � �

1 3 2 � � 1

5 � 5

Since 12 satisfies the original equation, the solution set is �12�. Now do Exercises 1–18

You can multiply each side of the equation in Example 1 by 6 to clear the fractions and get an equivalent equation, but multiplying an expression by a number to clear the fraction is not allowed. For example, multiplying

the expression � 1 6

� x � � 2 3

� by 6 to simplify it will change its value when x is

replaced with a number.

CAUTION

U Helpful Hint V

Note that the fractions in Example 1 will be eliminated if you multiply each side of the equation by any number divisible by both 2 and 3. For example, multiplying by 24 yields

12y � 24 � 8y � 24 4y � 48

y � 12.

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U2V Equations Involving Decimals When an equation involves decimal numbers, we can work with the decimal numbers or we can eliminate all of the decimal numbers by multiplying both sides by 10, or 100, or 1000, and so on. Multiplying a decimal number by 10 moves the decimal point one place to the right. Multiplying by 100 moves the decimal point two places to the right, and so on.

2-19 2.3 More Equations 103

E X A M P L E 2 An equation involving decimals Solve 0.3p � 8.04 � 12.6.

Solution The largest number of decimal places appearing in the decimal numbers of the equation is two (in the number 8.04). Therefore, we multiply each side of the equation by 100 because multiplying by 100 moves decimal points two places to the right:

0.3p � 8.04 � 12.6 Original equation

100(0.3p � 8.04) � 100(12.6) Multiplication property of equality

100(0.3p) � 100(8.04) � 100(12.6) Distributive property

30p � 804 � 1260

30p � 804 � 804 � 1260 � 804 Subtract 804 from each side.

30p � 456

� 3 3 0 0 p

� � � 4 3 5 0 6

� Divide each side by 30.

p � 15.2

You can use a calculator to check that

0.3(15.2) � 8.04 � 12.6.

The solution set is �15.2�. Now do Exercises 19–28

E X A M P L E 3 Another equation with decimals Solve 0.5x � 0.4(x � 20) � 13.4.

Solution First use the distributive property to remove the parentheses:

0.5x � 0.4(x � 20) � 13.4 Original equation

0.5x � 0.4x � 8 � 13.4 Distributive property

10(0.5x � 0.4x � 8) � 10(13.4) Multiply each side by 10.

5x � 4x � 80 � 134 Simplify.

9x � 80 � 134 Combine like terms.

9x � 80 � 80 � 134 � 80 Subtract 80 from each side.

9x � 54 Simplify.

x � 6 Divide each side by 9.

U Helpful Hint V

After you have used one of the prop- erties of equality on each side of an equation, be sure to simplify all expressions as much as possible before using another property of equality.

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If you multiply each side by 10 in Example 3 before using the distribu- tive property, be careful how you handle the terms in parentheses:

10 � 0.5x � 10 � 0.4(x � 20) � 10 � 13.4

5x � 4(x � 20) � 134

It is not correct to multiply 0.4 by 10 and also to multiply x � 20 by 10.

U3V Simplifying the Process It is very important to develop the skill of solving equations in a systematic way, writ- ing down every step as we have been doing. As you become more skilled at solving equations, you will probably want to simplify the process a bit. One way to simplify the process is by writing only the result of performing an operation on each side. Another way is to isolate the variable on the side where the variable has the larger coefficient, when the variable occurs on both sides. We use these ideas in Example 4 and in future examples in this text.

CAUTION

104 Chapter 2 Linear Equations and Inequalities in One Variable 2-20

Check 6 in the original equation:

0.5(6) � 0.4(6 � 20) � 13.4 Replace x by 6.

3 � 0.4(26) � 13.4

3 � 10.4 � 13.4

Since both sides of the equation have the same value, the solution set is �6�. Now do Exercises 29–32

E X A M P L E 4 Simplifying the process Solve each equation.

a) 2a � 3 � 0 b) 2k � 5 � 3k � 1

Solution a) Add 3 to each side, and then divide each side by 2:

2a � 3 � 0

2a � 3 Add 3 to each side.

a � � 3 2

� Divide each side by 2.

Check that �3 2

� satisfies the original equation. The solution set is ��32��. b) For this equation we can get a single k on the right by subtracting 2k from each side.

(If we subtract 3k from each side, we get �k, and then we need another step.)

2k � 5 � 3k � 1 5 � k � 1 Subtract 2k from each side. 4 � k Subtract 1 from each side.

Check that 4 satisfies the original equation. The solution set is �4�. Now do Exercises 33–48

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U4V Identities, Conditional Equations, and Inconsistent Equations It is easy to find equations that are satisfied by any real number that we choose as a replacement for the variable. For example, the equations

x � 2 � � 1 2

� x, x � x � 2x, and x � 1 � x � 1

are satisfied by all real numbers. The equation

� 5 x

� � � 5 x

�

is satisfied by any real number except 0 because division by 0 is undefined. All of these equations are called identities. Remember that the solution set for an

identity is not always the entire set of real numbers. There might be some exclusions because of undefined expressions.

2-21 2.3 More Equations 105

Identity

An equation that is satisfied by every real number for which both sides are defined is called an identity.

We cannot recognize that the equation in Example 5 is an identity until we have simplified each side.

E X A M P L E 5 Solving an identity Solve 7 � 5(x � 6) � 4 � 3 � 2(x � 5) � 3x � 28.

Solution We first use the distributive property to remove the parentheses:

7 � 5(x � 6) � 4 � 3 � 2(x � 5) � 3x � 28

7 � 5x � 30 � 4 � 3 � 2x � 10 � 3x � 28

41 � 5x � 41 � 5x Combine like terms.

This last equation is true for any value of x because the two sides are identical. So the solu- tion set to the original equation is the set of all real numbers or R.

Now do Exercises 49–50

If you get an equation in which both sides are identical, as in Example 5, there is no need to continue to simplify the equation. If you do continue, you will eventually get 0 � 0, from which you can still conclude that the equation is an identity.

The statement 2x � 4 � 10 is true only on condition that we choose x � 3. The equation x2 � 4 is satisfied only if we choose x � 2 or x � �2. These equations are called conditional equations.

CAUTION

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Every equation that we solved in Sections 2.1 and 2.2 is a conditional equation.

It is easy to find equations that are false no matter what number we use to replace the variable. Consider the equation

x � x � 1.

If we replace x by 3, we get 3 � 3 � 1, which is false. If we replace x by 4, we get 4 � 4 � 1, which is also false. Clearly, there is no number that will satisfy x � x � 1. Other examples of equations with no solutions include

x � x � 2, x � x � 5, and 0 � x � 6 � 7.

106 Chapter 2 Linear Equations and Inequalities in One Variable 2-22

Inconsistent Equation

An equation that has no solution is called an inconsistent equation.

The solution set to an inconsistent equation has no members. The set with no mem- bers is called the empty set, and it is denoted by the symbol .

E X A M P L E 6 Solving an inconsistent equation Solve 2 � 3(x � 4) � 4(x � 7) � 7x.

Solution Use the distributive property to remove the parentheses:

2 � 3(x � 4) � 4(x � 7) � 7x The original equation

2 � 3x � 12 � 4x � 28 � 7x Distributive property

14 � 3x � �28 � 3x Combine like terms on each side.

14 � 3x � 3x � �28 � 3x � 3x Add 3x to each side.

14 � �28 Simplify.

The last equation is not true for any x. So the solution set to the original equation is the empty set, . The equation is inconsistent.

Now do Exercises 51–68

Keep the following points in mind when solving equations.

Conditional Equation

A conditional equation is an equation that is satisfied by at least one real num- ber but is not an identity.

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The solution set to an identity is the set of all real numbers for which both sides of the equation are defined. The solution set to an inconsistent equation is the empty set, .

U5V Applications

2-23 2.3 More Equations 107

Recognizing Identities and Inconsistent Equations

If you are solving an equation and you get

1. an equation in which both sides are identical, the original equation is an identity.

2. an equation that is false, the original equation is an inconsistent equation.

Warm-Ups ▼

Fill in the blank. 1. If an equation involves fractions, we multiply each

side by the of all of the fractions.

2. If an equation involves decimals, we each side by a power of 10 to eliminate all decimals.

3. An is satisfied by all numbers for which both sides are defined.

4. A equation has at least one solution but is not an identity.

5. An equation has no solution.

True or false? 6. To solve �1

2 � x � �1

3 � � x � �1

6 �, multiply each side by 6.

7. The equation 0.2x � 0.03x � 8 is equivalent to 20x � 3x � 8.

8. The equation 5a � 3 � 0 is inconsistent.

9. The equation 2t � t is a conditional equation.

10. The equation w � 0.1w � 0.9w is an identity.

11. The equation �x x

� � 1 is an identity.

E X A M P L E 7 Discount Olivia got a 6% discount when she bought a new Xbox. If she paid $399.50 and x is the original price, then x satisfies the equation x � 0.06x � 399.50. Solve the equation to find the original price.

Solution We could multiply each side by 100, but in this case, it might be easier to just work with the decimals:

x � 0.06x � 399.50

0.94x � 399.50 1.00 � 0.06 � 0.94

x � �390 9 .9 .5 4 0

� � 425 Divide each side by 0.94.

Check that 425 � 0.06(425) � 399.50. The original price was $425.

Now do Exercises 87–90

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U1V Equations Involving Fractions

Solve each equation by first eliminating the fractions. See Example 1.

1. � 4 x

� � � 1 3 0 � � 0 2. �

1 x 5 � � �

1 6

� � 0

3. 3x � � 1 6

� � � 1 2

� 4. 5x � � 1 2

� � � 3 4

�

5. � 2 x

� � 3 � x � � 1 2

� 6. 13 � � 2 x

� � x � � 1 2

�

7. � 2 x

� � � 3 x

� � 20 8. � 2 x

� � � 3 x

� � 5

9. � w 2

� � � w 4

� � 12 10. � a 4

� � � a 2

� � �5

11. � 3 2 z � � �

2 3 z � � �10 12. �

3 4 m � � �

m 2

� � �5

13. � 1 3

� p � 5 � � 1 4

� p 14. � 1 2

� q � 6 � � 1 5

� q

15. � 1 6

� v � 1 � � 1 4

� v � 1

16. � 1 1 5 � k � 5 � �

1 6

� k � 10

17. � 1 2

� x � � 1 3

� � � 1 4

�x

18. � 1 3

� x � � 2 5

�x � � 5 6

�

U2V Equations Involving Decimals

Solve each equation by first eliminating the decimal numbers. See Examples 2 and 3.

19. x � 0.2x � 72

20. x � 0.1x � 63

21. 0.3x � 1.2 � 0.5x

22. 0.4x � 1.6 � 0.6x

23. 0.02x � 1.56 � 0.8x

24. 0.6x � 10.4 � 0.08x

25. 0.1a � 0.3 � 0.2a � 8.3

26. 0.5b � 3.4 � 0.2b � 12.4

27. 0.05r � 0.4r � 27

28. 0.08t � 28.3 � 0.5t � 9.5

29. 0.05y � 0.03(y � 50) � 17.5

30. 0.07y � 0.08(y � 100) � 44.5

31. 0.1x � 0.05(x � 300) � 105

32. 0.2x � 0.05(x � 100) � 35

U3V Simplifying the Process

Solve each equation. If you feel proficient enough, try simplifying the process, as described in Example 4.

33. 2x � 9 � 0 34. 3x � 7 � 0

35. �2x � 6 � 0 36. �3x � 12 � 0

37. � 5 z

� � 1 � 6 38. � 2 s

� � 2 � 5

39. � 2 c

� � 3 � �4 40. � b 3

� � 4 � �7

41. 3 � t � 6 42. �5 � y � 9

43. 5 � 2q � 3q

44. �4 � 5p � �4p

45. 8x � 1 � 9 � 9x

46. 4x � 2 � �8 � 5x

47. �3x � 1 � �1 � 2x

48. �6x � 3 � �7 � 5x

U4V Identities, Conditional Equations, and Inconsistent Equations

Solve each equation. Identify each as a conditional equation, an inconsistent equation, or an identity. See Examples 5 and 6. See Recognizing Identities and Inconsistent Equations on page 107.

49. x � x � 2x

50. 2x � x � x

51. a � 1 � a � 1

52. r � 7 � r 53. 3y � 4y � 12y

54. 9t � 8t � 7

55. �4 � 3(w � 1) � w � 2(w � 2) � 1

Exercises

U Study Tips V • What’s on the final exam? If your instructor thinks a problem is important enough for a test or quiz, it is probably important enough for

the final exam. You should be thinking of the final exam all semester. • Write all of the test and quiz questions on note cards, one to a card. To prepare for the final, shuffle the cards and try to answer the

questions in a random order.

2 .3

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2-25 2.3 More Equations 109

56. 4 � 5(w � 2) � 2(w � 1) � 7w � 4

57. 3(m � 1) � 3(m � 3)

58. 5(m � 1) � 6(m � 3) � 4 � m

59. x � x � 2

60. 3x � 5 � 0

61. 2 � 3(5 � x) � 3x

62. 3 � 3(5 � x) � 0

63. (3 � 3)(5 � z) � 0

64. (2 � 4 � 8)p � 0

65. � 0 x

� � 0

66. � 2 2 x � � x

67. x � x � x2

68. � 2 2 x x � � 1

Miscellaneous

Solve each equation.

69. 3x � 5 � 2x � 9

70. 5x � 9 � x � 4

71. x � 2(x � 4) � 3(x � 3) � 1

72. u � 3(u � 4) � 4(u � 5)

73. 23 � 5(3 � n) � �4(n � 2) � 9n

74. �3 � 4(t � 5) � �2(t � 3) � 11

75. 0.05x � 30 � 0.4x � 5

76. x � 0.08x � 460

77. �� 2 3

�a � 1 � 2

78. �� 3 4

�t � � 1 2

�

79. � 2 y

� � � 6 y

� � 20

80. � 3 5 w � � 1 � �

w 2

� � 1

81. 0.09x � 0.2(x � 4) � �1.46

82. 0.08x � 0.5(x � 100) � 73.2

83. 436x � 789 � �571

84. 0.08x � 4533 � 10x � 69

85. � 34

x 4

� � 235 � 292

86. 34(x � 98) � � 2 x

� � 453.5

U5V Applications

Solve each problem. See Example 7.

87. Sales commission. Danielle sold her house through an agent who charged 8% of the selling price. After the com- mission was paid, Danielle received $117,760. If x is the selling price, then x satisfies

x � 0.08x � 117,760.

Solve this equation to find the selling price.

88. Raising rabbits. Before Roland sold two female rabbits, half of his rabbits were female. After the sale, only one- third of his rabbits were female. If x represents his original number of rabbits, then

� 1 2

� x � 2 � � 1 3

�(x � 2).

Solve this equation to find the number of rabbits that he had before the sale.

89. Eavesdropping. Reginald overheard his boss complaining that his federal income tax for 2009 was $60,531.

a) Use the accompanying graph to estimate his boss’s taxable income for 2009.

b) Find his boss’s exact taxable income for 2009 by solving the equation

46,742 � 0.33(x � 208,850) � 60,531.

Figure for Exercise 89

90. Federal taxes. According to Bruce Harrell, CPA, the federal income tax for a class C corporation is found by solving a linear equation. The reason for the equation is that the amount x of federal tax is deducted before the state tax is figured, and the amount of state tax is deducted before the federal tax is figured. To find the amount of federal tax for a corporation with a taxable income of $200,000, for which the federal tax rate is 25% and the state tax rate is 10%, Bruce must solve

x � 0.25[200,000 � 0.10(200,000 � x)].

Solve the equation for Bruce.

Ta x

(t ho

us an

ds o

f $)

100

100 200 400300 Taxable income (thousands of $)

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110 Chapter 2 Linear Equations and Inequalities in One Variable 2-26

The formula C � � 5 9

� (F � 32) is used to find the Celsius temperature for a given Fahrenheit temperature. If we solve this formula for F, then we have a formula for finding Fahrenheit temperature for a given Celsius temperature.

2.4 Formulas and Functions

In this section, you will learn to rewrite formulas using the same properties of equality that we used to solve equations. You will also learn how to find the value of one of the variables in a formula when we know the value of all of the others.

U1V Solving for a Variable Most drivers know the relationship between distance, rate, and time. For example, if you drive 70 mph for 3 hours, then you will travel 210 miles. At 60 mph a 300-mile trip will take 5 hours. If a 400-mile trip took 8 hours, then you averaged 50 mph. The relationship between distance D, rate R, and time T is expressed by the formula

D � R � T.

A formula or literal equation is an equation involving two or more variables. To find the time for a 300-mile trip at 60 mph, you are using the formula in the

form T � �D R

�. The process of rewriting a formula for one variable in terms of the others

is called solving for a certain variable. To solve for a certain variable, we use the same techniques that we use in solving equations.

In This Section

U1V Solving for a Variable

U2V The Language of Functions

U3V Finding the Value of a Variable

U4V Applications

E X A M P L E 1 Solving for a certain variable Solve the formula D � RT for T.

Solution Since T is multiplied by R, dividing each side of the equation by R will isolate T:

D � RT Original formula

� D R

� � � R

R � T � Divide each side by R.

� D

R � � T Divide out (or cancel) the common factor R.

T � � D

R � It is customary to write the single variable on the left.

Now do Exercises 1–12

E X A M P L E 2 Solving for a certain variable Solve the formula C � �5

9 �(F � 32) for F.

Solution We could apply the distributive property to the right side of the equation, but it is simpler to proceed as follows:

C � � 5 9

� (F � 32)

� 9 5

� C � � 9 5

� � � 5 9

� (F � 32) Multiply each side by �9 5

�, the reciprocal of �5 9

�.

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2-27 2.4 Formulas and Functions 111

E X A M P L E 3 Expressing y as a function of x Find a formula that expresses y as a function of x if x � 2y � 6. Write the answer in the form y � mx � b where m and b are real numbers.

Solution x � 2y � 6 Original equation

2y � 6 � x Subtract x from each side.

� 1 2

� � 2y � � 1 2

� (6 � x) Multiply each side by �12�.

y � 3 � � 1 2

� x Distributive property

y � �� 1 2

� x � 3 Rearrange to get y � mx � b form.

The formula y � �� 1 2

� x � 3 expresses y as a function of x.

Now do Exercises 19–28

� 9 5

� C � F � 32 Simplify.

� 9 5

� C � 32 � F � 32 � 32 Add 32 to each side.

� 9 5

� C � 32 � F Simplify.

The formula is usually written as F � �9 5

�C � 32.

Now do Exercises 13–18

U2V The Language of Functions The formula D � RT is a rule for determining the distance D from the rate R and the time T. (In words, the rule is to multiply the rate and time to obtain the distance.) We say that D � RT expresses D as a function of R and T and that the formula is a func- tion. Distance is a function of rate and time. The formula T � �D

R � can be used to deter-

mine the time from the distance and rate. So this formula expresses time as a function of distance and rate. The formula C � �5

9 � (F � 32) expresses the Celsius temperature C

as a function of the Fahrenheit temperature F. The formula F � �9 5

�C � 32 expresses the Fahrenheit temperature as a function of the Celsius temperature.

Function

A function is a rule for determining uniquely the value of one variable a from the value(s) of one or more other variable(s). We say that a is a function of the other variable(s).

If y is a function of x, then there is only one y-value for any given x-value. The plus or minus symbol, , is sometimes used in a formula as in y � x. In this case, there are two possible y-values for a given x-value. Since y is not uniquely determined by x, y is not a function of x.

U Helpful Hint V

If we simply wanted to solve x � 2y � 6 for y, we could have written

y � �� 6 �

2 x

� or y � �� x

2 � 6 �.

However, in Example 3 we requested the form y � mx � b. This form is a popular form that we will study in detail in Chapter 3.

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E X A M P L E 4

Notice that in Example 3 we multiplied each side of the equation by �1 2

�, and so

we multiplied each term on the right-hand side by �1 2

�. Instead of multiplying by �1 2

�,

we could have divided each side of the equation by 2. We would then divide each term on the right side by 2. This idea is illustrated in Example 4.

112 Chapter 2 Linear Equations and Inequalities in One Variable 2-28

Expressing y as a function of x Find a formula that expresses y as a function of x if 2x � 3y � 9. Write the answer in the form y � mx �b where m and b are real numbers.

Solution 2x � 3y � 9 Original equation

�3y � �2x � 9 Subtract 2x from each side.

� �

�

3 3 y

� � � �2

�

x 3 � 9 � Divide each side by �3.

y � � �

�

2 3 x

� � � �

9 3 � By the distributive property, each term is divided by �3.

y � � 2 3

�x � 3 Simplify.

The formula y � �2 3

�x � 3 expresses y as a function of x.

Now do Exercises 29–40

E X A M P L E 5 Solving for a variable that appears on both sides Find a formula that expresses x as a function of b and d if 5x � b � 3x � d.

Solution First get all terms involving x onto one side and all other terms onto the other side:

5x � b � 3x � d Original formula

5x � 3x � b � d Subtract 3x from each side.

5x � 3x � b � d Add b to each side.

2x � b � d Combine like terms.

x � � b �

2 d

� Divide each side by 2.

The formula solved for x is x � �b �2 d

�. The formula x � �b �2 d

� expresses x as a function of b and d.

Now do Exercises 41–48

Note that in Example 4 we wrote the answer as y � �2 3

� x � 3 rather than y � �2

3 � x � (�3). If the form y � mx � b is requested, we may use a subtraction symbol

in place of the addition symbol when b is negative. When solving for a variable that appears more than once in the equation, we must

combine the terms to obtain a single occurrence of the variable. When a formula has been solved for a certain variable, that variable will not occur on both sides of the equation.

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If we simply add b to both sides and then divide by 5 in Example 5, we

get x � �3x � 5 b � d �. Since x appears on both sides, this formula is not

solved for x and does not express x as a function of b and d.

U3V Finding the Value of a Variable In many situations, we know the values of all variables in a formula except one. We use the formula to determine the unknown value.

CAUTION

2-29 2.4 Formulas and Functions 113

E X A M P L E 6 Finding the value of a variable in a formula If 2x � 3y � 9, find y when x � 6.

Solution Method 1: First solve the equation for y. Because we have already solved this equation for y in Example 4, we will not repeat that process in this example. We have

y � � 2 3

� x � 3.

Now replace x by 6 in this equation:

y � � 2 3

� (6) � 3

� 4 � 3 � 1 So when x � 6, we have y � 1.

Method 2: First replace x by 6 in the original equation, and then solve for y:

2x � 3y � 9 Original equation

2 � 6 � 3y � 9 Replace x by 6.

12 � 3y � 9 Simplify.

�3y � �3 Subtract 12 from each side.

y � 1 Divide each side by �3.

So when x � 6, we have y � 1.

Now do Exercises 49–58

It usually does not matter which method from Example 6 is used. However, if you want many y-values, it is best to have the equation solved for y. For example, com- pleting the y-column in the following table is straightforward if you have a formula that expresses y as a function of x:

x y

0 �3

3 �1

6 1

y � � 2 3

�x � 3

y � � 2 3

� (0) � 3 � �3

y � � 2 3

� (3) � 3 � �1

y � � 2 3

� (6) � 3 � 1

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In Example 8, we use the formula for the perimeter of a rectangle, P � 2L � 2W, which can be found inside the front cover of this book. The perimeter P is a function of the length L and the width W. For geometric problems it is usually best to draw a diagram as we do in Example 8.

114 Chapter 2 Linear Equations and Inequalities in One Variable 2-30

E X A M P L E 7 Finding the simple interest rate The principal is $400 and the time is 2 years. Find the simple interest rate for each of the following amounts of interest: $120, $60, $30.

Solution First solve the formula I � Prt for r:

Prt � I Simple interest formula

� P P r t t

� � � P I t

� Divide each side by Pt.

r � � P I t

� Simplify.

Now insert the values for P, t, and the three amounts of interest:

r � � 40

1 0 20

� 2 � � 0.15 � 15% Move the decimal point two places to the left.

r � � 40

6 0 0 � 2 � � 0.075 � 7.5%

r � � 40

3 0 0 � 2 � � 0.0375 � 3.75%

If the amount of interest is $120, $60, or $30, then the simple interest rate is 15%, 7.5%, or 3.75%, respectively.

Now do Exercises 67–70

U Helpful Hint V

All interest computation is based on simple interest. However, depositors do not like to wait 2 years to get inter- est as in Example 7. More often the time is �1

1 2 � year or �3

1 65 � year. Simple

interest computed every month is said to be compounded monthly. Simple interest computed every day is said to be compounded daily.

E X A M P L E 8 Using a geometric formula The perimeter of a rectangle is 36 feet. If the width is 6 feet, then what is the length?

Solution First, put the given information on a diagram as shown in Fig. 2.1. Substitute the given values into the formula for the perimeter of a rectangle and then solve for L. (We could solve for L first and then insert the given values.)

P � 2L � 2W Perimeter of a rectangle

36 � 2L � 2 � 6 Substitute 36 for P and 6 for W.

36 � 2L � 12 Simplify.

U4V Applications Example 7 involves the simple interest formula I � Prt, where I is the amount of interest, P is the principal or the amount invested, r is the annual interest rate, and t is the time in years. The amount of interest is a function of the principal, rate, and time. The interest rate is usually expressed as a percent, which must be converted to a decimal for computations.

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24 � 2L Subtract 12 from each side.

12 � L Divide each side by 2.

Check: If L � 12 and W � 6, then P � 2(12) � 2(6) � 36 feet. So we can be certain that the length is 12 feet.

Now do Exercises 71–74

If L is the list price or original price of an item and r is the rate of discount, then the amount of discount is the product of the list price and the rate of discount, rL. The sale price S is the list price minus the amount of discount. So S � L � rL. The sale price S is a function of the list price L and the rate of discount r. The rate of discount is usually expressed as a percent, which must be converted to a decimal for computations.

2-31 2.4 Formulas and Functions 115

W 6 ft

Figure 2.1

E X A M P L E 9 Finding the original price What was the original price of a stereo that sold for $560 after a 20% discount?

Solution Express 20% as the decimal 0.20 or 0.2, and use the formula S � L � rL:

Selling price � list price � amount of discount

560 � L � 0.2L

10(560) � 10(L � 0.2L) Multiply each side by 10.

5600 � 10L � 2L Remove the parentheses.

5600 � 8L Combine like terms.

� 56

8 00 � � �

8 8 L � Divide each side by 8.

700 � L

Since 20% of $700 is $140 and $700 � $140 � $560, we can be sure that the original price was $700. Note that if the discount is 20%, then the selling price is 80% of the list price. So we could have started with the equation 560 � 0.80L.

Now do Exercises 75–80

Warm-Ups ▼

Fill in the blank. 1. An equation with two or more variables is a or

equation.

2. To for a variable means to find an equivalent equation in which the variable is isolated.

3. If D � RT, then D is a of R and T.

4. The formula P � 2L � 2W is the formula for the of a rectangle.

5. The formula A � LW is the formula for the of a rectangle.

6. The formula C � �d is the formula for the of a circle.

True or false? 7. The formula D � R . T solved for T is T . R � D.

8. The formula a � b � 3a � m solved for a is a � 3a � m � b.

9. The formula A � LW solved for L is L � � W A

� .

10. The perimeter of a rectangle is the product of its length and width.

11. If x � �1 and y � �3x � 6, then y � 9.

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U1V Solving for a Variable

Solve each formula for the specified variable. See Examples 1 and 2.

1. D � RT for R 2. A � LW for W

3. C � �D for D 4. F � ma for a

5. I � Prt for P 6. I � Prt for t

7. F � � 9 5

� C � 32 for C

8. y � � 3 4

� x � 7 for x

9. A � � 1 2

� bh for h 10. A � � 1 2

� bh for b

11. P � 2L � 2W for L

12. P � 2L � 2W for W

13. A � � 1 2

� (a � b) for a

14. A � � 1 2

� (a � b) for b

15. S � P � Prt for r

16. S � P � Prt for t

17. A � � 1 2

� h(a � b) for a

18. A � � 1 2

� h(a � b) for b

U2V The Language of Functions

In each case find a formula that expresses y as a function of x. See Examples 3 and 4.

19. x � y � �9

20. 3x � y � �5

21. x � y � 6 � 0

22. 4x � y � 2 � 0

23. 2x � y � 2

24. x � y � �3

25. 3x � y � 4 � 0

26. �2x � y � 5 � 0

27. x � 2y � 4

28. 3x � 2y � 6

29. 2x � 2y � 1

30. 3x � 2y � �6

31. y � 2 � 3(x � 4)

32. y � 3 � �3(x � 1)

33. y � 1 � � 1 2

� (x � 2)

34. y � 4 � �� 2 3

� (x � 9)

35. � 1 2

� x � � 1 3

� y � �2

36. � 2 x

� � � 4 y

� � � 1 2

�

37. y � 2 � � 3 2

� (x � 3)

38. y � 4 � � 2 3

� (x � 2)

39. y � � 1 2

� � �� 1 4

��x � �12�� 40. y � �

1 2

� � �� 1 3

��x � �12��

Solve each equation for x. See Example 5.

41. 5x � a � 3x � b

42. 2c � x � 4x � c � 5b

43. 4(a � x) � 3(x � a) � 0

44. �2(x � b) � (5a � x) � a � b

45. 3x � 2(a � 3) � 4x � 6 � a

46. 2(x � 3w) � �3(x � w)

47. 3x � 2ab � 4x � 5ab

48. x � a � �x � a � 4b

Exercises

U Study Tips V • When studying for an exam, start by working the exercises in the Chapter Review. They are grouped by section so that you can go

back and review any topics that you have trouble with. • Never leave an exam early. Most papers turned in early contain careless errors that could be found and corrected. Every point counts.

2 .4

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2-33 2.4 Formulas and Functions 117

U3V Finding the Value of a Variable

For each equation that follows, find y given that x � 2. See Example 6.

49. y � 3x � 4 50. y � �2x � 5

51. 3x � 2y � �8 52. 4x � 6y � 8

53. � 3 2 x � � �

5 3 y � � 6 54. �

2 5 y � � �

3 4 x � � �

1 2

�

55. y � 3 � � 1 2

� (x � 6) 56. y � 6 � �� 3 4

� (x � 2)

57. y � 4.3 � 0.45(x � 8.6)

58. y � 33.7 � 0.78(x � 45.6)

Fill in the tables using the given formulas.

59. y � �3x � 30 60. y � 4x � 20

61. F � � 9 5

� C � 32 62. C � � 5 9

� (F � 32)

63. T � � 4 R 00 � 64. R � �

1 T 00 �

65. S � � n(n

2 �1) � 66. S ��

n(n �1) 6 (2n � 1) �

n S

T (hr) R (mph)

100

R (mph) T (hr)

100

F C

�40

C F

�10

�5

100

x y

�10

�5

x y

�10

U4V Applications

Solve each of the following problems. Some geometric formulas that may be helpful can be found inside the front cover of this text. See Examples 7–9.

67. Finding the rate. A loan of $5000 is made for 3 years. Find the interest rate for simple interest amounts of $600, $700, and $800.

68. Finding the rate. A loan of $1000 is made for 7 years. Find the interest rate for simple interest amounts of $420, $455, and $472.50.

69. Finding the time. Kathy paid $500 in simple interest on a loan of $2500. If the annual interest rate was 5%, then what was the time?

70. Finding the time. Robert paid $240 in simple interest on a loan of $1000. If the annual interest rate was 8%, then what was the time?

71. Finding the length. The area of a rectangle is 28 square yards. Find the length if the width is 2 yards, 3 yards, or 4 yards.

72. Finding the width. The area of a rectangle is 60 square feet. Find the width if the length is 10 feet, 16 feet, or 18 feet.

73. Finding the length. If it takes 600 feet of wire fencing to fence a rectangular feed lot that has a width of 75 feet, then what is the length of the lot?

74. Finding the depth. If it takes 500 feet of fencing to enclose a rectangular lot that is 104 feet wide, then how deep is the lot?

75. Finding MSRP. What was the manufacturer’s suggested retail price (MSRP) for a Lexus SC 430 that sold for $54,450 after a 10% discount?

76. Finding MSRP. What was the MSRP for a Hummer H1 that sold for $107,272 after an 8% discount?

77. Finding the original price. Find the original price if there is a 15% discount and the sale price is $255.

78. Finding the list price. Find the list price if there is a 12% discount and the sale price is $4400.

79. Rate of discount. Find the rate of discount if the discount is $40 and the original price is $200.

80. Rate of discount. Find the rate of discount if the discount is $20 and the original price is $250.

81. Width of a football field. The perimeter of a football field in the NFL, excluding the end zones, is 920 feet. How wide is the field? See the figure on the next page.

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118 Chapter 2 Linear Equations and Inequalities in One Variable 2-34

82. Perimeter of a frame. If a picture frame is 16 inches by 20 inches, then what is its perimeter?

83. Volume of a box. A rectangular box measures 2 feet wide, 3 feet long, and 4 feet deep. What is its volume?

84. Volume of a refrigerator. The volume of a rectangular refrigerator is 20 cubic feet. If the top measures 2 feet by 2.5 feet, then what is the height?

85. Radius of a pizza. If the circumference of a pizza is 8� inches, then what is the radius?

86. Diameter of a circle. If the circumference of a circle is 4� meters, then what is the diameter?

87. Height of a banner. If a banner in the shape of a triangle has an area of 16 square feet with a base of 4 feet, then what is the height of the banner?

88. Length of a leg. If a right triangle has an area of 14 square meters and one leg is 4 meters in length, then what is the length of the other leg?

Figure for Exercise 85

Figure for Exercise 84

2 ft

2.5 ft

x ft Receipe

Figure for Exercise 81

x yd

89. Length of the base. A trapezoid with height 20 inches and lower base 8 inches has an area of 200 square inches. What is the length of its upper base?

90. Height of a trapezoid. The end of a flower box forms the shape of a trapezoid. The area of the trapezoid is 300 square centimeters. The bases are 16 centimeters and 24 centi- meters in length. Find the height.

91. Fried’s rule. Doctors often prescribe the same drugs for children as they do for adults. The formula

d � 0.08aD

(Fried’s rule) expresses the child’s dosage d as a function of the adult dosage D and the child’s age a.

a) If a doctor prescribes 1000 milligrams of acetaminophen for an adult, then how many milligrams would he pre- scribe for an 8-year-old child?

b) If a doctor uses Fried’s rule to prescribe 200 milligrams of a drug to a child when he would prescribe 600 mil- ligrams to an adult, then how old is the child?

c) Use the accompanying bar graph to determine the age at which a child would get the same dosage as an adult.

Figure for Exercise 90

16 cm

24 cm

Figure for Exercise 87

4 ft

Division Champs

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2-35 2.4 Formulas and Functions 119

92. Cowling’s rule. Cowling’s rule is another function for determining the child’s dosage of a drug. For this rule, the formula

d � � D(a

2 �

4 1)

�

expresses the child’s dosage d as a function of the adult dosage D and the child’s age a.

a) If a doctor prescribes 1000 milligrams of acetaminophen for an adult, then how many milligrams would she pre- scribe for an eight-year-old child using Cowling’s rule?

b) If a doctor uses Cowling’s rule to prescribe 200 mil- ligrams of a drug to a child when she would prescribe 600 milligrams to an adult, then how old is the child?

93. Administering vancomycin. A patient is to receive 750 milligrams (desired dose) of the antibiotic vancomycin. However, vancomycin comes in a solution containing 1000 milligrams (available dose) of vancomycin per 5 milli- liters (quantity) of solution. The amount of solution to be given to the patient is a function of the desired dose, the available dose, and the quantity, given by the formula

Amount � � a d v e a s il i a re b d le

d d o o s s e e

� � quantity.

Find the amount of the solution that should be adminis- tered to the patient.

94. International communications. The global investment in telecom infrastructure since 1990 can be modeled by the function

I � 7.5t � 115,

where I is in billions of dollars and t is the number of years since 1990 (Fortune, www.fortune.com).

a) Use the formula to find the global investment in 2000.

b) Use the accompanying graph to estimate the year in which the global investment will reach $300 billion.

c) Use the formula to find the year in which the global investment will reach $300 billion.

Figure for Exercise 91

1 0

500

1000

2 3 4 5 6 7 8 9 101112

C hi

ld ’s

d os

ag e

(m g)

Age of child (yr)

Adult dose

Figure for Exercise 94

100

200

300

5 20 2510 15 Years since 1990

In ve

st m

en t (

bi lli

on s

of d

ol la

rs )

Photo for Exercise 95

95. The 2.4-meter rule. A 2.4-meter sailboat is a one-person boat that is about 13 feet in length, has a displacement of about 550 pounds, and a sail area of about 81 square feet. To compete in the 2.4-meter class, a boat must satisfy the formula

2.4 � ,

where L � length, F � freeboard, D � girth, and S � sail area. Solve the formula for L.

L � 2D � F�S ��

2.37

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120 Chapter 2 Linear Equations and Inequalities in One Variable 2-36

2.5 Translating Verbal Expressions into Algebraic Expressions You translated some verbal expressions into algebraic expressions in Section 1.6; in this section you will study translating in more detail.

In This Section

U1V Writing Algebraic Expressions

U2V Pairs of Numbers

U3V Consecutive Integers

U4V Using Formulas

U5V Writing Equations

U1V Writing Algebraic Expressions The following box contains a list of some frequently occurring verbal expressions and their equivalent algebraic expressions.

Translating Words into Algebra

Algebraic Verbal Phrase Expression

Addition: The sum of a number and 8 x � 8 Five is added to a number x � 5 Two more than a number x � 2 A number increased by 3 x � 3

Subtraction: Four is subtracted from a number x � 4 Three less than a number x � 3 The difference between 7 and a number 7 � x A number decreased by 2 x � 2

Mid-Chapter Quiz Sections 2.1 through 2.4 Chapter 2

Solve each equation. 1. x � 9 � �12 2. �

3 4

�m � � 1 2

�

3. �9x � 5 � 10x 4. 4a � 3 � 0

5. 8w � 5 � 6w � 4 6. 4(a � 3) � 8 � 48

7. 6 � 3(x � 2) � 4(x � 7)

8. � 3 2

�x � � 1 6

� � � 2 3

�

9. 0.8x � 120 � x � 70

10. 0.09x � 3.4 � 0.4x � 65.4

Identify each equation as a conditional equation, an inconsistent equation, or an identity. 11. 7x � 12x � �5x

12. 7x � 12x � �5

13. 7x � 12x � 6x

14. 7x � 12x � �5x � 4

Solve each equation for x.

15. ax � b � c

16. 5(x � a) � 2(x � b)

Miscellaneous. 17. What was the original price of a car that sold for $13,904

after a 12% discount?

18. If the perimeter of a rectangle is 48 yards and the length is 15 yards, then what is the width?

19. If x � 8 and 3x � 4y � 12, then what is y?

20. If the principal is $4000, the simple interest is $640, and the time is 2 years, then what is the simple interest rate?

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2-37 2.5 Translating Verbal Expressions into Algebraic Expressions 121

E X A M P L E 1

U Helpful Hint V

We know that x and 10 � x have a sum of 10 for any value of x. We can easily check that fact by adding:

x � 10 � x � 10

In general, it is not true that x and x � 10 have a sum of 10, because

x � x � 10 � 2x � 10.

For what value of x is the sum of x and x � 10 equal to 10?

U2V Pairs of Numbers There is often more than one unknown quantity in a problem, but a relationship between the unknown quantities is given. For example, if one unknown number is 5 more than another unknown number, we can use x to represent the smaller one and x � 5 to represent the larger one. If we use x to represent the larger unknown number, then x � 5 represents the smaller. Either way is correct.

If two numbers differ by 5, then one of them is 5 more than the other. So x and x � 5 can also be used to represent two numbers that differ by 5. Likewise, x and x � 5 could represent two numbers that differ by 5.

How would you represent two numbers that have a sum of 10? If one of the num- bers is 2, the other is certainly 10 � 2, or 8. Thus, if x is one of the numbers, then 10 � x is the other. The expressions

x and 10 � x

have a sum of 10 for any value of x.

Multiplication: The product of 5 and a number 5x Twice a number 2x

One-half of a number x

Five percent of a number 0.05x

Division: The ratio of a number to 6

The quotient of 5 and a number

Three divided by some number 3� x

5 � x

x � 6

1 � 2

Writing algebraic expressions Translate each verbal expression into an algebraic expression.

a) The sum of a number and 9

b) Eighty percent of a number

c) A number divided by 4

d) The result of a number subtracted from 5

e) Three less than a number

Solution a) If x is the number, then the sum of x and 9 is x � 9.

b) If w is the number, then eighty percent of the number is 0.80w.

c) If y is the number, then the number divided by 4 is � 4

y �.

d) If z is the number, then the result of subtracting z from 5 is 5 � z.

e) If a is the number, then 3 less than a is a � 3.

Now do Exercises 1–12

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For complementary angles, we use x and 90 � x for their degree measures. For supplementary angles, we use x and 180 � x. Complementary angles that share a common side form a right angle. Supplementary angles that share a common side form a straight angle or straight line.

122 Chapter 2 Linear Equations and Inequalities in One Variable 2-38

E X A M P L E 3 Degree measures Write algebraic expressions for each pair of angles shown.

a) b)

A C

30� ?

x ?

Degree Measures of Angles

Two angles are called complementary if the sum of their degree measures is 90°. Two angles are called supplementary if the sum of their degree measures is 180°. The sum of the degree measures of the three angles of any triangle is 180°.

Algebraic expressions for pairs of numbers Write algebraic expressions for each pair of numbers.

a) Two numbers that differ by 12

b) Two numbers with a sum of �8

Solution a) The expressions x and x � 12 represent two numbers that differ by 12. We can

check by subtracting:

x � (x � 12) � x � x � 12 � 12

Of course, x and x � 12 also differ by 12 because x � 12 � x � 12.

b) The expressions x and �8 � x have a sum of �8. We can check by addition:

x � (�8 � x) � x � 8 � x � �8

Now do Exercises 13–22

Pairs of numbers occur in geometry in discussing measures of angles. You will need the following facts about degree measures of angles.

E X A M P L E 2

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2-39 2.5 Translating Verbal Expressions into Algebraic Expressions 123

E X A M P L E 4 Expressions for integers Write algebraic expressions for the following unknown integers.

a) Two consecutive integers, the smallest of which is w.

b) Three consecutive even integers, the smallest of which is z.

c) Four consecutive odd integers, the smallest of which is y.

Solution a) Each integer is 1 larger than the preceding integer. So if w represents the smallest

of two consecutive integers, then w and w � 1 represent the integers.

b) Each even integer is 2 larger than the preceding even integer. So if z represents the smallest of three consecutive even integers, then z, z � 2, and z � 4 represent the three consecutive even integers.

c) Each odd integer is 2 larger than the preceding odd integer. So if y represents the smallest of four consecutive odd integers, then y, y � 2, y � 4, and y � 6 represent the four consecutive odd integers.

Now do Exercises 27–34

Solution a) Since the angles shown are complementary, we can use x to represent the degree

measure of the smaller angle and 90 � x to represent the degree measure of the larger angle.

b) Since the angles shown are supplementary, we can use x to represent the degree measure of the smaller angle and 180 � x to represent the degree measure of the larger angle.

c) If we let x represent the degree measure of angle B, then 180 � x � 30, or 150 � x , represents the degree measure of angle C.

Now do Exercises 23–26

U3V Consecutive Integers Note that each integer is one larger than the previous integer. For example, if x � 5, then x � 1 � 6 and x � 2 � 7. So if x is an integer, then x, x � 1, and x � 2 represent three consecutive integers. Each even (or odd) integer is two larger than the previous even (or odd) integer. For example, if x � 6, then x � 2 � 8, and x � 4 � 10. If x � 7, then x � 2 � 9, and x � 4 � 11. So x, x � 2, and x � 4 represent three consecutive even integers if x is even and three consecutive odd integers if x is odd.

The expressions x, x � 1, and x � 3 do not represent three consecutive odd integers no matter what x represents.

CAUTION

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124 Chapter 2 Linear Equations and Inequalities in One Variable 2-40

U5V Writing Equations To solve a problem using algebra, we describe or model the problem with an equation. In this section we write the equations only, and in Section 2.6 we write and solve them. Sometimes we must write an equation from the information given in the problem, and sometimes we use a standard model to get the equation. Some standard models are shown in the following box.

Uniform Motion Model

Distance � Rate � Time D � R � T

Percentage Models

What number is 5% of 40? x � 0.05 � 40 Ten is what percent of 80? 10 � x � 80 Twenty is 4% of what number? 20 � 0.04 � x

Selling Price and Discount Model

Discount � Rate of discount � Original price d � r � L Selling Price � Original price � Discount S � L � r � L

Summary of Algebraic Expressions for Pairs of Numbers

Verbal Phrase Algebraic Expressions

Two numbers that differ by 5 x and x � 5 Two numbers with a sum of 6 x and 6 � x Two consecutive integers x and x � 1 Two consecutive even integers x and x � 2 Two consecutive odd integers x and x � 2 Complementary angles x and 90 � x Supplementary angles x and 180 � x

U4V Using Formulas In writing expressions for unknown quantities, we often use standard formulas such as those given inside the front cover of this book.

E X A M P L E 5 Writing algebraic expressions using standard formulas Find an algebraic expression for

a) the distance if the rate is 30 miles per hour and the time is T hours.

b) the discount if the rate is 40% and the original price is p dollars.

Solution a) Using the formula D � RT, we have D � 30T. So 30T is an expression that

represents the distance in miles.

b) Since the discount is the rate times the original price, an algebraic expression for the discount is 0.40p dollars.

Now do Exercises 35–58

The following box contains a summary of some common verbal phrases and algebraic expressions for pairs of numbers.

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2-41 2.5 Translating Verbal Expressions into Algebraic Expressions 125

E X A M P L E 6 Writing equations Identify the variable and write an equation that describes each situation.

a) Find two numbers that have a sum of 14 and a product of 45.

b) A coat is on sale for 25% off the list price. If the sale price is $87, then what is the list price?

c) What percent of 8 is 2?

d) The value of x dimes and x � 3 quarters is $2.05.

Solution a) Let x � one of the numbers and 14 � x � the other number. Since their product is

45, we have

x(14 � x) � 45.

b) Let x � the list price and 0.25x � the amount of discount. We can write an equation expressing the fact that the selling price is the list price minus the discount:

List price � discount � selling price

x � 0.25x � 87

c) If we let x represent the percentage, then the equation is x � 8 � 2, or 8x � 2.

d) The value of x dimes at 10 cents each is 10x cents. The value of x � 3 quarters at 25 cents each is 25(x � 3) cents. We can write an equation expressing the fact that the total value of the coins is 205 cents:

Value of dimes � value of quarters � total value

10x � 25(x � 3) � 205

Now do Exercises 59–84

The value of the coins in Example 6(d) is either 205 cents or 2.05 dollars. If the total value is expressed in dollars, then all of the values must be expressed in dollars. So we could also write the equation as

0.10x � 0.25(x � 3) � 2.05.

CAUTION

U Helpful Hint V

At this point we are simply learning to write equations that model certain situations. Don’t worry about solving these equations now. In Section 2.6 we will solve problems by writing an equation and solving it.

Real Estate Commission Model

Commission � Rate of commission � Selling price Amount for owner � Selling price � Commission

Geometric Models for Perimeter

Perimeter of any figure � the sum of the lengths of the sides Rectangle: P � 2L � 2W Square: P � 4s

Geometric Models for Area

Rectangle: A � LW Square: A � s2

Parallelogram: A � bh Triangle: A � � 1 2

�bh

More geometric formulas can be found inside the front cover of this text.

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126 Chapter 2 Linear Equations and Inequalities in One Variable 2-42

Exercises

U Study Tips V • Almost everything that we do in algebra can be redone by another method or checked. So don’t close your mind to a new method or

checking. The answers will not always be in the back of the book. • When you take a test, work the problems that are easiest for you first. This will build your confidence. Make sure that you do not forget

to answer a question.

2 .5

U1V Writing Algebraic Expressions

Translate each verbal expression into an algebraic expression. See Example 1. See Translating Words into Algebra box on pages 120–121.

1. The sum of a number and 3

2. Two more than a number

3. Three less than a number

4. Four subtracted from a number

5. The product of a number and 5

6. Five divided by some number

7. Ten percent of a number

8. Eight percent of a number

9. The ratio of a number and 3

10. The quotient of 12 and a number

11. One-third of a number

12. Three-fourths of a number

U2V Pairs of Numbers

Write algebraic expressions for each pair of numbers. See Example 2.

13. Two numbers with a difference of 15 14. Two numbers that differ by 9 15. Two numbers with a sum of 6 16. Two numbers with a sum of 5 17. Two numbers such that one is 3 larger than the other

18. Two numbers such that one is 8 smaller than the other

19. Two numbers such that one is 5% of the other 20. Two numbers such that one is 40% of the other 21. Two numbers such that one is 30% more than the other

22. Two numbers such that one is 20% smaller than the other

Warm-Ups ▼

Fill in the blank. 1. Words such as “sum,” “plus,” “increased by,” and “more

than” indicate .

2. Words such as “product,” “twice,” and “percent of” indicate .

3. angles have degree measures with a sum of 90�.

4. angles have degree measures with a sum of 180�.

5. Distance is the of rate and time.

6. We can use x and x � 2 to represent consecutive or consecutive integers.

True or false? 7. For any value of x, x and x � 6 differ by 6.

8. For any value of a, a and 10 � a have a sum of 10.

9. If Jack ran x miles per hour for 3 hours, then he ran 3x miles.

10. If Jill ran x miles per hour for 10 miles, then she ran 10x hours.

11. Three consecutive odd integers can be represented by x, x � 1, and x � 3.

12. The value in cents of n nickels and d dimes is 0.05n � 0.10d

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2-43 2.5 Translating Verbal Expressions into Algebraic Expressions 127

Each of the following figures shows a pair of angles. Write algebraic expressions for the degree measures of each pair of angles. See Example 3.

23.

24.

25.

26.

U3V Consecutive Integers

Write algebraic expressions for the following unknown integers. See Example 4.

27. Two consecutive even integers, the smallest of which is n

28. Two consecutive odd integers, the smallest of which is x

29. Two consecutive integers 30. Three consecutive even integers

31. Three consecutive odd integers

32. Three consecutive integers

33. Four consecutive even integers

34. Four consecutive odd integers

U4V Using Formulas

Find an algebraic expression for the quantity in italics using the given information. See Example 5.

35. The distance, given that the rate is x miles per hour and the time is 3 hours

36. The distance, given that the rate is x � 10 miles per hour and the time is 5 hours

37. The discount, given that the rate is 25% and the original price is q dollars

38. The discount, given that the rate is 10% and the original price is t yen

39. The time, given that the distance is x miles and the rate is 20 miles per hour

40. The time, given that the distance is 300 kilometers and

the rate is x � 30 kilometers per hour

41. The rate, given that the distance is x � 100 meters and

the time is 12 seconds

42. The rate, given that the distance is 200 feet and the time

is x � 3 seconds

43. The area of a rectangle with length x meters and width 5 meters

44. The area of a rectangle with sides b yards and b � 6 yards

45. The perimeter of a rectangle with length w � 3 inches and width w inches

46. The perimeter of a rectangle with length r centimeters and width r � 1 centimeters

47. The width of a rectangle with perimeter 300 feet and length x feet

48. The length of a rectangle with area 200 square feet and

width w feet

49. The length of a rectangle, given that its width is x feet and its length is 1 foot longer than twice the width

50. The length of a rectangle, given that its width is w feet and its length is 3 feet shorter than twice the width

Figure for Exercise 25

Lake Ashley

Konecnyburg

?60�

Dugo City Morrisville

Figure for Exercise 26

Figure for Exercise 23

Figure for Exercise 24

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128 Chapter 2 Linear Equations and Inequalities in One Variable 2-44

51. The area of a rectangle, given that the width is x meters and the length is 5 meters longer than the width

52. The perimeter of a rectangle, given that the length is x yards and the width is 10 yards shorter

53. The simple interest, given that the principal is x � 1000, the rate is 18%, and the time is 1 year

54. The simple interest, given that the principal is 3x , the rate is 6%, and the time is 1 year

55. The price per pound of peaches, given that x pounds

sold for $16.50

56. The rate per hour of a mechanic who gets $480 for

working x hours

57. The degree measure of an angle, given that its complemen- tary angle has measure x degrees

58. The degree measure of an angle, given that its supplementary angle has measure x degrees

U5V Writing Equations

Identify the variable and write an equation that describes each situation. Do not solve the equation. See Example 6.

59. Two numbers differ by 5 and have a product of 8.

60. Two numbers differ by 6 and have a product of �9.

61. Herman’s house sold for x dollars. The real estate agent received 7% of the selling price and Herman received $84,532.

62. Gwen sold her car on consignment for x dollars. The saleswoman’s commission was 10% of the selling price and Gwen received $6570.

63. What percent of 500 is 100? 64. What percent of 40 is 120? 65. The value of x nickels and x � 2 dimes is $3.80.

66. The value of d dimes and d � 3 quarters is $6.75.

67. The sum of a number and 5 is 13.

68. Twelve subtracted from a number is �6.

69. The sum of three consecutive integers is 42.

70. The sum of three consecutive odd integers is 27.

71. The product of two consecutive integers is 182.

72. The product of two consecutive even integers is 168.

73. Twelve percent of Harriet’s income is $3000.

74. If 9% of the members buy tickets, then we will sell 252 tickets to this group.

75. Thirteen is 5% of what number?

76. Three hundred is 8% of what number?

77. The length of a rectangle is 5 feet longer than the width, and the area is 126 square feet.

78. The length of a rectangle is 1 yard shorter than twice the width, and the perimeter is 298 yards.

79. The value of n nickels and n � 1 dimes is 95 cents.

80. The value of q quarters, q � 1 dimes, and 2q nickels is 90 cents.

81. The measure of an angle is 38° smaller than the measure of its supplementary angle.

82. The measure of an angle is 16° larger than the measure of its complementary angle.

83. Target heart rate. For a cardiovascular workout, fitness experts recommend that you reach your target heart rate and stay at that rate for at least 20 minutes (HealthStatus, www.healthstatus.com). To find your target heart rate, find the sum of your age and your resting heart rate, and then subtract that sum from 220. Find 60% of that result and add it to your resting heart rate.

a) Write an equation with variable r expressing the fact that the target heart rate for 30-year-old Bob is 144.

b) Judging from the accompanying graph, does the target heart rate for a 30-year-old increase or decrease as the resting heart rate increases?

84. Adjusting the saddle. The saddle height on a bicycle should be 109% of the rider’s inside leg measurement L (www.harriscyclery.com). See the figure. Write an equation expressing the fact that the saddle height for Brenda is 36 in.

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2-45 2.5 Translating Verbal Expressions into Algebraic Expressions 129

Miscellaneous Translate each verbal expression into an algebraic expression. Do not simplify.

85. The sum of 6 and x

86. w less than 12

87. m increased by 9

88. q decreased by 5

89. t multiplied by 11

90. 10 less than the square of y

91. 5 times the difference between x and 2

92. The sum of two-thirds of k and 1

93. m decreased by the product of 3 and m

94. 7 increased by the quotient of x and 2

95. The ratio of 8 more than h and h

96. The product of 5 and the total of r and 3

97. 5 divided by the difference between y and 9

98. The product of n and the sum of n and 6

99. The quotient of 8 less than w and twice w

100. 3 more than one-third of the square of b

Figure for Exercise 84

109% of the inside leg measurement

101. 9 less than the product of v and �3

102. The total of 4 times the cube of t and the square of b

103. x decreased by the quotient of x and 7

104. Five-eighths of the sum of y and 3

105. The difference between the square of m and the total of m and 7

106. The product of 13 and the total of t and 6

107. x increased by the difference between 9 times x and 8

108. The quotient of twice y and 8

109. 9 less than the product of 13 and n

110. The product of s and 5 more than s 111. 6 increased by one-third of the sum of x and 2

112. x decreased by the difference between 5x and 9

113. The sum of x divided by 2 and x

114. Twice the sum of 6 times n and 5

Given that the area of each figure is 24 square feet, use the dimensions shown to write an equation expressing this fact. Do not solve the equation.

115.

116.

117.

118.

Figure for Exercise 83

50 60 70 80 130

140

150

Resting heart rate

Ta rg

et h

ea rt

r at

Target heart rate for 30-year-old

x + 3

h � 2

w � 4

y � 2

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130 Chapter 2 Linear Equations and Inequalities in One Variable 2-46

In This Section

U1V Number Problems

U2V General Strategy for Solving Verbal Problems

U3V Geometric Problems

U4V Uniform Motion Problems

2.6 Number, Geometric, and Uniform Motion Applications In this section, we apply the ideas of Section 2.5 to solving problems. Many of the problems can be solved by using arithmetic only and not algebra. However, remember that we are not just trying to find the answer; we are trying to learn how to apply algebra. So even if the answer is obvious to you, set the problem up and solve it by using algebra as shown in the examples.

U1V Number Problems Algebra is often applied to problems involving time, rate, distance, interest, or discount. Number problems do not involve any physical situation; we simply find some num- bers that satisfy some given conditions. These problems can provide good practice for solving more complex problems.

E X A M P L E 1

Strategy for Solving Problems

1. Read the problem as many times as necessary. Guessing the answer and checking it will help you understand the problem.

2. If possible, draw a diagram to illustrate the problem.

3. Choose a variable and write what it represents.

4. Write algebraic expressions for any other unknowns in terms of that variable.

5. Write an equation that describes the situation.

A consecutive integer problem The sum of three consecutive integers is 48. Find the integers.

Solution If x represents the smallest of the three consecutive integers, then x, x � 1, and x � 2 represent the three consecutive integers. Since the sum of x, x � 1, and x � 2 is 48, we write that fact as an equation and solve it:

x � (x � 1) � (x � 2) � 48

3x � 3 � 48 Combine like terms.

3x � 45 Subtract 3 from each side.

x � 15 Divide each side by 3.

x � 1 � 16 If x is 15, then x � 1 is 16 and x � 2 is 17.

x � 2 � 17

Because 15 � 16 � 17 � 48, the three consecutive integers that have a sum of 48 are 15, 16, and 17.

Now do Exercises 1–8

U2V General Strategy for Solving Verbal Problems You should use the following steps as a guide for solving problems.

U Helpful Hint V

Making a guess can be a good way to get familiar with the problem. For example, let’s guess that the answers to Example 1 are 20, 21, and 22. Since 20 � 21 � 22 � 63, these are not the correct numbers. But now we realize that we should use x, x � 1, and x � 2 and that the equation should be

x � x � 1 � x � 2 � 48.

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U3V Geometric Problems For geometric problems, always draw the figure and label it. Common geometric for- mulas are given in Section 2.5 and inside the front cover of this text. The perimeter of any figure is the sum of the lengths of all of the sides of the figure. The perimeter for a square is given by P � 4s, for a rectangle P � 2L � 2W, and for a triangle P � a � b � c. You can use these formulas or simply remember that the sum of the lengths of all sides is the perimeter.

2-47 2.6 Number, Geometric, and Uniform Motion Applications 131

6. Solve the equation.

7. Answer the original question.

8. Check your answer in the original problem (not the equation).

E X A M P L E 2

E X A M P L E 3 Complementary angles In Fig. 2.3, the angle formed by the guy wire and the ground is 3.5 times as large as the angle formed by the guy wire and the antenna. Find the degree measure of each of these angles.

Solution Let x � the degree measure of the smaller angle, and let 3.5x � the degree measure of the larger angle. Since the antenna meets the ground at a 90° angle, the sum of the degree

A perimeter problem The length of a rectangular piece of property is 1 foot less than twice the width. If the perimeter is 748 feet, find the length and width.

Solution Let x � the width. Since the length is 1 foot less than twice the width, 2x � 1 � the length. Draw a diagram as in Fig. 2.2. We know that 2L � 2W � P is the formula for the perimeter of a rectangle. Substituting 2x � 1 for L and x for W in this formula yields an equation in x:

2L � 2W � P

2(2x � 1) � 2(x) � 748 Replace L by 2x � 1 and W by x.

4x � 2 � 2x � 748 Remove the parentheses.

6x � 2 � 748 Combine like terms.

6x � 750 Add 2 to each side.

x � 125 Divide each side by 6.

If x � 125, then 2x �1 � 2(125) �1 � 249. Check by computing the perimeter:

P � 2L � 2W � 2(249) � 2(125) � 748

So the width is 125 feet and the length is 249 feet.

Now do Exercises 9–14

Example 3 involves the degree measures of angles. For this problem, the figure is given.

U Helpful Hint V

To get familiar with the problem, guess that the width is 50 ft. Then the length is 2 � 50 � 1 or 99. The perimeter would be

2(50) � 2(99) � 298,

which is too small. But now we realize that we should let x be the width, 2x � 1 be the length, and we should solve

2x � 2(2x � 1) � 748.

2x � 1

Figure 2.2

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132 Chapter 2 Linear Equations and Inequalities in One Variable 2-48

measures of the other two angles of the right triangle is 90°. (They are complementary angles.) So we have the following equation:

x � 3.5x � 90

4.5x � 90 Combine like terms.

x � 20 Divide each side by 4.5.

3.5x � 70 Find the other angle.

Check: 70° is 3.5 � 20° and 20° � 70° � 90°. So the smaller angle is 20°, and the larger angle is 70°.

Now do Exercises 15–16

U4V Uniform Motion Problems Problems involving motion at a constant rate are called uniform motion problems. In uniform motion problems, we often use an average rate when the actual rate is not constant. For example, you can drive all day and average 50 miles per hour, but you are not driving at a constant 50 miles per hour.

E X A M P L E 4 Finding the rate Bridgette drove her car for 2 hours on an icy road. When the road cleared up, she increased her speed by 35 miles per hour and drove 3 more hours, completing her 255-mile trip. How fast did she travel on the icy road?

Solution It is helpful to draw a diagram and then make a table to classify the given information. Remember that D � RT.

3.5x

Figure 2.3

Icy road Clear road

2 hrs x mph

3 hrs x � 35 mph

255 mi

Rate Time Distance

Icy road x 2 hr 2x mi

Clear road x � 35 3 hr 3(x � 35) mi mi � hr

mi � hr

The equation expresses the fact that her total distance traveled was 255 miles:

Icy road distance � clear road distance � total distance

2x � 3(x � 35) � 255

2x � 3x � 105 � 255

5x � 105 � 255

5x � 150

x � 30

x � 35 � 65

U Helpful Hint V

To get familiar with the problem, guess that she traveled 20 mph on the icy road and 55 mph (20 � 35) on the clear road. Her total distance would be

20 � 2 � 55 � 3 � 205 mi.

Of course this is not correct, but now you are familiar with the problem.

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2-49 2.6 Number, Geometric, and Uniform Motion Applications 133

E X A M P L E 5 Finding the time Pierce drove from Allentown to Baker, averaging 55 miles per hour. His journey back to Allentown using the same route took 3 hours longer because he averaged only 40 miles per hour. How long did it take him to drive from Allentown to Baker? What is the distance between Allentown and Baker?

Solution Draw a diagram and then make a table to classify the given information. Remember that D � RT.

If she drove at 30 miles per hour for 2 hours on the icy road, she went 60 miles. If she drove at 65 miles per hour for 3 hours on the clear road, she went 195 miles. Since 60 � 195 � 255, we can be sure that her speed on the icy road was 30 mph.

Now do Exercises 17–20

In the next uniform motion problem we find the time.

Allentown

Bakerx hr at 55 mph

x � 3 hr at 40 mph

Rate Time Distance

Going 55 x hr 55x mi

Returning 40 x � 3 hr 40(x � 3) mi mi � hr

mi � hr

We can write an equation expressing the fact that the distance either way is the same:

Distance going � distance returning

55x � 40(x � 3)

55x � 40x � 120

15x � 120

x � 8

The trip from Allentown to Baker took 8 hours. The distance between Allentown and Baker is 55 � 8, or 440 miles.

Now do Exercises 21–22

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U1V Number Problems Show a complete solution to each problem. See Example 1.

1. Consecutive integers. Find two consecutive integers whose sum is 79.

2. Consecutive odd integers. Find two consecutive odd integers whose sum is 56.

3. Consecutive integers. Find three consecutive integers whose sum is 141.

4. Consecutive even integers. Find three consecutive even integers whose sum is 114.

5. Consecutive odd integers. Two consecutive odd integers have a sum of 152. What are the integers?

6. Consecutive odd integers. Four consecutive odd integers have a sum of 120. What are the integers?

7. Consecutive integers. Find four consecutive integers whose sum is 194.

8. Consecutive even integers. Find four consecutive even integers whose sum is 340.

U3V Geometric Problems Show a complete solution to each problem. See Examples 2 and 3. See the Strategy for Solving Problems box on pages 130–131.

9. Olympic swimming. If an Olympic swimming pool is twice as long as it is wide and the perimeter is 150 meters, then what are the length and width?

Exercises

U Study Tips V • Make sure you know how your grade in this course is determined. How much weight is given to tests, homework, quizzes, and projects?

Does your instructor give any extra credit? • You should keep a record of all of your scores and compute your own final grade.

2 .6

Warm-Ups ▼

Fill in the blank. 1. motion is motion at a constant rate.

2. When solving a problem you should draw a figure and label it.

3. If x and x � 10 are angles, then

x � x � 10 � 90.

4. If x and x – 45 are angles, then

x � x – 45 � 180.

5. If x is an even integer, then x � 2 is an integer.

6. If x is an odd integer, then x � 2 is an integer.

True or false? 7. The first step in solving a word problem is to write the

equation.

8. You should always write down what the variable represents.

9. Diagrams and tables are used as aids in solving word problems.

10. If x is an odd integer, then x � 1 is also an odd integer.

11. The degree measures of two complementary angles can be represented by x and 90 � x.

12. The degree measures of two supplementary angles can be represented by x and x � 180.

134 Chapter 2 Linear Equations and Inequalities in One Variable 2-50

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2-51 2.6 Number, Geometric, and Uniform Motion Applications 135

10. Wimbledon tennis. If the perimeter of a tennis court is 228 feet and the length is 6 feet longer than twice the width, then what are the length and width?

11. Framed. Julia framed an oil painting that her uncle gave her. The painting was 4 inches longer than it was wide, and it took 176 inches of frame molding. What were the dimen- sions of the picture?

12. Industrial triangle. Geraldo drove his truck from Indianapolis to Chicago, then to St. Louis, and then back to Indianapolis. He observed that the second side of his triangular route was 81 miles short of being twice as long as the first side and that the third side was 61 miles longer than the first side. If he traveled a total of 720 miles, then how long is each side of this triangular route?

13. Triangular banner. A banner in the shape of an isosceles triangle has a base that is 5 inches shorter than either of the equal sides. If the perimeter of the banner is 34 inches, then what is the length of the equal sides?

Figure for Exercise 12

x 2x � 81

x � 61

Chicago

Indianapolis

St. Louis

Figure for Exercise 10

x2x � 6

Figure for Exercise 9

w 2w

14. Border paper. Dr. Good’s waiting room is 8 feet longer than it is wide. When Vincent wallpapered Dr. Good’s waiting room, he used 88 feet of border paper. What are the dimen- sions of Dr. Good’s waiting room?

15. Roof truss design. An engineer is designing a roof truss as shown in the accompanying figure. Find the degree measure of the angle marked w.

16. Another truss. Another truss is shown in the accompanying figure. Find the degree measure of the angle marked z.

Figure for Exercise 16

z � 6

3z z

Figure for Exercise 15

w 2w

2w � 40

Figure for Exercise 14

xx � 8

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136 Chapter 2 Linear Equations and Inequalities in One Variable 2-52

U4V Uniform Motion Problems

Show a complete solution to each problem. See Examples 4 and 5.

17. Highway miles. Bret drove for 4 hours on the freeway, and then decreased his speed by 20 miles per hour and drove for 5 more hours on a country road. If his total trip was 485 miles, then what was his speed on the freeway?

18. Walking and running. On Saturday morning, Lynn walked for 2 hours and then ran for 30 minutes. If she ran twice as fast as she walked and she covered 12 miles altogether, then how fast did she walk?

19. Driving all night. Kathryn drove her rig 5 hours before dawn and 6 hours after dawn. If her average speed was 5 miles per hour more in the dark and she covered 630 miles altogether, then what was her speed after dawn?

20. Commuting to work. On Monday, Roger drove to work in 45 minutes. On Tuesday he averaged 12 miles per hour more, and it took him 9 minutes less to get to work. How far does he travel to work?

21. Head winds. A jet flew at an average speed of 640 mph from Los Angeles to Chicago. Because of head winds the jet averaged only 512 mph on the return trip, and the return trip took 48 minutes longer. How many hours was the flight from Chicago to Los Angeles? How far is it from Chicago to Los Angeles?

22. Ride the Peaks. Penny’s bicycle trip from Colorado Springs to Pikes Peak took 1.5 hours longer than the return trip to Colorado Springs. If she averaged 6 mph on the way to Pikes Peak and 15 mph for the return trip, then how long was the ride from Colorado Springs to Pikes Peak?

Miscellaneous Solve each problem.

23. Perimeter of a frame. The perimeter of a rectangular frame is 64 in. If the width of the frame is 8 in. less than the length, then what are the length and width of the frame?

Figure for Exercise 17

x mph on freeway for 4 hours

x � 20 mph on country road for 5 hours

24. Perimeter of a box. The width of a rectangular box is 20% of the length. If the perimeter is 192 cm, then what are the length and width of the box?

25. Isosceles triangle. An isosceles triangle has two equal sides. If the shortest side of an isosceles triangle is 2 ft less than one of the equal sides and the perimeter is 13 ft, then what are the lengths of the sides?

26. Scalene triangle. A scalene triangle has three unequal sides. The perimeter of a scalene triangle is 144 m. If the first side is twice as long as the second side and the third side is 24 m longer than the second side, then what are the measures of the sides?

27. Angles of a scalene triangle. The largest angle in a scalene triangle is six times as large as the smallest. If the middle angle is twice the smallest, then what are the degree measures of the three angles?

28. Angles of a right triangle. If one of the acute angles in a right triangle is 38°, then what are the degree measures of all three angles?

29. Angles of an isosceles triangle. One of the equal angles in an isosceles triangle is four times as large as the smallest angle in the triangle. What are the degree measures of the three angles?

30. Angles of an isosceles triangle. The measure of one of the equal angles in an isosceles triangle is 10° larger than twice the smallest angle in the triangle. What are the degree measures of the three angles?

31. Super Bowl score. The 1977 Super Bowl was played in the Rose Bowl in Pasadena. In that football game the Oakland Raiders scored 18 more points than the Minnesota Vikings. If the total number of points scored was 46, then what was the final score for the game?

32. Top payrolls. Payrolls for the three highest paid baseball teams (the Yankees, Mets, and Cubs) for 2009 totaled $485 million (www.usatoday.com). If the team payroll for the Yankees was $52 million greater than the payroll for the Mets and the payroll for the Mets was $14 million greater than the payroll for the Cubs, then what was the 2009 payroll for each team?

33. Idabel to Lawton. Before lunch, Sally drove from Idabel to Ardmore, averaging 50 mph. After lunch she continued on to Lawton, averaging 53 mph. If her driving time after lunch was 1 hour less than her driving time before lunch and the total trip was 256 miles, then how many hours did she drive before lunch? How far is it from Ardmore to Lawton?

34. Norfolk to Chadron. On Monday, Chuck drove from Norfolk to Valentine, averaging 47 mph. On Tuesday, he continued on to Chadron, averaging 69 mph. His driving

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2-53 2.7 Discount, Investment, and Mixture Applications 137

time on Monday was 2 hours longer than his driving time on Tuesday. If the total distance from Norfolk to Chadron is 326 miles, then how many hours did he drive on Monday? How far is it from Valentine to Chadron?

35. Golden oldies. Joan Crawford, John Wayne, and James Stewart were born in consecutive years (Doubleday Almanac). Joan Crawford was the oldest of the three, and James Stewart was the youngest. In 1950, after all three had their birthdays, the sum of their ages was 129. In what years were they born?

36. Leading men. Bob Hope was born 2 years after Clark Gable and 2 years before Henry Fonda (Doubleday Almanac). In 1951, after all three of them had their birthdays, the sum of their ages was 144. In what years were they born?

37. Trimming a garage door. A carpenter used 30 ft of molding in three pieces to trim a garage door. If the long piece was 2 ft longer than twice the length of each shorter piece, then how long was each piece?

38. Fencing dog pens. Clint is constructing two adjacent rectangular dog pens. Each pen will be three times as long as it is wide, and the pens will share a common long side. If Clint has 65 ft of fencing, what are the dimensions of each pen?

Figure for Exercise 38

Figure for Exercise 37

In This Section

U1V Discount Problems

U2V Commission Problems

U3V Investment Problems

U4V Mixture Problems

2.7 Discount, Investment, and Mixture Applications

In this section, we continue our study of applications of algebra. The problems in this section involve percents.

U1V Discount Problems When an item is sold at a discount, the amount of the discount is usually described as being a percentage of the original price. The percentage is called the rate of discount. Multiplying the rate of discount and the original price gives the amount of the discount.

E X A M P L E 1 Finding the original price Ralph got a 12% discount when he bought his new 2010 Corvette Coupe. If the amount of his discount was $6606, then what was the original price of the Corvette?

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U2V Commission Problems A salesperson’s commission for making a sale is often a percentage of the selling price. Commission problems are very similar to other problems involving percents. The commission is found by multiplying the rate of commission and the selling price.

138 Chapter 2 Linear Equations and Inequalities in One Variable 2-54

E X A M P L E 2

E X A M P L E 3

Finding the original price When Susan bought her new car, she also got a discount of 12%. She paid $17,600 for her car. What was the original price of Susan’s car?

Solution Let x represent the original price for Susan’s car. The amount of discount is 12% of x, or 0.12x. We can write an equation expressing the fact that the original price minus the dis- count is the price Susan paid.

Original price � discount � sale price

x � 0.12x � 17,600

0.88x � 17,600 1.00x � 0.12x � 0.88x

x � � 17 0 , . 6 8 0 8 0

� Divide each side by 0.88.

x � 20,000

Check: 12% of $20,000 is $2400, and $20,000 � $2400 � $17,600. The original price of Susan’s car was $20,000.

Now do Exercises 3–4

U Helpful Hint V

To get familiar with the problem, guess that the original price was $30,000. Then her discount is 0.12(30,000) or $3600. The price she paid would be 30,000 � 3600 or $26,400, which is incorrect.

Real estate commission Sarah is selling her house through a real estate agent whose commission rate is 7%. What should the selling price be so that Sarah can get the $83,700 she needs to pay off the mortgage?

Solution Let x represent the original price. The discount is found by multiplying the 12% rate of dis- count and the original price:

Rate of discount � original price � amount of discount

0.12x � 6606

x � � 6 0 6 .1 0 2 6

� Divide each side by 0.12.

x � 55,050

To check, find 12% of $55,050. Since 0.12 � 55,050 � 6606, the original price of the Corvette was $55,050.

Now do Exercises 1–2

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Solution Let x be the selling price. The commission is 7% of x (not 7% of $83,700). Sarah receives the selling price less the sales commission:

Selling price � commission � Sarah’s share

x � 0.07x � 83,700

0.93x � 83,700 1.00x � 0.07x � 0.93x

x � � 83 0 , . 7 9 0 3 0

�

x � 90,000

Check: 7% of $90,000 is $6300, and $90,000 � $6300 � $83,700. So the house should sell for $90,000.

Now do Exercises 5–8

2-55 2.7 Discount, Investment, and Mixture Applications 139

E X A M P L E 4 Diversified investing Ruth Ann invested some money in a certificate of deposit with an annual yield of 9%. She invested twice as much in a mutual fund with an annual yield of 10%. Her interest from the two investments at the end of the year was $232. How much was invested at each rate?

Solution When there are many unknown quantities, it is often helpful to identify them in a table. Since the time is 1 year, the amount of interest is the product of the interest rate and the amount invested.

Since the total interest from the investments was $232, we can write the following equation:

CD interest � mutual fund interest � total interest

0.09x � 0.10(2x) � 232

0.09x � 0.20x � 232

0.29x � 232

x � � 0 2 . 3 2 2 9

�

x � 800 2x � 1600

U Helpful Hint V

To get familiar with the problem, guess that she invested $1000 at 9% and $2000 at 10%. Then her earnings in 1 year would be

0.09(1000) � 0.10(2000)

or $290, which is close but incorrect.

Interest Rate Amount Invested Interest for 1 Year

CD 9% x 0.09x

Mutual fund 10% 2x 0.10(2x)

U3V Investment Problems The interest on an investment is a percentage of the investment, just as the sales com- mission is a percentage of the sale amount. However, in investment problems we must often account for more than one investment at different rates. So it is a good idea to make a table, as in Example 4.

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140 Chapter 2 Linear Equations and Inequalities in One Variable 2-56

To check, we find the total interest:

0.09(800) � 0.10(1600) � 72 � 160 � 232

So Ruth Ann invested $800 at 9% and $1600 at 10%.

Now do Exercises 9–12

U4V Mixture Problems Mixture problems are concerned with the result of mixing two quantities, each of which contains another substance. Notice how similar the following mixture prob- lem is to the last investment problem.

E X A M P L E 5 Mixing milk How many gallons of milk containing 4% butterfat must be mixed with 80 gallons of 1% milk to obtain 2% milk?

Solution It is helpful to draw a diagram and then make a table to classify the given information.

� �

x gal milk 4% fat

80 gal milk 1% fat

x � 80 gal milk 2% fat

U Helpful Hint V

To get familiar with the problem, guess that we need 100 gal of 4% milk. Mixing that with 80 gal of 1% milk would produce 180 gal of 2% milk. Now the two milks separately have

0.04(100) � 0.01(80)

or 4.8 gal of fat. Together the amount of fat is 0.02(180) or 3.6 gal. Since these amounts are not equal, our guess is incorrect.

Percentage of Fat Amount of Milk Amount of Fat

4% milk 4% x 0.04x

1% milk 1% 80 0.01(80)

2% milk 2% x � 80 0.02(x � 80)

The equation expresses the fact that the total fat from the first two types of milk is the same as the fat in the mixture:

Fat in 4% milk � fat in 1% milk � fat in 2% milk

0.04x � 0.01(80) � 0.02(x � 80)

0.04x � 0.8 � 0.02x � 1.6 Simplify.

100(0.04x � 0.8) � 100(0.02x � 1.6) Multiply each side by 100.

4x � 80 � 2x � 160 Distributive property

2x � 80 � 160 Subtract 2x from each side.

2x � 80 Subtract 80 from each side.

x � 40 Divide each side by 2.

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U1V Discount Problems

Show a complete solution to each problem. See Examples 1 and 2.

1. Close-out sale. At a 25% off sale, Jose saved $80 on a 19-inch Panasonic TV. What was the original price of the television?

2. Nice tent. A 12% discount on a Walrus tent saved Melanie $75. What was the original price of the tent?

3. Circuit city. After getting a 20% discount, Robert paid $320 for a Pioneer CD player for his car. What was the original price of the CD player?

4. Chrysler Sebring. After getting a 15% discount on the price of a new Chrysler Sebring convertible, Helen paid $27,000. What was the original price of the convertible to the nearest dollar?

2 .7Exercises

U Study Tips V • Find out what kinds of help are available for commuting students, online students, and on-campus students. • Sometimes a minor issue can be resolved very quickly and you can get back on the path to success.

2-57 2.7 Discount, Investment, and Mixture Applications 141

To check, calculate the total fat:

2% of 120 gallons � 0.02(120) � 2.4 gallons of fat

0.04(40) � 0.01(80) � 1.6 � 0.8 � 2.4 gallons of fat

So we mix 40 gallons of 4% milk with 80 gallons of 1% milk to get 120 gallons of 2% milk.

Now do Exercises 13–16

In mixture problems, the solutions might contain fat, alcohol, salt, or some other sub- stance. We always assume that the substance neither appears nor disappears in the process. For example, if there are 3 grams of salt in one glass of water and 2 grams in another, then there are exactly 5 grams in a mixture of the two.

Warm-Ups ▼

Fill in the blank. 1. The of discount is a percentage.

2. The is the amount by which a price is reduced.

3. The of the original price and the rate of discount is the discount.

4. A helps us to organize information given in a word problem.

5. An interest is a percentage.

True or false? 6. If Jim gets a 12% commission for selling a $1000

Wonder Vac, then his commission is $120.

7. If Bob earns a 5% commission on an $80,000 motorhome sale, then Bob earns $400.

8. If Sue gets a 20% discount on a TV with a list price of x dollars, then Sue pays 0.8x dollars.

9. If you get a 6% discount on a Chevy Volt for which the MSRP is x dollars, then your discount is 0.6x dollars.

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142 Chapter 2 Linear Equations and Inequalities in One Variable 2-58

U2V Commission Problems

Show a complete solution to each problem. See Example 3.

5. Selling price of a home. Kirk wants to get $115,000 for his house. The real estate agent gets a commission equal to 8% of the selling price for selling the house. What should the selling price be?

6. Horse trading. Gene is selling his palomino at an auction. The auctioneer’s commission is 10% of the selling price. If Gene still owes $810 on the horse, then what must the horse sell for so that Gene can pay off his loan?

7. Sales tax collection. Merilee sells tomatoes at a roadside stand. Her total receipts including the 7% sales tax were $462.24. What amount of sales tax did she collect?

8. Toyota Corolla. Gwen bought a new Toyota Corolla. The selling price plus the 8% state sales tax was $15,714. What was the selling price?

U3V Investment Problems

Show a complete solution to each problem. See Example 4.

9. Wise investments. Wiley invested some money in the Berger 100 Fund and $3000 more than that amount in the Berger 101 Fund. For the year he was in the fund, the 100 Fund paid 18% simple interest and the 101 Fund paid 15% simple interest. If the income from the two investments totaled $3750 for 1 year, then how much did he invest in each fund?

10. Loan shark. Becky lent her brother some money at 8% simple interest, and she lent her sister twice as much at twice the interest rate. If she received a total of 20 cents interest, then how much did she lend to each of them?

11. Investing in bonds. David split his $25,000 inheritance between Fidelity Short-Term Bond Fund with an annual yield of 5% and T. Rowe Price Tax-Free Short-Intermediate Fund with an annual yield of 4%. If his total income for 1 year on the two investments was $1140, then how much did he invest in each fund?

Photo for Exercise 5

12. High-risk funds. Of the $50,000 that Natasha pocketed on her last real estate deal, $20,000 went to charity. She invested part of the remainder in Dreyfus New Leaders Fund with an annual yield of 16% and the rest in Templeton Growth Fund with an annual yield of 25%. If she made $6060 on these investments in 1 year, then how much did she invest in each fund?

U4V Mixture Problems

Show a complete solution to each problem. See Example 5.

13. Mixing milk. How many gallons of milk containing 1% butterfat must be mixed with 30 gallons of milk containing 3% butterfat to obtain a mixture containing 2% butterfat?

14. Acid solutions. How many gallons of a 5% acid solution should be mixed with 30 gallons of a 10% acid solution to obtain a mixture that is 8% acid?

15. Alcohol solutions. Gus has on hand a 5% alcohol solution and a 20% alcohol solution. He needs 30 liters of a 10% alcohol solution. How many liters of each solution should he mix together to obtain the 30 liters?

16. Adjusting antifreeze. Angela needs 20 quarts of 50% antifreeze solution in her radiator. She plans to obtain this by mixing some pure antifreeze with an appropriate amount of a 40% antifreeze solution. How many quarts of each should she use?

Miscellaneous Solve each problem.

17. Registered voters. If 60% of the registered voters of Lancaster County voted in the November election and 33,420 votes were cast, then how many registered voters are there in Lancaster County?

Figure for Exercise 16

40% solution ? qts

50% solution 20 qts

� � 100%

antifreeze ? qts

Figure for Exercise 13

x gal 1% fat �

30 gal 3% fat �

x � 30 gal 2% fat

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2-59 2.7 Discount, Investment, and Mixture Applications 143

18. Tough on crime. In a random sample of voters, 594 respondents said that they favored passage of a $33 billion crime bill. If the number in favor of the crime bill was 45% of the number of voters in the sample, then how many voters were in the sample?

19. Ford Taurus. At an 8% sales tax rate, the sales tax on Peter’s new Ford Taurus was $1200. What was the price of the car?

20. Taxpayer blues. Last year, Faye paid 24% of her income to taxes. If she paid $9600 in taxes, then what was her income?

21. Making a profit. A retail store buys shirts for $8 and sells them for $14. What percent increase is this?

22. Monitoring AIDS. If 28 new AIDS cases were reported in Landon County this year and 35 new cases were reported last year, then what percent decrease in new cases is this?

23. High school integration. Wilson High School has 400 students, of whom 20% are African American. The school board plans to merge Wilson High with Jefferson High. This one school will then have a student population that is 44% African American. If Jefferson currently has a student population that is 60% African American, then how many students are at Jefferson?

24. Junior high integration. The school board plans to merge two junior high schools into one school of 800 students in which 40% of the students will be Caucasian. One of the schools currently has 58% Caucasian students; the other has only 10% Caucasian students. How many students are in each of the two schools?

25. Hospital capacity. When Memorial Hospital is filled to capacity, it has 18 more people in semiprivate rooms (two patients to a room) than in private rooms. The room rates are $200 per day for a private room and $150 per day for a semiprivate room. If the total receipts for rooms is $17,400 per day when all are full, then how many rooms of each type does the hospital have?

Photo for Exercise 17

26. Public relations. Memorial Hospital is planning an advertising campaign. It costs the hospital $3000 each time a television ad is aired and $2000 each time a radio ad is aired. The administrator wants to air 60 more television ads than radio ads. If the total cost of airing the ads is $580,000, then how many ads of each type will be aired?

27. Mixed nuts. Cashews sell for $4.80 per pound, and pista- chios sell for $6.40 per pound. How many pounds of pistachios should be mixed with 20 pounds of cashews to get a mixture that sells for $5.40 per pound?

28. Premium blend. Premium coffee sells for $6.00 per pound, and regular coffee sells for $4.00 per pound. How many pounds of each type of coffee should be blended to obtain 100 pounds of a blend that sells for $4.64 per pound?

29. Nickels and dimes. Candice paid her library fine with 10 coins consisting of nickels and dimes. If the fine was $0.80, then how many of each type of coin did she use?

30. Dimes and quarters. Jeremy paid for his breakfast with 36 coins consisting of dimes and quarters. If the bill was $4.50, then how many of each type of coin did he use?

31. Cooking oil. Crisco Canola Oil is 7% saturated fat. Crisco blends corn oil that is 14% saturated fat with Crisco Canola Oil to get Crisco Canola and Corn Oil, which is 11% saturated fat. How many gallons of corn oil must Crisco mix with 600 gallons of Crisco Canola Oil to get Crisco Canola and Corn Oil?

32. Chocolate ripple. The Delicious Chocolate Shop makes a dark chocolate that is 35% fat and a white chocolate that is 48% fat. How many kilograms of dark chocolate should be mixed with 50 kilograms of white chocolate to make a ripple blend that is 40% fat?

33. Hawaiian Punch. Hawaiian Punch is 10% fruit juice. How much water would you have to add to one gallon of Hawaiian Punch to get a drink that is 6% fruit juice?

34. Diluting wine. Arestaurant manager has 2 liters of white wine that is 12% alcohol. How many liters of white grape juice

should he add to get a drink that is 10% alcohol?

35. Bargain hunting. A smart shopper bought 5 pairs of shorts and 8 tops for a total of $108. If the price of a pair of shorts was twice the price of a top, then what was the price of each type of clothing?

36. VCRs and CDs. The manager of a stereo shop placed an order for $10,710 worth of VCRs at $120 each and CD players at $150 each. If the number of VCRs she ordered was three times the number of CD players, then how many of each did she order?

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144 Chapter 2 Linear Equations and Inequalities in One Variable 2-60

In This Section

U1V Inequalities

U2V Graphing Inequalities

U3V Graphing Compound Inequalities

U4V Checking Inequalities

U5V Writing Inequalities

2.8 Inequalities

In Chapter 1, we defined inequality in terms of the number line. One number is greater than another number if it lies to the right of the other number on the number line. In this section, you will study inequality in greater depth.

U1V Inequalities The symbols used to express inequality and their meanings are given in the following box.

U Helpful Hint V

A good way to learn inequality sym- bols is to notice that the inequality symbol always points at the smaller number. This observation will help you read an inequality such as �2 � x. Reading right to left, we say that x is greater than �2. It is usually easier to understand an inequality if you read the variable first.

Inequality Symbols

Symbol Meaning � Is less than Is less than or equal to Is greater than � Is greater than or equal to

U Calculator Close-Up V

A graphing calculator can determine whether an inequality is correct. Use the inequality symbols from the TEST menu to enter the inequality.

When ENTER is pressed, the calculator returns a 1 if the inequality is correct or a 0 if the inequality is incorrect.

Verifying inequalities Determine whether each of the following statements is correct.

a) 3 � 4 b) �1 � �2 c) �2 0

d) 0 � 0 e) 2(�3) � 8 9 f) (�2)(�5) 10

Solution a) Locate 3 and 4 on the number line shown in Fig. 2.6. Because 3 is to the left of 4

on the number line, 3 � 4 is correct.

b) Locate �1 and �2 on the number line shown in Fig. 2.6. Because �1 is to the right of �2, on the number line, �1 � �2 is not correct.

c) Because �2 is to the left of 0 on the number line, �2 0 is correct.

d) Because 0 is equal to 0, 0 � 0 is correct.

e) Simplify the left side of the inequality to get 2 9, which is not correct.

f ) Simplify the left side of the inequality to get 10 10, which is correct.

Now do Exercises 1–16

Figure 2.6

0 1 2 3 4�2�3 �1

E X A M P L E 1

The statement a � b means that a is to the left of b on the number line as shown in Fig. 2.4. The statement c d means that c is to the right of d on the number line, as shown in Fig. 2.5. Of course, a � b has the same meaning as b a. The statement a b means that either a is to the left of b or a corresponds to the same point as b on the number line. The statement a b has the same meaning as the statement b � a.

Figure 2.5Figure 2.4

0 1 2 3

c d (or d � c) d c

�2�3 �10 1 2 3

a � b (or b a) a b

�4 �1�2�3

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2-61 2.8 Inequalities 145

U2V Graphing Inequalities If a is a fixed real number, then any real number x located to the right of a on the num- ber line satisfies x � a. The set of real numbers located to the right of a on the number line is the solution set to x � a. This solution set is written in set-builder notation as {x | x � a}, or more simply in interval notation as (a, �). We graph the inequality by graphing the solution set (a, �). Recall from Chapter 1 that a bracket means that an endpoint is included in an interval and a parenthesis means that an endpoint is not included in an interval. Remember also that � is not a number. It simply indicates that there is no end to the interval.

U3V Graphing Compound Inequalities A statement involving more than one inequality is a compound inequality. We will study one type of compound inequality here and see other types in Section 8.1.

If a and b are real numbers and a � b, then the compound inequality a � x � b

means that a � x and x � b. Reading x first makes a � x � b clearer: “x is greater than a and x is less than b.”

If x is greater than a and less than b, then x is between a and b. So the solution set to a � x � b is the interval (a, b).

E X A M P L E 2 Graphing inequalities State the solution set to each inequality in interval notation and sketch its graph.

a) x � 5 b) �2 � x c) x � 10

Solution a) All real numbers less than 5 satisfy x � 5. The solution set is the interval (��, 5)

and the graph of the solution set is shown in Fig. 2.7.

b) The inequality �2 � x indicates that x is greater than �2. The solution set is the interval (�2, �) and the graph of the inequality is shown in Fig. 2.8.

c) All real numbers greater than or equal to 10 satisfy x � 10. The solution set is the interval [10, �) and the graph is shown in Fig. 2.9.

Now do Exercises 17–28

Figure 2.9Figure 2.8

0–10 10 20 300�2�4 2 4 6

U Helpful Hint V

A person in debt has a negative net worth. If Bob’s net worth is �$8000 and Mary’s net worth is �$3000, then Bob certainly has the greater debt, but we write

�8000 � �3000

because �8000 lies to the left of �3000 on the number line.

0 1 2 3 4 65

Figure 2.7

E X A M P L E 3 Graphing compound inequalities State the solution set to each inequality in interval notation and sketch its graph.

a) 2 � x � 3 b) �2 � x � 1

Solution a) All real numbers between 2 and 3 satisfy 2 � x � 3. The solution set is the interval

(2, 3), and the graph of the solution set is shown in Fig. 2.10 on the next page.

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146 Chapter 2 Linear Equations and Inequalities in One Variable 2-62

E X A M P L E 4 Checking inequalities Determine whether the given number satisfies the inequality following it.

a) 0, 2x � 3 � �5 b) �4, x � 5 � 2x 1 c) 1 3 3 , 6 � 3x � 5 � 14

Solution a) Replace x by 0 in the inequality and simplify:

2x � 3 � �5 2 � 0 � 3 � �5

�3 � �5 Incorrect

Since this last inequality is incorrect, 0 is not a solution to the inequality.

b) Replace x by �4 and simplify:

x � 5 � 2x 1 �4 � 5 � 2(�4) 1

�9 � �7 Incorrect

Since this last inequality is incorrect, �4 is not a solution to the inequality.

c) Replace x by 1 3 3 and simplify:

6 � 3x � 5 � 14

6 � 3 � 1 3 3 � 5 � 14

6 � 13 � 5 � 14 6 � 8 � 14 Correct

Since 8 is greater than 6 and less than 14, this inequality is correct. So 1 3 3 satisfies

the original inequality.

Now do Exercises 47–64

U Calculator Close-Up V

To check 13�3 in 6 � 3x � 5 � 14

we check each part of the compound inequality separately.

Because both parts of the compound inequality are correct, 13�3 satisfies the compound inequality.

b) The real numbers that satisfy �2 � x � 1 are between �2 and 1, including �2 but not including 1. So the solution set is the interval [�2, 1), and the graph of this compound inequality is shown in Fig. 2.11.

Now do Exercises 29–36

We write a � x � b only if a � b, and we write a � x � b only if a � b. Similar rules hold for � and �. So 4 � x � 9 and �6 � x � �8 are correct uses of this notation, but 5 � x � 2 is not correct. Also, the inequalities should not point in opposite directions as in 5 � x � 7.

U4V Checking Inequalities In Examples 2 and 3 we determined the solution sets to some inequalities. In Section 2.9, more complicated inequalities will be solved by using steps similar to those used for solving equations. In Example 4, we determine whether a given num- ber satisfies an inequality of the type that we will be solving in Section 2.9.

CAUTION

Figure 2.11Figure 2.10

�1�2�3 0 1 20 1 2 3 4

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2-63 2.8 Inequalities 147

U5V Writing Inequalities Inequalities occur in applications, just as equations do. Certain verbal phrases indicate inequalities. For example, if you must be at least 18 years old to vote, then you can vote if you are 18 or older. The phrase “at least” means “greater than or equal to.” If an elevator has a capacity of at most 20 people, then it can hold 20 people or fewer. The phrase “at most” means “less than or equal to.”

E X A M P L E 5 Writing inequalities Write an inequality that describes each situation.

a) Lois plans to spend at most $500 on a washing machine including the 9% sales tax.

b) The length of a certain rectangle must be 4 meters longer than the width, and the perimeter must be at least 120 meters.

c) Fred made a 76 on the midterm exam. To get a B, the average of his midterm and his final exam must be between 80 and 90.

Solution a) If x is the price of the washing machine, then 0.09x is the amount of sales tax.

Since the total must be less than or equal to $500, the inequality is

x 0.09x � 500.

b) If W represents the width of the rectangle, then W 4 represents the length. Since the perimeter (2W 2L) must be greater than or equal to 120, the inequality is

2(W ) 2(W 4) � 120.

c) If we let x represent Fred’s final exam score, then his average is x 2

76 .

To indicate that the average is between 80 and 90, we use the compound inequality

80 � x

2 76

� 90.

Now do Exercises 73–85

In Example 5(b) you are given that L is 4 meters longer than W. So L � W 4, and you can use W 4 in place of L. If you knew only that L was longer than W, then you would know only that L � W.

CAUTION

Warm-Ups ▼

Fill in the blank. 1. The symbols �, �, �, and � are symbols.

2. To graph x � a on a number line we use a at a.

3. To graph x � a on a number line we use a at a.

4. A inequality involves more than one inequality.

5. If a � x � b, then x is a and b.

True or false? 6. –2 � �2

7. –5 � �6 � 7

8. �3 � �2 � �1

9. The inequalities x � 7 and 7 � x have the same graph.

10. The graph of x � �3 includes the point at �3.

11. The number �3 is a solution to �2 � x.

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U1V Inequalities

Determine whether each of the following statements is true. See Example 1.

1. �5 � �8 2. �6 � �3 3. �3 � 5 4. �6 � 0 5. 4 � 4 6. �3 � �3 7. �6 � �5 8. �2 � �9 9. �4 � �3

10. �5 � �10 11. (�3)(4) � 1 � 0 � 3 12. 2(4) � 6 � �3(5) � 1 13. �4(5) � 6 � 5(�6) 14. 4(8) � 30 � 7(5) � 2(17) 15. 7(4) � 12 � 3(9) � 2 16. �3(4) � 12 � 2(3) � 6

U2V Graphing Inequalities

State the solution set to each inequality in interval notation and sketch its graph. See Example 2.

17. x � 3

18. x � �7

19. x � �2

20. x � 4

21. �1 � x

22. 0 � x

23. �2 � x

24. �5 � x

25. x �

26. x � �

27. x � 5.3

28. x � �3.4

U3V Graphing Compound Inequalities

State the solution set to each inequality in interval notation and sketch its graph. See Example 3.

29. �3 � x � 1

30. 0 � x � 5

31. 3 � x � 7

32. �3 � x � �1

33. �5 � x � 0

2 3

1 2

Exercises

U Study Tips V • Be careful not to spend too much time on a single problem when taking a test. If a problem seems to be taking too much time, you

might be on the wrong track. Be sure to finish the test. • Before you take a test on this chapter, work the test given in this book at the end of this chapter. This will give you a good idea of your

test readiness.

2 .8

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2-65 2.8 Inequalities 149

34. �2 � x � 2

35. 40 � x � 100

36. 0 � x � 600

For each graph, write the corresponding inequality and the solution set to the inequality using interval notation.

37.

38.

39.

40.

41.

42.

43.

44.

45.

46.

U4V Checking Inequalities

Determine whether the given number satisfies the inequality following it. See Example 4.

47. �9, �x � 3

48. 5, �3 � �x

49. �2, 5 � x

50. 4, 4 � x

51. �6, 2x � 3 � �11

52. 4, 3x � 5 � 7

53. 3, �3x 4 � �7

54. �4, �5x 1 � �5

5�5 �2�3�4 �1 0 1 2 3 4

0 2 8�2�4�6 4 6

5�5 �2�3�4 �1 0 1 2 3 4

50 1 2 3 4�5�4�3�2�1

5�5�4�3�2�1 0 1 2 3 4

�6�5�4�3�2�1 0 1 2 3 4

0 1 2 3 4�4�3�2�1 5 6

0 1 2 3 4 5�3�2�1 76

55. 0, 3x � 7 � 5x � 7

56. 0, 2x 6 � 4x � 9

57. 2.5, �10x 9 � 3(x 3)

58. 1.5, 2x � 3 � 4(x � 1)

59. �7, �5 � x � 9

60. �9, �6 � x � 40

61. �2, �3 � 2x 5 � 9

62. �5, �3 � �3x � 7 � 8

63. �3.4, �4.25x � 13.29 � 0.89

64. 4.8, 3.25x � 14.78 � 1.3

For each inequality, determine which of the numbers �5.1, 0, and 5.1 satisfies the inequality.

65. x � �5 66. x � 0 67. 5 � x 68. �5 � x 69. 5 � x � 7 70. 5 � �x � 7 71. �6 � �x � 6 72. �5 � x � 0.1 � 5

U5V Writing Inequalities

Write an inequality to describe each situation. Do not solve. See Example 5.

73. Sales tax. At an 8% sales tax rate, Susan paid more than $1500 sales tax when she purchased her new Camaro. Let p represent the price of the Camaro.

74. Internet shopping. Carlos paid less than $1000 including $40 for shipping and 9% sales tax when he bought his new computer. Let p represent the price of the computer.

75. Fine dining. At Burger Brothers the price of a hamburger is twice the price of an order of French fries, and the price of a Coke is $0.25 more than the price of the fries. Burger Brothers advertises that you can get a complete meal (burger, fries, and Coke) for under $2.00. Let p represent the price of an order of fries.

76. Cats and dogs. Willow Creek Kennel boards only cats and dogs. One Friday night there were twice as many dogs as cats in the kennel and at least 30 animals spent the night there. Let d represent the number of dogs.

77. Barely passing. Travis made 44 and 72 on the first two tests in algebra and has one test remaining. The average on the three tests must be at least 60 for Travis to pass the course. Let s represent his score on the last test.

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150 Chapter 2 Linear Equations and Inequalities in One Variable 2-66

Figure for Exercise 83

G ir

th (

in .)

160

120

130

110

100 100 15 20 25

Height (in.)

140

150

78. Ace the course. Florence made 87 on her midterm exam in psychology. The average of her midterm and her final must be at least 90 to get an A in the course. Let s represent her

score on the final.

79. Coast to coast. On Howard’s recent trip from Bangor to San Diego, he drove for 8 hours each day and traveled between 396 and 453 miles each day. Let R represent his average speed for each day.

80. Mother’s Day present. Bart and Betty are looking at color televisions that range in price from $399.99 to $579.99. Bart can afford more than Betty and has agreed to spend $100 more than Betty when they purchase this gift for their mother. Let b represent Betty’s portion of the gift.

81. Positioning a ladder. Write an inequality in the variable x for the degree measure of the angle at the base of the ladder shown in the figure, given that the angle at the base must be between 60° and 70°.

82. Building a ski ramp. Write an inequality in the variable x for the degree measure of the smallest angle of the triangle shown in the figure, given that the degree measure of the smallest angle is at most 30°.

83. Maximum girth. United Parcel Service defines the girth of a box as the sum of the length, twice the width, and twice the height. The maximum girth that UPS will ship is 130 in.

a) If a box has a length of 45 in. and a width of 30 in., then what inequality must be satisfied by the height?

Figure for Exercise 82

x � 8

x ?

Figure for Exercise 81

b) The accompanying graph shows the girth of a box with a length of 45 in., a width of 30 in., and height of h in. Use the graph to estimate the maximum height that is allowed for this box.

Figure for Exercise 84

B at

tin g

av er

ag e

0.6

0.3

0.4

0.2

0.1

500 0

100 150 200 Number of hits in 337 at bats

0.5

84. Batting average. Near the end of the season a professional baseball player has 93 hits in 317 times at bat for an average of 93�317 or 0.293. He gets a $1 million bonus if his season average is over 0.300. He estimates that he will bat 20 more times before the season ends. Let x represent the number of hits in the last 20 at bats of the season.

a) Write an inequality that must be satisfied for him to get the bonus.

b) Use the accompanying graph to estimate the number of hits in 337 at bats that will put his average over 0.300.

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2-67 2.9 Solving Inequalities and Applications 151

Solve.

85. Bicycle gear ratios. The gear ratio r for a bicycle is defined by the formula

r � N n w ,

where N is the number of teeth on the chainring (by the pedal), n is the number of teeth on the cog (by the wheel), and w is the wheel diameter in inches (Cycling, Burkett and Darst). The accompanying chart gives uses for the various gear ratios. A bicycle with a 27-inch-diameter wheel has 50 teeth on the chainring and 17 teeth on the cog. Find the gear ratio and indicate what this gear ratio is good for.

Ratio Use

r � 90 hard pedaling on level ground

70 � r � 90 moderate effort on level ground

50 � r � 70 mild hill climbing

35 � r � 50 long hill climbing with load

Figure for Exercise 85

Math at Work Body Mass Index

Medical professionals say that two-thirds of all Americans are overweight and excess weight has about the same effect on life expectancy as smoking. How can you tell if you are overweight or normal? Body mass index (BMI) can help you decide. To determine BMI divide your weight in kilograms by the square of your height in meters. Don’t know your weight and height in the metric sys- tem? Then use the formula BMI � 703W/H2, where W is your weight in pounds and H is your height in inches.

If 23 � BMI � 25, then you are probably not overweight. If BMI � 26, then you are probably overweight and are statistically likely to have a lower life expectancy. According to the National Heart, Lung, and Blood Institute, you are over- weight if 25 � BMI � 29.9 and obese if BMI � 30. If your BMI is between 17 and 22, your life span might be longer than average. Men are usually happy with a BMI between 23 and 25 and women like to see their BMI between 20 and 22. However, BMI does not distinguish between muscle and fat and can wrongly suggest that a person with a short muscular build is overweight. Also, the BMI does not work well for children, because normal varies with age.

If you want to learn more about body mass index or don’t want to do the calculations yourself, then check out any of the numerous websites that discuss BMI and even have online BMI calculators. Just do a search for body mass index.

In This Section

U1V Rules for Inequalities

U2V Solving Inequalities

U3V Applications

2.9 Solving Inequalities and Applications

To solve equations, we write a sequence of equivalent equations that ends in a very simple equation whose solution is obvious. In this section, you will learn that the procedure for solving inequalities is the same. However, the rules for performing operations on each side of an inequality are slightly different from the rules for equations.

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152 Chapter 2 Linear Equations and Inequalities in One Variable 2-68

U1V Rules for Inequalities Equivalent inequalities are inequalities that have exactly the same solutions. Inequalities such as x � 3 and x 2 � 5 are equivalent because any number that is larger than 3 certainly satisfies x 2 � 5 and any number that satisfies x 2 � 5 must certainly be larger than 3.

We can get equivalent inequalities by performing operations on each side of an inequality just as we do for solving equations. If we start with the inequality 6 � 10 and add 2 to each side, we get the true statement 8 � 12. Examine the results of per- forming the same operation on each side of 6 � 10.

U Helpful Hint V

You can think of an inequality like a seesaw that is out of balance.

If the same weight is added to or sub- tracted from each side, it will remain in the same state of imbalance.

50 � 20

Addition Property of Inequality

If we add the same number to each side of an inequality, we get an equivalent inequality. If a � b, then a c � b c.

The addition property of inequality also enables us to subtract the same number from each side of an inequality because subtraction is defined in terms of addition.

Perform these operations on each side:

Add 2 Subtract 2 Multiply by 2 Divide by 2

Start with 6 � 10 8 � 12 4 � 8 12 � 20 3 � 5

All of the resulting inequalities are correct. Now if we repeat these operations using �2, we get the following results.

Perform these operations on each side:

Add �2 Subtract �2 Multiply by �2 Divide by �2

Start with 6 � 10 4 � 8 8 � 12 �12 � �20 �3 � �5

Notice that the direction of the inequality symbol is the same for all of the results except the last two. When we multiplied each side by �2 and when we divided each side by �2, we had to reverse the inequality symbol to get a correct result. These tables illustrate the rules for solving inequalities.

The multiplication property of inequality also enables us to divide each side of an inequality by a nonzero number because division is defined in terms of multiplication. So if we multiply or divide each side by a negative number, the inequality symbol is reversed.

U Helpful Hint V

Changing the signs of numbers changes their relative position on the number line. For example, 3 lies to the left of 5 on the number line, but �3 lies to the right of �5. So 3 � 5, but �3 � �5. Since multiplying and dividing by a negative cause sign changes, these operations reverse the inequality.

Multiplication Property of Inequality

If we multiply each side of an inequality by the same positive number, we get an equivalent inequality. If a � b and c � 0, then ac � bc. If we multiply each side of an inequality by the same negative number and reverse the inequality symbol, we get an equivalent inequality. If a � b and c � 0, then ac � bc.

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2-69 2.9 Solving Inequalities and Applications 153

E X A M P L E 1 Writing equivalent inequalities Write the appropriate inequality symbol in the blank so that the two inequalities are equivalent.

a) x 3 � 9, x _____ 6 b) �2x � 6, x _____ �3

Solution a) If we subtract 3 from each side of x 3 � 9, we get the equivalent inequality x � 6.

b) If we divide each side of �2x � 6 by �2, we get the equivalent inequality x � �3.

Now do Exercises 1–10

E X A M P L E 2 Isolating the variable on the left side Solve the inequality 4x � 5 � 19. State the solution set using interval notation and sketch its graph.

Solution 4x � 5 � 19 Original inequality

4x � 5 5 � 19 5 Add 5 to each side.

4x � 24 Simplify.

x � 6 Divide each side by 4.

Since the last inequality is equivalent to the first, it has the same solution set as the first. So the solution set to 4x � 5 � 19 is (6, �). The graph is shown in Fig. 2.12.

Now do Exercises 11–12Figure 2.12 4 5 6 7 81 2 3

We use the properties of inequality just as we use the properties of equality. However, when we multiply or divide each side by a negative number, we must reverse the inequality symbol.

U2V Solving Inequalities To solve inequalities, we use the properties of inequality to isolate x on one side.

CAUTION

U Calculator Close-Up V

You can use the TABLE feature of a graphing calculator to numerically support the solu- tion to the inequality 4x � 5 � 19 in Example 2. Use the Y � key to enter the equation y1 � 4x � 5.

Next, use TBLSET to set the table so that the values of x start at 4.5 and the change in x is 0.5.

Notice that when x is larger than 6, y1 (or 4x � 5) is larger than 19. The table verifies or supports the algebraic solu- tion, but it should not replace the algebraic method.

Finally, press TABLE to see lists of x-values and the corre- sponding y-values.

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154 Chapter 2 Linear Equations and Inequalities in One Variable 2-70

Rewriting 3 2

� x as x � 3 2

in Example 3 is not “reversing the inequality.”

Multiplying or dividing each side of 3 2

� x by a negative number would reverse the inequality. For example, multiplying by �1 yields �

3 2

� �x. In Example 4, we divide each side of an inequality by a negative number and reverse the inequality symbol.

Remember that 5 � x is equivalent to x � 5. So the variable can be isolated on the right side of an inequality as shown in Example 3.

E X A M P L E 3 Isolating the variable on the right side Solve the inequality 5x � 2 � 7x � 5. State the solution set using interval notation and sketch its graph.

Solution 5x � 2 � 7x � 5 Original inequality

5x � 2 � 5x � 7x � 5 � 5x Subtract 5x from each side.

�2 � 2x � 5 Simplify.

3 � 2x Add 5 to each side.

3 2

� x Divide each side by 2.

Note that 3 2

� x is equivalent to x � 3 2

. The solution set is the interval � 32 , �� and the graph is shown in Fig. 2.13. Notice that 3

2 is halfway between 1 and 2 on the number line.

Now do Exercises 13–16

Figure 2.13

0 1 2 3�1�2�3

3 2

E X A M P L E 4 Reversing the inequality symbol Solve 5 � 5x � 1 2(5 � x). State the solution set in interval notation and sketch its graph.

Solution

5 � 5x � 1 2(5 � x) Original inequality

5 � 5x � 11 � 2x Simplify the right side.

5 � 3x � 11 Add 2x to each side.

�3x � 6 Subtract 5 from each side.

x � �2 Divide each side by �3, and reverse the inequality.

The solution set is the interval [�2, �) and the graph is shown in Fig. 2.14.

Now do Exercises 17–38

Figure 2.14

0 1�1�6 �5 �4 �3 �2

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2-71 2.9 Solving Inequalities and Applications 155

We can use the rules for solving inequalities on the compound inequalities that we studied in Section 2.8.

There are many negative numbers in Example 5, but the inequality was not reversed, since we did not multiply or divide by a negative number. An inequality is reversed only if you multiply or divide by a negative number.

CAUTION

E X A M P L E 5 Solving a compound inequality Solve �9 � 2

3 x � 7 � 5. State the solution set in interval notation and sketch its graph.

Solution

�9 � 2 3 x � 7 � 5 Original inequality

�9 7 � 2 3 x � 7 7 � 5 7 Add 7 to each part.

�2 � 2 3 x � 12 Simplify.

3 2

(�2) � 3 2

� 2 3 x �

3 2

� 12 Multiply each part by 32 .

�3 � x � 18 Simplify.

Since the last compound inequality is equivalent to the first, the solution set is [�3, 18). The graph is shown in Fig. 2.15.

Now do Exercises 39–42

Figure 2.15

0 3 6 9 12 15 18�3

E X A M P L E 6 Reversing inequality symbols in a compound inequality Solve �3 � 5 � x � 5. State the solution set in interval notation and sketch its graph.

Solution �3 � 5 � x � 5 Original inequality

�3 � 5 � 5 � x � 5 � 5 � 5 Subtract 5 from each part.

�8 � �x � 0 Simplify.

(�1)(�8) � (�1)(�x) � (�1)(0) Multiply each part by �1, reversing the inequality symbols.

8 � x � 0

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156 Chapter 2 Linear Equations and Inequalities in One Variable 2-72

U3V Applications Example 7 shows how inequalities can be used in applications.

E X A M P L E 7 Averaging test scores Mei Lin made a 76 on the midterm exam in history. To get a B, the average of her midterm and her final exam must be between 80 and 90. For what range of scores on the final exam will she get a B?

Solution Let x represent the final exam score. Her average is then x

2 76

. The inequality

expresses the fact that the average must be between 80 and 90:

80 � x

76 � 90

2(80) � 2� x 2 76

� � 2(90) Multiply each part by 2. 160 � x 76 � 180 Simplify.

160 � 76 � x 76 � 76 � 180 � 76 Subtract 76 from each part.

84 � x � 104 Simplify.

The last inequality indicates that Mei Lin’s final exam score must be between 84 and 104.

Now do Exercises 59–74

U Helpful Hint V

Remember that all inequality sym- bols in a compound inequality must point in the same direction. We usually have them all point to the left so that the numbers are increasing in size as you go from left to right in the inequality.

Warm-Ups ▼

Fill in the blank. 1. inequalities have the same solution set.

2. According to the property of inequality, adding the same number to both sides of an inequality produces an equivalent inequality.

3. According to the property of inequality, the inequality symbol is reversed when multiplying by a negative number and not reversed when multiplying by a positive number.

True or false? 4. The inequality 2x � 18 is equivalent to x � 9.

5. The inequality x � 5 � 0 is equivalent to x � 5.

6. The inequality �2x � 6 is equivalent to x � �3.

7. The statement “x is at most 7” is written as x � 7.

8. The statement “x is not more than 85” is written as x � 85.

9. The inequality �3 � x � �9 is equivalent to �9 � x ��3

It is customary to write 8 � x � 0 with the smallest number on the left:

0 � x � 8

Since the last compound inequality is equivalent to the first, the solution set is [0, 8]. The graph is shown in Fig. 2.16.

Now do Exercises 43–52

Figure 2.16

9–1 876543210

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Exercises

U Study Tips V • Do some review on a regular basis. The Making Connections exercises at the end of each chapter can be used to review, compare, and

contrast different concepts that you have studied. • No one covers every topic in this text. Be sure you know what you are responsible for.

2 .9

U1V Rules for Inequalities

Write the appropriate inequality symbol in the blank so that the two inequalities are equivalent. See Example 1.

1. x � 7 � 0 2. x � 6 � 0 x __ �7 x __ 6

3. 9 � 3w 4. 10 � 5z w __ 3 z __ 2

5. �x � 8 6. �x � �3 x __ �8 x __ 3

7. �4k � �4 8. �9t � 27 k __ 1 t __�3

9. �� 1 2

� y � 4 10. �� 1 3

� x � 4

y __ �8 x __ �12

U2V Solving Inequalities

Solve each inequality. State the solution set in interval notation and sketch its graph. See Examples 2–4.

11. x � 3 � 0

12. x � 9 � �8

13. �3 � w � 1

14. 9 � w � 12

15. 8 � 2b

16. 35 � 7b

17. �8z � 4

18. �4y � �10

19. 3y � 2 � 7

20. 2y � 5 � �9

21. 3 � 9z � 6

22. 5 � 6z � 13

23. 6 � �r � 3

24. 6 � 12 � r

25. 5 � 4p � �8 � 3p

26. 7 � 9p � 11 � 8p

27. �� 5 6

� q � �20

28. �� 2 3

� q � �4

29. 1 � � 1 4

�t � � 1 8

�

30. � 1 6

� � � 1 3

�t � 0

31. 0.1x � 0.35 � 0.2

32. 1 � 0.02x � 0.6

33. 2x � 5 � x � 6

34. 3x � 4 � 2x � 9

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158 Chapter 2 Linear Equations and Inequalities in One Variable 2-74

35. x � 4 � 2(x 3)

36. 2x 3 � 3(x � 5)

37. 0.52x � 35 � 0.45x 8

38. 8455(x � 3.4) � 4320

Solve each compound inequality. State the solution set in interval notation and sketch its graph. See Examples 5 and 6.

39. 5 � x � 3 � 7

40. 2 � x � 5 � 6

41. 3 � 2v 1 � 10

42. �3 � 3v 4 � 7

43. �4 � 5 � k � 7

44. 2 � 3 � k � 8

45. �2 � 7 � 3y � 22

46. �1 � 1 � 2y � 3

47. 5 � 2

u � 3 � 17

48. �4 � 3

u � 1 � 11

49. �2 � � 2 3

50. 0 � 3 �

2m � 9

51. 0.02 � 0.54 � 0.0048x � 0.05

4m � 4

52. 0.44 � 34.55

12 �

4.5 22.3x

� 0.76

Solve each inequality. State the solution set in interval notation and sketch its graph.

53. 1 2

x � 1 � 4 � 1 3

54. 4

y �

2 �

55. 1 2

�x � 14 � � 14 �6x � 12 �

56. � 1 2

�z � 25 � � 23 � 34 z � 65 � 57.

1 3

� 1 4

x � 1 6

� 1 7 2

58. � 3 5

� 1 5

� 1 2 5 w � �

1 3

U3V Applications

Solve each of the following problems by using an inequality. See Example 7.

59. Boat storage. The length of a rectangular boat storage shed must be 4 meters more than the width, and the perimeter must be at least 120 meters. What is the range of values for the width?

60. Fencing a garden. Elka is planning a rectangular garden that is to be twice as long as it is wide. If she can afford to buy at most 180 feet of fencing, then what are the possible values for the width?

61. Car shopping. Harold Ivan is shopping for a new car. In addition to the price of the car, there is a 5% sales tax and a $144 title and license fee. If Harold Ivan decides that he will spend less than $9970 total, then what is the price range for the car?

Photo for Exercise 60

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2-75 2.9 Solving Inequalities and Applications 159

62. Car selling. Ronald wants to sell his car through a broker who charges a commission of 10% of the selling price. Ronald still owes $11,025 on the car. Ronald must get enough to at least pay off the loan. What is the range of the selling price?

63. Microwave oven. Sherie is going to buy a microwave in a city with an 8% sales tax. She has at most $594 to spend. In what price range should she look?

64. Dining out. At Burger Brothers the price of a hamburger is twice the price of an order of French fries, and the price of a Coke is $0.40 more than the price of the fries. Burger Brothers advertises that you can get a complete meal (burger, fries, and Coke) for under $4.00. What is the price range of an order of fries?

65. Averaging test scores. Tilak made 44 and 72 on the first two tests in algebra and has one test remaining. For Tilak to pass the course, the average on the three tests must be at least 60. For what range of scores on his last test will Tilak pass the course?

66. Averaging income. Helen earned $400 in January, $450 in February, and $380 in March. To pay all of her bills, she must average at least $430 per month. For what income in April would her average for the 4 months be at least $430?

67. Going for a C. Professor Williams gives only a midterm exam and a final exam. The semester average is

computed by taking 13 of the midterm exam score plus 2 3

of the final exam score. To get a C, Stacy must have a semester average between 70 and 79 inclusive. If Stacy scored only 48 on the midterm, then for what range of scores on the final exam will Stacy get a C?

68. Different weights. Professor Williamson counts his midterm as 23 of the grade and his final as

1 3 of the grade.

Wendy scored only 48 on the midterm. What range of scores on the final exam would put Wendy’s average between 70 and 79 inclusive? Compare to the previous exercise.

69. Average driving speed. On Halley’s recent trip from Bangor to San Diego, she drove for 8 hours each day and traveled between 396 and 453 miles each day. In what range was her average speed for each day of the trip?

70. Driving time. On Halley’s trip back to Bangor, she drove at an average speed of 55 mph every day and traveled between 330 and 495 miles per day. In what range was her daily driving time?

71. Sailboat navigation. As the sloop sailed north along the coast, the captain sighted the lighthouse at points A and B as shown in the figure. If the degree measure of the angle at the lighthouse is less than 30°, then what are the possible values for x?

72. Flight plan. A pilot started at point A and flew in the direction shown in the diagram for some time. At point B she made a 110° turn to end up at point C, due east of where she started. If the measure of angle C is less than 85°, then what are the possible values for x?

73. Bicycle gear ratios. The gear ratio r for a bicycle is defined by the formula

r � ,

where N is the number of teeth on the chainring (by the pedal), n is the number of teeth on the cog (by the wheel), and w is the wheel diameter in inches (www.sheldonbrown.com/gears).

a) If the wheel has a diameter of 27 in. and there are 12 teeth on the cog, then for what number of teeth on the chainring is the gear ratio between 60 and 80?

b) If a bicycle has 48 teeth on the chainring and 17 teeth on the cog, then for what diameter wheel is the gear ratio between 65 and 70?

c) If a bicycle has a 26-in.-diameter wheel and 40 teeth on the chainring, then for what number of teeth on the cog is the gear ratio less than 75?

74. Virtual demand. The weekly demand (the number bought by consumers) for the Acme Virtual Pet is given by the formula

d � 9000 � 60p

where p is the price for each in dollars.

a) What is the demand when the price is $30 each? b) In what price range will the demand be above 6000?

Nw n

Figure for Exercise 72

110�

Figure for Exercise 71

85�

B ?

North

Lighthouse

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160 Chapter 2 Linear Equations and Inequalities in One Variable 2-76

Wrap-Up

Summary

Equations Examples

Linear equation An equation of the form ax � b with a � 0 3x � 7

Identity An equation that is satisfied by every number for x x � 2x which both sides are defined

Conditional An equation that has at least one solution but is 5x � 10 � 0 equation not an identity

Inconsistent An equation that has no solution x � x 1 equation

Equivalent Equations that have exactly the same solutions 2x 1 � 5 equations 2x � 4

Properties of If the same number is added to or subtracted x � 5 � �9 equality from each side of an equation, the resulting x � �4

equation is equivalent to the original equation.

If each side of an equation is multiplied or 9x � 27 divided by the same nonzero number, the x � 3 resulting equation is equivalent to the original equation.

Solving equations 1. Remove parentheses by using the distributive 2(x � 3) � �7 3(x � 1) property and then combine like terms to simplify 2x � 6 � �10 3x each side as much as possible. �x � 6 � �10

2. Use the addition property of equality to get like �x � �4 terms from opposite sides onto the same side so that x � 4 they may be combined. Check:

3. The multiplication property of equality is 2(4 � 3) � �7 3(4 � 1) generally used last. 2 � 2

4. Check that the solution satisfies the original equation.

Formulas and Functions Examples

Formula An equation involving two or more variables D � RT

Solving for a Rewrite the formula so that the specified variable Solve for R. specified variable is isolated on the left side and does not occur on the R � D

right side.

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2-77 Chapter 2 Review Exercises 161

Functions A function is a rule for determining uniquely the D is a function of value of one variable a from the value(s) of one or R and T. more other variable(s). D � RT is a function.

Applications

Steps in solving 1. Read the problem. applied problems 2. If possible, draw a diagram to illustrate the problem.

3. Choose a variable and write down what it represents. 4. Represent any other unknowns in terms of

that variable. 5. Write an equation that describes the situation. 6. Solve the equation. 7. Answer the original question. 8. Check your answer by using it to solve the

original problem (not the equation).

Inequalities Examples

Properties Addition, subtraction, multiplication, and division �3x 1 � 7 of inequality may be performed on each side of an inequality, �3x � 6

just as we do in solving equations, with one x � �2 exception. When multiplying or dividing by a negative number, the inequality symbol is reversed.

7. An equation involving two or more variables is a(n) equation or .

8. If the value of y can be determined from the value of x, then y is a of x.

9. Angles whose degree measures total 90 are angles.

10. Angles whose degree measures total 180 are angles.

11. Motion at a constant rate is motion.

12. A statement that uses �, �, �, or � is an .

13. Inequalities that have the same solution set are .

Enriching Your Mathematical Word Power

Fill in the blank.

1. A(n) is a sentence that expresses the equality of two algebraic expressions.

2. A(n) equation has the form ax � b with a � 0.

3. An is satisfied by all real numbers for which both sides are defined.

4. A equation has at least one solution but is not an identity.

5. A(n) equation has no solutions.

6. Equations that have the same solution are equations.

Review Exercises

2.1 The Addition and Multiplication Properties of Equality Solve each equation and check your answer.

1. x � 23 � 12 2. 14 � 18 y

3. 2 3

u � �4 4. � 3 8

r � 15

5. �5y � 35 6. �12 � 6h

7. 6m � 13 5m 8. 19 � 3n � �2n

2.2 Solving General Linear Equations Solve each equation and check your answer.

9. 2x � 5 � 9

10. 5x � 8 � 38

11. 3p � 14 � �4p

12. 36 � 9y � 3y

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162 Chapter 2 Linear Equations and Inequalities in One Variable 2-78

13. 2z 12 � 5z � 9

14. 15 � 4w � 7 � 2w

15. 2(h � 7) � �14

16. 2(t � 7) � 0

17. 3(w � 5) � 6(w 2) � 3

18. 2(a � 4) 4 � 5(9 � a)

2.3 More Equations Solve each equation. Identify each equation as a conditional equation, an inconsistent equation, or an identity.

19. 2(x � 7) � 5 � 5 � (3 � 2x)

20. 2(x � 7) 5 � �(9 � 2x)

21. 2(w � w) � 0

22. 2y � y � 0

23. 3 3 r

r � 1

24. 3 3 t

� 1

25. 1 2

a � 5 � 1 3

a � 1

26. 1 2

b � 1 2

� 1 4

27. 0.06q 14 � 0.3q � 5.2

28. 0.05(z 20) � 0.1z � 0.5

29. 0.05(x 100) 0.06x � 115

30. 0.06x 0.08(x 1) � 0.41

Solve each equation.

31. 2x 1 2

� 3x 1 4

32. 5x � 1

3 � 6x �

33. 2

x �

4 �

34. 1

3 �

x �

2 �

35. 5

6 x � �

36. � 2

3 x �

37. � 1

2 (x � 10) �

4 x

38. � 1

3 (6x � 9) � 23

39. 3 � 4(x � 1) 6 � �3(x 2) � 5

40. 6 � 5(1 � 2x) 3 � �3(1 � 2x) � 1

41. 5 � 0.1(x � 30) � 18 0.05(x 100)

42. 0.6(x � 50) � 18 � 0.3(40 � 10x)

2.4 Formulas and Functions Solve each equation for x.

43. ax b � 0

44. mx e � t

45. ax � 2 � b

46. b � 5 � x

47. LWx � V

48. 3xy � 6

49. 2x � b � 5x

50. t � 5x � 4x

In each case find a formula that expresses y as a function of x. Write the answer in the form y � mx b where m and b are real numbers.

51. 5x 2y � 6

52. 5x � 3y 9 � 0

53. y � 1 � � 1

2 (x � 6)

54. y 6 � 1

2 (x 8)

55. 1

2 x

4 y � 4

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2-79 Chapter 2 Review Exercises 163

56. � 3

y � 1

Find the value of y in each formula if x � �3.

57. y � 3x � 4

58. 2x � 3y � �7

59. 5xy � 6

60. 3xy � 2x � �12

61. y � 3 � �2(x � 4)

62. y 1 � 2(x � 5)

Fill in the tables using the given formulas.

63. y � �5x 10

64. y � 2x � 4

65. y � 2 3

x � 1

66. y � 10x 100

x y

�20

�10

x y

�3

x y

�1

2.5 Translating Verbal Expressions into Algebraic Expressions

Translate each verbal expression into an algebraic expression.

67. The sum of a number and 9

68. The product of a number and 7

69. Two numbers that differ by 8

70. Two numbers with a sum of 12

71. Sixty-five percent of a number

72. One-half of a number

Identify the variable, and write an equation that describes each situation. Do not solve the equation.

73. One side of a rectangle is 5 feet longer than the other, and the area is 98 square feet.

74. One side of a rectangle is one foot longer than twice the other side, and the perimeter is 56 feet.

75. By driving 10 miles per hour slower than Jim, Barbara travels the same distance in 3 hours as Jim does in 2 hours.

76. Gladys and Ned drove 840 miles altogether, with Gladys averaging 5 miles per hour more in her 6 hours at the wheel than Ned did in his 5 hours at the wheel.

77. The sum of three consecutive even integers is 90.

78. The sum of two consecutive odd integers is 40.

79. The three angles of a triangle have degree measures of t, 2t, and t � 10.

80. Two complementary angles have degree measures p and 3p � 6.

2.6–7 Applications Solve each problem.

81. Odd integers. If the sum of three consecutive odd integers is 237, then what are the integers?

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164 Chapter 2 Linear Equations and Inequalities in One Variable 2-80

82. Even integers. Find two consecutive even integers that have a sum of 450.

83. Driving to the shore. Lawanda and Betty both drive the same distance to the shore. By driving 15 miles per hour faster than Betty, Lawanda can get there in 3 hours while Betty takes 4 hours. How fast does each of them drive?

84. Rectangular lot. The length of a rectangular lot is 50 feet more than the width. If the perimeter is 500 feet, then what are the length and width?

85. Combined savings. Wanda makes $6000 more per year than her husband does. Wanda saves 10% of her income for retirement, and her husband saves 6%. If together they save $5400 per year, then how much does each of them make per year?

86. Layoffs looming. American Products plans to lay off 10% of its employees in its aerospace division and 15% of its employees in its agricultural division. If altogether 12% of the 3000 employees in these two divisions will be laid off, then how many employees are in each division?

2.8 Inequalities Determine whether the given number is a solution to the inequality following it.

87. 3, �2x 5 � x � 6

88. �2, 5 � x � 4x 3

89. �1, �2 � 6 4x � 0

90. 0, 4x 9 � 5(x � 3)

For each graph write the corresponding inequality and the solution set to the inequality using interval notation.

91. �1 0 1 2 3 4 7�2 85 6

Figure for Exercise 83

Betty 4 hours x mph

City

Lawanda 3 hours x � 15 mph Sand

Water

92.

93.

94.

95.

96.

97.

98.

2.9 Solving Inequalities and Applications Solve each inequality. State the solution set in interval notation and sketch its graph.

99. x 2 � 1

100. x � 3 � 7

101. 3x � 5 � x 1

102. 5x � 5 � 9 � 2x

103. � 3

4 x � 3

104. � 2

3 x � 10

105. 3 � 2x � 11

106. 5 � 3x � 35

�4�5 �2�1�3 0 1 2 3 4 5

�7�8 �5�6 �4�3 �1 0 1 2�2

�5�6 �4�3 �1 0 1 2 3 4�2

�4�5 �2�1�3 0 1 2 3 4 5

�1 0 1 2 3 4 7�2 85 6

�4�3�5 �1 0 1 2 3 4�2 5

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2-81 Chapter 2 Review Exercises 165

107. �3 � 2x � 1 � 9

108. 2 � 3x 2 � 8

109. 0 � 1 � 2x � 5

110. �5 � 3 � 4x � 7

111. �1 � � 1

112. �3 � 4 �

x � 2

113. 1

3 �

x �

114. � 3

8 � �

4 x

8 �

Miscellaneous Use an equation, inequality, or formula to solve each problem.

115. Plasma TV discount. Nexus got a 14% discount when he bought a new plasma television. If the amount of the discount was $392, then what was the original price of the television?

116. Laptop discount. Zeland got a 12% discount on a new laptop computer. If he paid $1166 for the laptop, then what was the original price?

117. Rug commission. Caroline sold an antique rug through a broker who got 8% of the selling price as a commission. If Caroline got $7820 for the rug after the broker’s commission, then what was the selling price of the rug?

118. Buyer’s premium. Brittany paid $95,920 for a 1966 Mustang at a classic car auction where there is a 9% buyer’s premium. This means that the buyer pays the bid price plus 9% of the bid price. What was the bid price?

119. Long-term yields. The annual yield on a 30-year treasury bond is 5.375%. Use the simple interest formula to find the amount of interest earned during the first year on a $10,000 bond.

2x � 3

120. High interest rate. Eddie wrote a $280 check to a check holding company, which gave him $260 in cash. After two weeks, the company will cash his $280 check. Find the annual simple interest rate for this loan. Note that the time is a fraction of a year.

121. Combined videos. The owners of ABC Video discovered that they had no movies in common with XYZ Video and bought XYZ’s entire stock. Although XYZ had 200 titles, they had no children’s movies, while 60% of ABC’s titles were children’s movies. If 40% of the movies in the combined stock are children’s movies, then how many movies did ABC have before the merger?

122. Living comfortably. Gary has figured that he needs to take home $30,400 a year to live comfortably. If the government gets 24% of Gary’s income, then what must his income be for him to live comfortably?

123. Bracing a gate. The diagonal brace on a rectangular gate forms an angle with the horizontal side with degree measure x and an angle with the vertical side with degree measure 2x � 3. Find x.

124. Digging up the street. A contractor wants to install a pipeline connecting point A with point C on opposite sides of a road as shown in the figure below. To save money, the contractor has decided to lay the pipe to point B and then under the road to point C. Find the measure of the angle marked x in the figure.

125. Perimeter of a triangle. One side of a triangle is 1 foot longer than the shortest side, and the third side is twice as long as the shortest side. If the perimeter is less than 25 feet, then what is the range of the length of the shortest side?

126. Restricted hours. Alana makes $5.80 per hour working in the library. To keep her job, she must make at least $116 per week; but to keep her scholarship, she must not earn more than $145 per week. What is the range of the number of hours per week that she may work?

Figure for Exercise 124

A B

20� x

50�

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166 Chapter 2 Linear Equations and Inequalities in One Variable 2-82

Chapter 2 Test

Solve each equation.

1. �10x � 6 4x � �4x 8

2. 5(2x � 3) � x 3

3. � 2

3 x 1 � 7

4. x 0.06x � 742

5. x � 0.03x � 0.97

6. 6x � 7 � 0

7. 1 2

x � 1 3

� 1 4

x 1 6

8. 2(x 6) � 2x � 5

9. x 7x � 8x

Solve for the indicated variable.

10. 2x � 3y � 9 for y

11. m � aP � w for a

For each graph write the corresponding inequality and the solution set to the inequality using interval notation.

12.

13.

Solve each inequality. State the solution set in interval notation and sketch its graph.

14. 4 � 3(w � 5) � �2w

15. 1 � 1 �

2x � 5

8�2 10�1 2 3 4 5 6 7

50 1 2 3 4�5�4�3�2�1

16. 1 � 3x � 2 � 7

17. � 2

3 y � 4

Write a complete solution to each problem.

18. The perimeter of a rectangle is 72 meters. If the width is 8 meters less than the length, then what is the width of the rectangle?

19. a) What formula expresses the area of a triangle A as a function of its base b and height h?

b) Find a formula that expresses the height of a triangle as a function of its area and base.

c) If the area of a triangle is 54 square inches and the base is 12 inches, then what is the height?

20. How many liters of a 20% alcohol solution should Maria mix with 50 liters of a 60% alcohol solution to obtain a 30% solution?

21. Brandon gets a 40% discount on loose diamonds where he works. The cost of the setting is $250. If he plans to spend at most $1450, then what is the price range (list price) of the diamonds that he can afford?

22. If the degree measure of the smallest angle of a triangle is one-half of the degree measure of the second largest angle and one-third of the degree measure of the largest angle, then what is the degree measure of each angle?

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2-83 Chapter 2 Making Connections 167

Simplify each expression.

1. 3x 5x 2. 3x � 5x

3. 4. 5 � 4(3 � x)

5. 3x 8 � 5(x � 1)

6. (�6)2 � 4(�3)2 7. 32 � 23

8. 4(�7) � (�6)(3) 9. �2x � x � x

10. (�1)(�1)(�1)(�1)(�1)

Evaluate each expression if x � �2 and y � 3.

11. 5x 4x 12. 9x 13. (y � x)(y x) 14. y2 � x2

15. (x � y)2 16. x2 � 2xy y2

17. (2x y)2 18. 4x2 4xy y2

Write the interval notation for each set.

19. The real numbers less than 2 20. The real numbers greater than �6 21. The real numbers greater than or equal to 5 22. The real numbers less than or equal to �1 23. The real numbers between 2 and 6 inclusive 24. The real numbers greater than 4 and less than 8

Perform the following operations.

25. 1 2

1 6

26. 1 2

� 1 3

27. 5 3

� 1 1 5 28.

2 3

� 5 6

29. 6 � � 53 1 2

� 30. 15� 23 � 1 2 5 �

31. 4 � � 2 x

1 4

� 32. 12� 56 x � 3 4

� Find the solution set to each equation or inequality.

33. x � 1 2

� 1 6

34. x 1 3

� 1 2

35. x � 1 2

� 1 6

36. x 1 3

� 1 2

37. 3 5

x � 1 1 5 38.

3 2

x � 5 6

39. � 3 5

x � 1 1 5 40. �

3 2

x � 5 6

41. 5 3

x 1 2

� 1 42. 2 3

x � 1 2 5 � 2

43. 2 x

1 4

� 1 2

44. 5 6

x � 3 4

� 1 5 2

4x 2

45. 3x 5x � 8 46. 3x 5x � 8x

47. 3x 5x � 7x 48. 3x 5 � 8

49. 3x 5x � 7x 50. 3x 5x � 8x

51. 3x 1 � 7 52. 5 � 4(3 � x ) � 1

53. 3x 8 � 5(x � 1) 54. x � 0.05x � 190

55. 5 � 3x � 11 56. 19 � 3 8x

57. 0 � x

5 3

� 3

58. 1 � 7

1 �

2 x

� 4

Solve the problem.

59. Linear Depreciation. In computing income taxes, a company is allowed to depreciate a $20,000 computer system over five years. Using linear depreciation, the value V of the computer system at any year t from 0 through 5 is given by

V � C � C �

5 S

where C is the initial cost of the system and S is the scrap value of the system. a) What is the value of the computer system after two

years if its scrap value is $4000? b) If the value of the system after three years is claimed

to be $14,000, then what is the scrap value of the company’s system?

c) If the accompanying graph models the depreciation of the system, then what is the scrap value of the system?

MakingConnections A Review of Chapters 1–2

Figure for Exercise 59

0 1 2 3 4 5 Year

V al

ue (

th ou

sa nd

s of

d ol

la rs

)

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168 Chapter 2 Linear Equations and Inequalities in One Variable 2-84

1. Visible squares. How many squares are visible in each of the following diagrams?

a) b)

c) d)

2. Baker’s dilemma. A baker needs 8 cups of flour. He sends his apprentice to the flour bin with a scoop that holds 6 cups and a scoop that holds 11 cups. How can the apprentice measure 8 cups of flour with these scoops?

resulting expression is 100. For example,

98 7 � 6 � 5 4 3 � 2 1 � 100.

4. Four threes. Check out these equations:

3 3

�

3 3

� 1, 3 3

3 3

� 2, (3 � 3)3 3 � 3.

Using exactly four 3’s write arithmetic expressions whose values are 4, 5, 6, and so on. How far can you go?

5. Palindrome time. A palindrome is a sequence of words or numbers that reads the same forward or backward. For example, “A TOYOTA” is a palindrome and 14341 is a palindromic number. How many times per day does a digital clock display a palindromic number? Of course the answer depends on the format in which the digital clock displays the time. First, state precisely the type of digital clock display you are using, and then count the palin- dromic numbers for that type of display.

6. Reversible products. Find the product of 32 and 46. Now reverse the digits and find the product of 23 and 64. The products are the same. Does this happen with any pair of two-digit numbers? Find two other pairs of two-digit numbers (with different digits) that have this property.

7. Running late. Alice, Bea, Carl, and Don all have an 8 o’clock class. Alice’s watch is 8 minutes fast, but she thinks it is 4 minutes slow. Bea’s watch is 8 minutes slow, but she thinks it is 8 minutes fast. Carl’s watch is 4 minutes slow, but he thinks it is 8 minutes fast. Don’s watch is 4 minutes fast, but he thinks it is 8 minutes slow. Each student leaves so they will get to class at exactly 8 o’clock. Each student assumes the correct time is what they think it is by their watch. Who is late to class and by how much?

8. Automorphic numbers. Automorphic numbers are inte- gers whose squares end in the given integer. Since 12 � 1 and 62 � 36, both 1 and 6 are automorphic. Find the next four automorphic numbers.

CriticalThinking For Individual or Group Work Chapter 2

These exercises can be solved by a variety of techniques, which may or may not require algebra. So be creative and think critically. Explain all answers. Answers are in the Instructor’s Edition of this text.

Photo for Exercise 2

3. Totaling one hundred. Start with the sequence of digits 987654321. Place any number of plus or minus signs between the digits in the sequence so that the value of the

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