Algebra Writing 300 words total Due today

150 words

Can a linear equation and a linear inequality be solved in the same way? Explain why. What makes them different?

150 words

What are the four steps for solving a problem? Should any other factors be accounted for when solving a problem? Should any factors be accounted for when explaining how to solve a problem? Explain your answer.

Please use page 111 of the attached text for the second question

Source: J. Williams, The USA Today

Weather Almanac

2.1 Introduction to

Equations

2.2 Linear Equations
2.3 Introduction to

Problem Solving
2.4 Formulas
2.5 Linear Inequalities

Mathematics is a unique subject that is essential for describing, or modeling,
events in the real world. For example, ultraviolet light from the sun is responsible for
both tanning and burning exposed skin. Mathematics lets us use numbers to describe the
intensity of ultraviolet light. The table shows the maximum ultraviolet intensity mea-
sured in milliwatts per square meter for various latitudes and dates.

2 Linear Equations and Inequalities

Education is not the
filling of a pail, but the
lighting of a fire.

— WILLIAM BUTLER YEATS

If a student from Chicago, located at a latitude of 42°, spends spring break in Hawaii
with a latitude of 20°, the sun’s ultraviolet rays in Hawaii will be approximately 24999 � 2.5
times as intense as they are in Chicago. Equations can be used to describe, or model, the
intensity of the sun at various latitudes. In this chapter we will focus on linear equations
and the related concept of linear inequalities.

Latitude Mar. 21 June 21 Sept. 21 Dec. 2

0� 325 254 325 27

10� 311 275 280 220

20� 249 292 256 14

30� 179 248 182 80

40� 99 199 127 3

50� 57 143 75

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90 CHAPTER 2 LINEAR EQUATIONS AND INEQUALITIES

2.1 Introduction to Equations
Basic Concepts ● Equations and Solutions ● The Addition Property of Equality ●
The Multiplication Property of Equalit

A LOOK INTO MATH N The Global Positioning System (GPS) consists of 24 satellites that travel around Earth in
nearly circular orbits. GPS can be used to determine locations and velocities of cars, air-
planes, and hikers with an amazing degree of accuracy. New cars often come equipped
with GPS, and their drivers can determine their cars’ locations to within a few feet. To cre-
ate GPS, thousands of equations were solved, and mathematics was essential in finding
their solutions. In this section we discuss many of the basic concepts needed to solve equa-
tions. (Source: J. Van Sickle, GPS for Land Surveyors.)

Basic Concepts
N REAL-WORLD CONNECTION Suppose that during a storm it rains 2 inches before noon and

1 inch per hour thereafter until 5 P.M. Table 2.1 lists the total rainfall R after various elapsed
times x, where x = 0 corresponds to noon.

NEW VOCABULARY

n Solution
n Solution set
n Equivalent equations

TABLE 2.1 Rainfall x Hours Past Noon

Elapsed Time: x (hours) 0 1 2 3 4 5

Total Rainfall: R (inches) 2 3 4 5 6 7

The data suggest that the total rainfall R in inches is 2 more than the elapsed time x. A
formula that models, or describes, the rainfall x hours past noon is given by

R = x + 2.

For example, 3 hours past noon, or at 3 P.M.,

R = 3 + 2 = 5

inches of rain have fallen. Even though x = 4.5 does not appear in the table, we can calcu-
late the amount of rainfall at 4:30 P.M. with the formula as

R = 4.5 + 2 = 6.5 inches.

The advantage that a formula has over a table of values is that a formula can be used to cal-
culate the rainfall at any time x, not just at the times listed in the table.

Equations and Solutions
In the rainfall example above, how can we determine when 6 inches of rain have fallen?
From Table 2.1 the solution is 4, or 4 P.M. To find this solution without the table, we can
solve the equation

x + 2 = 6.

An equation can be either true or false. For example, the equation 1 + 2 = 3 is true,
whereas the equation 1 + 2 = 4 is false. When an equation contains a variable, the equa-
tion may be true for some values of the variable and false for other values of the variable.
Each value of the variable that makes the equation true is called a solution to the equation,

READING CHECK

• What is a solution to an
equation?

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912.1 INTRODUCTION TO EQUATIONS

and the set of all solutions is called the solution set. Solving an equation means finding all
of its solutions. Because 4 + 2 = 6, the solution to the equation

x + 2 =

is 4, and the solution set is {4}. Note that braces {} are used to denote a set. Table 2.2
shows examples of equations and their solution sets. Note that some equations, such as the
third one in the table, can have more than one solution.

TABLE 2.2 Equations and Solution Sets

Equation Solution Set True Equation(s)

3 – x = 1 {2} 3 – 2 = 1

10 – 4y = 6 {1} 10 – 4(1) = 6

x2 = 4 {�2, 2} (�2)2 = 4 and 22 = 4

Many times we cannot solve an equation simply by looking at it. In these situations
we must use a step-by-step procedure. During each step an equation is transformed into a
different but equivalent equation. Equivalent equations are equations that have the same
solution set. For example, the equations

x + 2 = 5 and x = 3

are equivalent equations because the solution set for both equations is {3}.

When solving equations, it is often helpful to transform a more complicated equation
into an equivalent equation that has an obvious solution, such as x = 3. The addition prop-
erty of equality and the multiplication property of equality can be used to transform an
equation into an equivalent equation that is easier to solve.

The Addition Property of Equality
When solving an equation, we have to apply the same operation to each side of the equa-
tion. For example, one way to solve the equation

x + 2 = 5

is to add -2 to each side. This step results in isolating the x on one side of the equation.

x + 2 = 5 Given equation
x + 2 � (�2) = 5 � (�2) Add -2 to each side.
x + 0 = 3 Addition of real numbers
x = 3 Additive identity

MAKING CONNECTIONS

Equations and Expressions

Although the words “equation” and “expression” occur frequently in mathematics, they are
not interchangeable. An equation always contains an equals sign but an expression never
contains an equals sign. We often want to solve an equation, whereas an expression can
sometimes be simplified. Furthermore, the equals sign in an equation separates two expres-
sions. For example, 3x – 5 = x + 1 is an equation where 3x – 5 and x + 1 are each
expressions.

READING CHECK

• What is the difference
between an equation and
an expression?

STUDY TIP

The addition property of
equality is used to solve
equations throughout the
remainder of the text. Be sure
that you have a firm under-
standing of this important
property.

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Beginning and Intermediate Algebra with Applications & Visualization, Third edition, by Gary K. Rockswold and Terry A. Krieger. Published by Addison Wesley.
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92 CHAPTER 2 LINEAR EQUATIONS AND INEQUALITIES

These four equations (on the bottom of the previous page) are equivalent, but the solution is
easiest to see in the last equation. When �2 is added to each side of the given equation, the
addition property of equality is used.

READING CHECK

• Why do we use the addi-
tion property of equality?

ADDITION PROPERTY OF EQUALITY

If a, b, and c are real numbers, then

a = b is equivalent to a + c = b + c.

That is, adding the same number to each side of an equation results in an equivalent
equation.

NOTE: Because any subtraction problem can be changed to an addition problem, the addi-
tion property of equality also works for subtraction. That is, if the same number is sub-
tracted from each side of an equation, the result is an equivalent equation.

EXAMPLE 1 Using the addition property of equality

Solve each equation.
(a) x + 10 = 7 (b) t – 4 = 3 (c) 12 = –

3
4 + y

Solution
(a) When solving an equation, we try to isolate the variable on one side of the equation. If

we add -10 to (or subtract 10 from) each side of the equation, we find the value of x.

x + 10 = 7 Given equation
x + 10 � 10 = 7 � 10 Subtract 10 from each side.
x + 0 = -3 Addition of real numbers
x = -3 Additive identity

The solution is -3.
(b) To isolate the variable t, add 4 to each side.

t – 4 = 3 Given equation
t – 4 � 4 = 3 � 4 Add 4 to each side.
t + 0 = 7 Addition of real numbers
t = 7 Additive identity

The solution is 7.
(c) To isolate the variable y, add 34 to each side.

2
= –

4
+ y Given equation

2
�

3
4

= –

4
�

3
4

+ y Add 34 to each side.

4
= 0 + y Addition of real numbers

4
= y Additive identity

The solution is 54.

Now Try Exercises 17, 19, 23

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932.1 INTRODUCTION TO EQUATIONS

CHECKING A SOLUTION To check a solution, substitute it in the given equation to find
out if a true statement results. To check the solution for Example 1(c), substitute 54 for y in
the given equation. Note that a question mark is placed over the equals sign when a solution
is being checked.

1
2
= –
3

4
+ y Given equation

2
� –

4
+

5
4

Replace y with 54.

2
� 2

4
Add fractions.

2
=

2
✓

The answer checks.

The answer of 54 checks because the resulting equation is true.

CRITICAL THINKING

When you are checking a
solution, why do you substi-
tute your answer in the given
equation?

EXAMPLE 2 Solving an equation and checking a solution

Solve the equation -5 + y = 3 and then check the solution.

Solution
Isolate y by adding 5 to each side.

-5 + y = 3 Given equation
-5 � 5 + y = 3 � 5 Add 5 to each side.
0 + y = 8 Addition of real numbers
y = 8 Additive identity

The solution is 8. To check this solution, substitute 8 for y in the given equation.

-5 + y = 3 Given equation
-5 + 8 � 3 Replace y with 8.
3 = 3 ✓ Add; the answer checks.

Now Try Exercise 21

MAKING CONNECTIONS

Equations and Scales

Think of an equation as an old-fashioned scale, where two pans must balance, as shown in
the figure. If the two identical golden weights balance the pans, then adding identical red
weights to each pan results in the pans remaining balanced. Similarly, removing (subtract-
ing) identical weights from each side will also keep the pans balanced.

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94 CHAPTER 2 LINEAR EQUATIONS AND INEQUALITIES

The Multiplication Property of Equality
We can illustrate the multiplication property of equality by considering a formula that con-
verts yards to feet. Because there are 3 feet in 1 yard, the formula F = 3Y computes F, the
number of feet in Y yards. For example, if Y = 5 yards, then F = 3 # 5 = 15 feet.

Now consider the reverse, converting 27 feet to yards. The answer to this conversion
corresponds to the solution to the equation

27 = 3Y.

To find the solution, multiply each side of the equation by the reciprocal of 3, or 13.

27 = 3Y Given equation

1
3
# 27 = 1

3
# 3 # Y Multiply each side by 13.

9 = 1 # Y Multiplication of real numbers
9 = Y Multiplicative identity

Thus 27 feet are equivalent to 9 yards.

STUDY TIP

The multiplication property
of equality is used to solve
equations throughout the
remainder of the text. Be sure
that you have a firm under-
standing of this important
property.

READING CHECK

• Why do we use the multiplication property of equality?

MULTIPLICATION PROPERTY OF EQUALITY

If a, b, and c are real numbers with c � 0, then

a = b is equivalent to ac = bc.

That is, multiplying each side of an equation by the same nonzero number results in
an equivalent equation.

NOTE: Because any division problem can be changed to a multiplication problem, the mul-
tiplication property of equality also works for division. That is, if each side of an equation is
divided by the same nonzero number, the result is an equivalent equation.

EXAMPLE 3 Using the multiplication property of equality

Solve each equation.
(a) 13 x = 4 (b) -4y = 8 (c) 5 =

3
4 �

Solution
(a) We start by multiplying each side of the equation by 3, the reciprocal of 13.

3
x = 4 Given equation

3 # 1
3

x = 3 # 4 Multiply each side by 3.
1 # x = 12 Multiplication of real numbers
x = 12 Multiplicative identity

The solution is 12.

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952.1 INTRODUCTION TO EQUATIONS

(b) The coefficient of the y-term is -4, so we can either multiply each side of the equation
by – 14 or divide each side by �4. This step will make the coefficient of y equal to 1.

-4y = 8 Given equation

-4y
�4

=
8

�4
Divide each side by -4.

y = -2 Simplify fractions.

The solution is -2.
(c) To change the coefficient of � from 34 to 1, multiply each side of the equation by

4
3, the

reciprocal of 34.

5 =
3

4
� Given equation

4
3
# 5 = 4

3
# 3

4
� Multiply each side by 43.

3
= 1 # � Multiplication of real numbers

3
= � Multiplicative identity

The solution is 203

Now Try Exercises 37, 43,

EXAMPLE 4 Solving an equation and checking a solution

Solve the equation 34 = –
3
7 t and then check the solution.

Solution
Multiply each side of the equation by �73, the reciprocal of –

3
7.

4
= –

7
t Given equation

�
7
3
# 3

4
= �

7
3
# a – 3

7
b t Multiply each side by – 73.

–
7

4
= 1 # t Multiplication of real numbers

–
7

4
= t Multiplicative identity

The solution is – 74. To check this answer, substitute �
7
4 for t in the given equation.

3
4
= –
3
7
t Given equation

4
� –

7
# a� 7

4
b Replace t with – 74.

4
=

4
✓ Multiply; the answer checks.

Now Try Exercise 47

96 CHAPTER 2 LINEAR EQUATIONS AND INEQUALITIES

N REAL-WORLD CONNECTION Twitter is a microblogging Web site that is used to post short
messages called “tweets” on the Internet. In its early years, Twitter’s popularity increased
dramatically and new accounts were added at an amazing rate. People from around the
world began posting millions of tweets every day. (Source: Twitter.)

EXAMPLE 5 Analyzing Twitter account dat

In the early months of 2010, Twitter added 0.3 million new accounts every day.
(a) Write a formula that gives the number of new Twitter accounts T added in x days.
(b) At this rate, how many days would be needed to add 18 million new accounts?

Solution
(a) In 1 day 0.3 # 1 = 0.3 million new accounts were added, in 2 days 0.3 # 2 = 0.6 mil-

lion new accounts were added, and in x days 0.3 # x = 0.3x new accounts were added.
So the formula is T = 0.3 x, where x is in days and T is in millions.

(b) To find the number of days needed for Twitter to add 18 million new accounts, replace
the variable T in the formula with 18 and solve the resulting equation.

T = 0.3x Formula from part

(a)

18 = 0.3x Replace T with 18.

3
=

0.3

0.3
Divide each side by 0.3.

60 = x Simplify.

At this rate, it takes 60 days to add 18 million accounts.

Now Try Exercise 59

2.1 Putting It All Togethe

Equation An equation is a mathematical state-
ment that two expressions are equal.
An equation can be either true or false.

The equation 2 + 3 = 5 is true.
The equation 1 + 3 = 7 is false.

Solution A value for a variable that makes an
equation a true statement

The solution to x + 5 = 20 is 15, and
the solutions to x2 = 9 are -3 and 3.

Solution Set The set of all solutions to an equation The solution set to x + 5 = 20 is {15},
and the solution set to x2 = 9 is{-3, 3}.

CONCEPT COMMENTS EXAMPLES

Equivalent Equations Two equations are equivalent if they
have the same solution set.

The equations

2 x = 14 and x = 7

are equivalent because the solution set to
both equations is {7}.

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CONCEPT COMMENTS EXAMPLES

Addition Property of
Equality

The equations

a = b and a + c = b +

are equivalent. This property is used
to solve equations.

To solve x – 3 = 8, add 3 to each side
of the equation.

x – 3 + 3 = 8 + 3
x = 11

The solution is 11.

Multiplication Property
of Equality

When c � 0, the equations

a = b and a # c = b # c
are equivalent. This property is used
to solve equations.

To solve 15 x = 10, multiply each side of
the equation by 5.

5 # 1
5

x = 5 # 10
x = 50

The solution is 50.

Checking a Solution Substitute the solution for the variable
in the given equation and then sim-
plify each side to see if a true state-
ment results.

To show that 8 is a solution to

x + 12 = 20,

substitute 8 for x.

8 + 12 � 20
20 = 20 True

972.1 INTRODUCTION TO EQUATIONS

2.1 Exercises

CONCEPTS AND VOCABULARY

1. Each value of a variable that makes an equation true
is called a(n) _____.

2. The equation 1 + 3 = 4 is (true/false).

3. The equation 2 + 3 = 6 is (true/false).

4. The _____ is the set of all solutions to an equation.

5. To solve an equation, find all _____.

6. Equations with the same solution set are called _____
equations.

7. If a = b, then a + c = _____.

8. Because any subtraction problem can be changed to
an addition problem, the addition property of equality
also works for _____.

9. If a = b and c � 0, then ac = _____ .

10. Because any division problem can be changed to a
multiplication problem, the multiplication property of
equality also works for _____.

11. To solve an equation, transform the equation into a(n)
_____ equation that is easier to solve.

12. To check a solution, substitute it for the variable in
the _____ equation.

THE ADDITION PROPERTY OF EQUALITY

13. To solve x – 22 = 4, add _____ to each side.

14. To solve 56 =
1
6 + x, add _____ to each side.

15. To solve x + 3 = 13, subtract _____ from each side.

16. To solve 34 =
1
4 + x, subtract _____ from each side.

98 CHAPTER 2 LINEAR EQUATIONS AND INEQUALITIES

Exercises 17–30: Solve the equation. Check your answer.

17. x + 5 = 0 18. x + 3 = 7

19. a – 12 = -3 20. a – 19 = -11

21. 9 = y – 8 22. 97 = -23 + y

23. 15 = z –
3
2 24.

3
4 + z = –

1
2

25. t – 0.8 = 4.3 26. y – 1.23 = -0.02

27. 4 + x = 1 28. 16 + x = -2

29. 1 = 13 + y 30.
7
2 = -2 + y

31. Thinking Generally To solve x – a = b for x,
add _____ to each side.

32. Thinking Generally To solve x + a = b for x,
subtract _____ from each side.

THE MULTIPLICATION PROPERTY OF EQUALITY

33. To solve 5x = 4, multiply each side by _____.

(c) Use your formula to calculate the total rainfall at
3 P.M. Does the answer agree with the value in
your table from part (a)?

(d) How much rain has fallen by 2:15 P.M.?

56. Cold Weather A furnace is turned on at midnight
when the temperature inside a cabin is 0� F. The
cabin warms at a rate of 10� F per hour until 7 A.M.
(a) Make a table that shows the cabin temperature

T in degrees Fahrenheit, x hours past midnight,
ending at 7 A.M.

(b) Write a formula that calculates T.
(c) Use your formula to calculate the temperature at

5 A.M. Does the answer agree with the value in
your table from part (a)?

(d) Find the cabin temperature at 2:45 A.M.

57. Football Field A football field is 300 feet long.
(a) Write a formula that gives the length L in feet of

x football fields.
(b) Use your formula to write an equation whose solu-

tion gives the number of football fields in 870 feet.

58. Acreage An acre equals 43,560 square feet.
(a) Write a formula that converts A acres to S square

feet.
(b) Use your formula to write an equation whose

solution gives the number of acres in 871,200
square feet.

59. Twitter Accounts (Refer to Example 5.) In the early
months of 2010, Twitter added 0.3 million new
accounts every day. At this rate, how many days
would be needed to add 15 million new Twitter
accounts? (Source: Twitter.)

60. Web Site Visits If a Web site was increasing its num-
ber of visitors by 14,000 every day, how many days
would it take for the site to gain a total of 98,000 new
visitors?

61. Online Exploration The city of Winnipeg is located
in the province of Manitoba in Canada.
(a) Use the Internet to find the latitude of Winnipeg

to the nearest degree.

(b) Use the table on page 89 to determine how many

times as intense the sun’s ultraviolet rays are at
the equator (latitude 0°) on March 21 compared
to the sun’s intensity in Winnipeg. Round your
answer to 1 decimal place.

62. Online Exploration Columbus is a city located in
the center of Ohio. It is the state’s capital city.
(a) Use the Internet to find the latitude of Columbus

to the nearest degree.

34. To solve 43 y = 8, multiply each side by _____.

35. To solve 6 x = 11, divide each side by _____.

36. To solve 0.2 x = 4, divide each side by _____.

Exercises 37–52: Solve the equation. Check your answer.

37. 5x = 15 38. -2 x = 8

39. -7x = 0 40. 25x = 0

41. -35 = -5a 42. -32 = -4a

43. -18 = 3a 44. -70 = 10a

45. 12 x =
3
2 46.

3
4 x =

5
8

47. 12 =
2
5 z 48. –

3
4 = –

1
8 z

49. 0.5t = 3.5 50. 2.2t = -9.9

51. -1.7 = 0.2 x 52. 6.4 = 1.6 x

53. Thinking Generally To solve 1a # x = b for x, multi-
ply each side by _____ .

54. Thinking Generally To solve ax = b for x, where
a � 0, divide each side by _____ .

APPLICATIONS

55. Rainfall On a stormy day it rains 3 inches before
noon and 12 inch per hour thereafter until 6 P.M.
(a) Make a table that shows the total rainfall R in

inches, x hours past noon, ending at 6 P.M.
(b) Write a formula that calculates R.

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992.2 LINEAR EQUATIONS

(b) Use the table on page 89 to determine how many
times as intense the sun’s ultraviolet rays are in
Limon, Costa Rica, (latitude 10°N) on June 21
compared to the sun’s intensity in Columbus.
Round your answer to 1 decimal place.

63. Cost of a Car When the cost of a car is multiplied by
0.07 the result is $1750. Find the cost of the car.

64. Raise in Salary If an employee’s salary is multiplied
by 1.06, which corresponds to a 6% raise, the result is
$58,300. Find the employee’s current salary.

WRITING ABOUT MATHEMATICS

65. A student solves an equation as follows.

x + 30 = 64
x � 64 + 30
x � 94

Identify the student’s mistake. What is the solution?

66. What is a good first step for solving the equation
a
b x = 1, where a and b are natural numbers? What is
the solution? Explain your answers.

2.2 Linear Equations
Basic Concepts ● Solving Linear Equations ● Applying the Distributive Property ●
Clearing Fractions and Decimals ● Equations with No Solutions or Infinitely Many Solutions

A LOOK INTO MATH N Billions of dollars are spent each year to solve equations that lead to the creation of better
products. If our society could not solve equations, we would not have HDTV, high-speed
Internet, satellites, fiber optics, CAT scans, smart phones, or accurate weather forecasts. In
this section we discuss linear equations and some of their applications. Linear equations
can always be solved by hand.

Basic Concepts
Suppose that a bicyclist is 5 miles from home, riding away from home at 10 miles per hour,
as shown in Figure 2.1. The distance between the bicyclist and home for various elapsed
times is shown in Table 2.3.NEW VOCABULARY

n Linear equation
n Identity
n Contradiction

TABLE 2.3 Distance from Home

Elapsed Time (hours) 0 1 2 3

Distance (miles) 5 15 25 35

10 miles 10 miles 10 miles

The bicyclist is moving at a constant speed, so the distance increases by 10 miles every
hour. The distance D from home after x hours can be calculated by the formula

D = 10 x + 5.

For example, after 2 hours the distance is

D = 10(2) + 5 = 25 miles.

Table 2.3 verifies that the bicyclist is 25 miles from home after 2 hours. However, the
table is less helpful if we want to find the elapsed time when the bicyclist is 18 miles from
home. To answer this question, we could begin by substituting 18 for D in the formula to
obtain the equation

18 = 10 x + 5.

Figure 2.1 Distance from Home

5 mi

100 CHAPTER 2 LINEAR EQUATIONS AND INEQUALITIES

The equation 18 = 10x + 5 can be written in a different form by applying the addition
property of equality. Subtracting 18 from each side gives an equivalent equation.

18 � 18 = 10x + 5 � 18 Subtract 18 from each side.
0 = 10x – 13 Simplify.
10 x – 13 = 0

Rewrite the equation.

Even though these steps did not result in a solution to the equation 18 = 10x + 5, apply-
ing the addition property of equality allowed us to rewrite the equation as 10 x – 13 = 0,
which is an example of a linear equation. (See Example 3(a) for a solution to the equation
10x – 13 = 0.) Linear equations can model applications in which things move or change
at a constant rate.

LINEAR EQUATION IN ONE VARIABLE

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

a x + b = 0,

where a and b are constants with a � 0.

If an equation is linear, writing it in the form a x + b = 0 should not require any prop-
erties or processes other than the following.

• using the distributive property to clear any parentheses
• combining like terms
• applying the addition property of equality

For example, the equation 18 = 10 x + 5 is linear because applying the addition property
of equality results in 10 x – 13 = 0, as shown above.

Table 2.4 gives examples of linear equations and values for a and b.

NOTE: An equation cannot be written in the form a x + b = 0 if after clearing parentheses
and combining like terms, any of the following statements are true.

1. The variable has an exponent other than 1.
2. The variable appears in a denominator of a fraction.
3. The variable appears under the symbol 1 or within an absolute value.

TABLE 2.4 Linear Equations

Equation In a x + b = 0 Form a

x = 1 x – 1 = 0 1 -1

-5x + 4 = 3 -5x + 1 = 0 -5 1

2.5x = 0 2.5x + 0 = 0 2.5 0

READING CHECK

• Name three things that tell
you that an equation is not
a linear equation.

EXAMPLE 1 Determining whether an equation is linear

Determine whether the equation is linear. If the equation is linear, give values for a and b
that result when the equation is written in the form ax + b = 0.
(a) 4 x + 5 = 0 (b) 5 = – 34 x (c) 4x

2 + 6 = 0 (d) 3x + 5 = 0

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Beginning and Intermediate Algebra with Applications & Visualization, Third edition, by Gary K. Rockswold and Terry A. Krieger. Published by Addison Wesley.
Copyright © 2013 by Pearson Education, Inc.

1012.2 LINEAR EQUATIONS

Solution
(a) The equation is linear because it is in the form a x + b = 0 with a = 4 and b = 5.
(b) The equation can be rewritten as follows.

5 = –
3

4
x Given equation

3
4

x + 5 =
3
4

x + a – 3
4

xb Add 34 x to each side.

4
x + 5 = 0 Additive inverse

The given equation is linear because it can be written in the form a x + b = 0 with
a = 34 and b = 5.

NOTE: If 5 had been subtracted from each side, the result would be 0 = – 34 x – 5,
which is an equivalent linear equation with a = – 34 and b = -5.

(d) The equation is not linear because it cannot be written in the form a x + b = 0. The
variable appears in the denominator of a fraction.

Now Try Exercises 9, 11, 13, 15

Solving Linear Equations
Every linear equation has exactly one solution. Showing that this is true is left as an exercise
(see Exercise 59). Solving a linear equation means finding the value of the variable that
makes the equation true.

SOLVING LINEAR EQUATIONS NUMERICALLY One way to solve a linear equation
is to make a table of values. A table provides an organized way of checking possible values
of the variable to see if there is a value that makes the equation true. For example, if we
want to solve the equation

2 x – 5 = �7,

we substitute various values for x in the left side of the equation. If one of these values
results in �7, then the value makes the equation true and is the solution. In the next exam-
ple a table of values is used to solve this equation.

EXAMPLE 2 Using a table to solve an equation

Complete Table 2.5 for the given values of x. Then solve the equation 2 x – 5 = -7.

TABLE 2.5

x -3 -2 -1 0 1 2 3

2 x – 5 -11

Solution
To complete the table, substitute x = -2, -1, 0, 1, 2, and 3 into the expression 2 x – 5.
For example, if x = -2, then 2 x – 5 = 2(�2) – 5 = -9. The other values shown in
Table 2.6 on the next page can be found similarly.

102 CHAPTER 2 LINEAR EQUATIONS AND INEQUALITIES

From the table, 2 x – 5 equals �7 when x = �1. So the solution to 2 x – 5 = -7 is -1.

Now Try Exercise 25

TABLE 2.6

x -3 -2 �1 0 1 2 3

2 x – 5 -11 -9 �7 -5 -3 -1 1

SOLVING LINEAR EQUATIONS SYMBOLICALLY Although tables can be used to
solve some linear equations, the process of creating a table that contains the solution can
take a significant amount of time. For example, the solution to the equation 9x – 4 = 0 is
4
9. However, creating a table that reveals this solution would be quite challenging.

The following strategy, which involves the addition and multiplication properties of
equality, is a method for solving linear equations symbolically.

TECHNOLOGY NOTE

Graphing Calculators
and Tables
Many graphing calculators
have the capability to make
tables. Table 2.6 is shown
in the accompanying
figure.

Y1�2X�5

X Y1
�3 �11
�2 �9
�1 �7
0 �5
1 �3
2 �1
3 1

CALCULATOR HELP
To make a table on a calculator, see
Appendix A (pages AP-2 and AP-3). SOLVING A LINEAR EQUATION SYMBOLICALLY

STEP 1: Use the distributive property to clear any parentheses on each side of the
equation. Combine any like terms on each side.

STEP 2: Use the addition property of equality to get all of the terms containing the
variable on one side of the equation and all other terms on the other side of
the equation. Combine any like terms on each side.

STEP 3: Use the multiplication property of equality to isolate the variable by multi-
plying each side of the equation by the reciprocal of the number in front of
the variable (or divide each side by that number).

STEP 4: Check the solution by substituting it in the given equation.

When a linear equation does not contain parentheses, we can start with the second step
in the strategy shown above. This is the case for the equations in the next example.

STUDY TIP

We will be solving equations
throughout the remainder
of the text. Spend a little
extra time practicing these
steps so that they are famil-
iar and easy to recall when
needed later.

EXAMPLE 3 Solving linear equations

Solve each linear equation. Check the answer for part (b).
(a) 10 x – 13 = 0 (b) 12 x + 3 = 6 (c) 5x + 7 = 2 x + 3

Solution
(a) First, isolate the x-term on the left side of the equation by adding 13 to each side.

10 x – 13 = 0 Given equation
10 x – 13 � 13 = 0 � 13 Add 13 to each side. (Step 2)
10 x = 13 Add the real numbers.

To obtain a coefficient of 1 on the x-term, divide each side by 10.

10 x

10
=

10
Divide each side by 10. (Step 3)

x =
13

10
Simplify.

The solution is 1310.

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Beginning and Intermediate Algebra with Applications & Visualization, Third edition, by Gary K. Rockswold and Terry A. Krieger. Published by Addison Wesley.
Copyright © 2013 by Pearson Education, Inc.

1032.2 LINEAR EQUATIONS

(b) Start by subtracting 3 from each side.

2
x + 3 = 6 Given equation

2
x + 3 � 3 = 6 � 3 Subtract 3 from each side. (Step 2)

2
x = 3 Subtract the real numbers.

2 #

1
2

x = 2 # 3 Multiply each side by 2. (Step 3)
x = 6 Multiply the real numbers.

The solution is 6. To check it, substitute 6 for x in the equation, 12 x + 3 = 6.

2
# 6 + 3 � 6 Replace x with 6. (Step 4)

3 + 3 � 6 Multiply.
6 = 6 ✓ Add; the answer checks.

(c) Since this equation has two x-terms, we need to get all x-terms on one side of the equa-
tion and all real numbers on the other side. To do this, begin by subtracting 2x from
each side.

5x + 7 = 2x + 3 Given equation
5x � 2x + 7 = 2x � 2x + 3 Subtract 2x from each side. (Step 2)
3x + 7 = 3 Combine like terms.
3x + 7 � 7 = 3 � 7 Subtract 7 from each side. (Step 2)
3x = -4 Simplify.

3
=

-4
3

Divide each side by 3. (Step 3)

x = –
4

3
Simplify the fractions.

The solution is – 43.

Now Try Exercises 31, 33, 37

N REAL-WORLD CONNECTION In recent years, the number of worldwide Internet users has
increased at a constant rate. Recall that linear equations are often used to model situations
that exhibit a constant rate of change. In the next example we solve a linear equation that
models Internet use.

EXAMPLE 4 Estimating numbers of worldwide Internet users

The number of Internet users I in millions during year x can be approximated by the formula

I = 241x – 482,440,

where x Ú 2007. Estimate the year when there were 2210 million (2.21 billion) Internet
users. (Source: Internet World Stats.)

104 CHAPTER 2 LINEAR EQUATIONS AND INEQUALITIES

Solution
Let I = 2210 in the formula I = 241x – 482,440 and solve for x.

2210 = 241x – 482,440 Equation to be solved
484,650 = 241x Add 482,440 to each side.

484,650

241
=

241x

241
Divide each side by 241.

484,650

241
= x Simplify.

x � 2011 Approximate.

During 2011 the number of Internet users reached 2210 million.

Now Try Exercise 73READING CHECK

• In the strategy for solving
linear equations symboli-
cally, which step involves
the use of the distributive
property?

Applying the Distributive Property
Sometimes the distributive property is helpful in solving linear equations. The next example
demonstrates how to apply the distributive property in such situations. Use of the distribu-
tive property appeared in Step 1 of the strategy for solving linear equations discussed earlier.

EXAMPLE 5 Applying the distributive property

Solve each linear equation. Check the answer for part (a).
(a) 4(x – 3) + x = 0 (b) 2(3� – 4) + 1 = 3(� + 1)

Solution
(a) Begin by applying the distributive property.

4(x – 3) + x = 0 Given equation
4x – 12 + x = 0 Distributive property (Step 1)
5x – 12 = 0 Combine like terms.
5x – 12 � 12 = 0 � 12 Add 12 to each side. (Step 2)
5x = 12 Add the real numbers.

5
=

5
Divide each side by 5. (Step 3)

x =
12

5
Simplify.

To see if 125 is the solution, substitute
12
5 for x in the equation 4(x – 3) + x = 0.

4a 12
5

– 3b + 12
5

� 0 Replace x with 125 . (Step 4)

4¢ 12
5

–
15

5
≤ + 12

5
� 0 Common denominator

4¢ – 3
5
≤ + 12

5
� 0 Subtract within parentheses.

–
12

5
+

5
� 0 Multiply.

0 = 0 ✓ Add; the answer checks.

(b) Begin by applying the distributive property to each side of the equation. Then get all
�-terms on the left side and terms containing only real numbers on the right side.

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Beginning and Intermediate Algebra with Applications & Visualization, Third edition, by Gary K. Rockswold and Terry A. Krieger. Published by Addison Wesley.
Copyright © 2013 by Pearson Education, Inc.

1052.2 LINEAR EQUATIONS

2(3� – 4) + 1 = 3(� + 1) Given equation
6� – 8 + 1 = 3� + 3 Distributive property (Step 1)
6� – 7 = 3� + 3 Add the real numbers.
3� – 7 = 3 Subtract 3� from each side. (Step 2)
3� = 10 Add 7 to each side. (Step 2)

3�
3

=
10

3
Divide each side by 3. (Step 3)

� =
10

3
Simplify.

The solution is 103 .

Now Try Exercises 43, 45

Clearing Fractions and Decimals
Some people prefer to do calculations without fractions or decimals. For this reason, clearing
an equation of fractions or decimals before solving it can be helpful. To clear fractions or
decimals, multiply each side of the equation by the least common denominator (LCD).

READING CHECK

• Why do some people
prefer to clear fractions or
decimals from an equation?

EXAMPLE 6 Clearing fractions from linear equations

Solve each linear equation.
(a) 17 x –

5
7 =

3
7 (b)

2
3 x –

1
6 = x

Solution
(a) Multiply each side of the equation by the LCD 7 to clear (remove) fractions.

7
x –

7
=

7
Given equation

7a 1
7

x –
5

7
b = 7 # 3

7
Multiply each side by 7.

7
1
# 1

7
x –

7
1
# 5

7
=

7
1
# 3

7
Distributive property

x – 5 = 3 Simplify.
x = 8 Add 5 to each side.

The solution is 8.
(b) The LCD for 3 and 6 is 6. Multiply each side of the equation by 6.

3
x –

6
= x Given equation

6a 2
3

x –
1

6
b = 6 # x Multiply each side by 6.

6
1
# 2

3
x –

6
1
# 1

6
= 6 # x Distributive property

4 x – 1 = 6 x Simplify.
-1 = 2 x Subtract 4 x from each side.

–
1

2
= x Divide each side by 2.

The solution is – 12.

Now Try Exercises 53, 55I
S

B
N

1-
25

6-
49

08
2-

2
Beginning and Intermediate Algebra with Applications & Visualization, Third edition, by Gary K. Rockswold and Terry A. Krieger. Published by Addison Wesley.
Copyright © 2013 by Pearson Education, Inc.

106 CHAPTER 2 LINEAR EQUATIONS AND INEQUALITIES

EXAMPLE 7 Clearing decimals from a linear equation

Solve each linear equation.
(a) 0.2 x – 0.7 = 0.4 (b) 0.01x – 0.42 = -0.

2 x

Solution
(a) The least common denominator for 0.2, 0.7, and 0.4 1or 210, 710, and 4102 is 10. Multiply

each side by 10. When multiplying by 10, move the decimal point 1 place to the right.

0.2 x – 0.7 = 0.4 Given equation
10(0.2 x – 0.7) = 10(0.4) Multiply each side by 10.
10(0.2 x) – 10(0.7) = 10(0.4) Distributive property
2 x – 7 = 4 Simplify.
2 x = 11 Add 7 to each side.

x =
11

2
Divide each side by 2.

The solution is 112 , or 5.5.
(b) The least common denominator for 0.01, 0.42, and 0.2 1or 1100, 42100, and 2102 is 100.

Multiply each side by 100. To do this move the decimal point 2 places to the right.

0.01x – 0.42 = -0.2 x Given equation
100(0.01x – 0.42) = 100(-0.2 x) Multiply each side by 100.
100(0.01x) – 100(0.42) = 100(-0.2 x) Distributive property
x – 42 = -20 x Simplify.
x – 42 � 20 x � 42 = -20 x � 20 x � 42 Add 20 x and 42.
21x = 42 Combine like terms.
x = 2 Divide each side by 21.

The solution is 2.

Now Try Exercises 49, 51

Equations with No Solutions or Infinitely
Many Solutions
Some equations that appear to be linear are not because when they are written in the form
ax + b = 0 the value of a is 0 and no x-term appears. This type of equation can have no
solutions or infinitely many solutions. An example of an equation that has no solutions is

x = x + 1

because a number x cannot equal itself plus 1. If we attempt to solve this equation by sub-
tracting x from each side, we obtain the equation 0 = 1, which has no x-term and is always
false. An equation that is always false has no solutions.

An example of an equation with infinitely many solutions is

5x = 2x + 3x,

because the equation simplifies to

5x = 5x,

which is true for any real number x. If 5x is subtracted from each side the result is 0 = 0,
which has no x-term and is always true. When an equation containing a variable is always
true, it has infinitely many solutions.

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1-256-49082-2
Beginning and Intermediate Algebra with Applications & Visualization, Third edition, by Gary K. Rockswold and Terry A. Krieger. Published by Addison Wesley.
Copyright © 2013 by Pearson Education, Inc.

1072.2 LINEAR EQUATIONS

NOTE: An equation that is always true is called an identity, and an equation that is always
false is called a contradiction.

EXAMPLE 8 Determining numbers of solutions

Determine whether the equation has no solutions, one solution, or infinitely many solutions.
(a) 3x = 2(x + 1) + x (b) 2 x – (x + 1) = x – 1 (c) 5x = 2(x – 4)

Solution
(a) Start by applying the distributive property.

3x = 2(x + 1) + x Given equation
3x = 2 x + 2 + x Distributive property
3x = 3x + 2 Combine like terms.
0 = 2 Subtract 3x from each side.

The equation 0 = 2 is always false, so the given equation is a contradiction with no
solutions.

(b) Start by applying the distributive property.

2 x – (x + 1) = x – 1 Given equation
2 x – x – 1 = x – 1 Distributive property
x – 1 = x – 1 Combine like terms.
x = x Add 1 to each side.
0 = 0 Subtract x from each side.

The equation 0 = 0 is always true, so the given equation is an identity that has infi-
nitely many solutions. Note that the solution set contains all real numbers.

5x = 2(x – 4) Given equation
5x = 2 x – 8 Distributive property
3x = -8 Subtract 2x from each side.

x = –
8

3
Divide each side by 3.

Thus there is one solution.

Now Try Exercises 61, 63, 67

CRITICAL THINKING

What must be true about
b and d for the equation

bx – 2 = dx + 7

to have no solutions?
What must be true about
b and d for this equation
to have exactly one solu-
tion?

MAKING CONNECTIONS

Number of Solutions

When solving the general form ax + b = 0, where a and b can be any real numbers, the
resulting equivalent equation will indicate whether the given equation has no solutions, one
solution, or infinitely many solutions.

No Solutions: The result is an equation such as 4 = 0 or 3 = 2, which is always false
for any value of the variable.

One Solution: The result is an equation such as x = 1 or x = -12, which is true for
only one value of the variable.

Infinitely Many
Solutions: The result is an equation such as 0 = 0 or -3 = -3, which is always

true for any value of the variable.

READING CHECK

• How can you tell when an
equation will have no solu-
tions, one solution, or infi-
nitely many solutions?

108 CHAPTER 2 LINEAR EQUATIONS AND INEQUALITIES

2.2 Putting It All Together

Linear Equation Can be written as

ax + b = 0,

where a � 0; has one solution

The equation 5x – 8 = 0 is linear, with
a = 5 and b = -8.

The equation 2×2 + 4 = 0 is not linear.

Solving Linear Equations
Numerically

To solve a linear equation numeri-
cally, complete a table for various val-
ues of the variable and then select the
solution from the table, if possible.

The solution to 2 x – 4 = �2 is 1.

Solving Linear Equations
Symbolically

Use the addition and multiplication
properties of equality to isolate the
variable. See the four-step approach to
solving a linear equation on page 102.

5x – 8 = 0 Given equation
5x = 8 Add 8 to each side.

x =
8

5
Divide each side by 5.

CONCEPT COMMENTS EXAMPLES

x -1 0 1

2 x – 4 -6 -4 �2

Equations with No
Solutions

Some equations that appear to be
linear have no solutions. Solving will
result in an equivalent equation that is
always false.

The equation

x =

x + 5

has no solutions because a number can-
not equal itself plus 5.

Equations with Infinitely
Many Solutions

Some equations that appear to be
linear have infinitely many solutions.
Solving will result in an equivalent
equation that is always true.

The equation

2x = x + x

has infinitely many solutions because the
equation is true for all values of x.

2.2 Exercises

CONCEPTS AND VOCABULARY

1. Linear equations can model applications in which
things move or change at a(n) _____ rate.

2. A linear equation can be written in the form _____
with a � 0.

3. How many solutions does a linear equation in one
variable have?

4. When a table of values is used to solve a linear equa-
tion, the equation is being solved _____.

5. What two properties of equality are frequently used
to solve linear equations?

6. To clear fractions or decimals from an equation, mul-
tiply each side by the _____ .

7. If solving an equation results in 0 = 4, how many
solutions does it have?

8. If solving an equation results in 0 = 0, how many
solutions does it have?

IDENTIFYING LINEAR EQUATIONS

Exercises 9–22: (Refer to Example 1.) Is the equation
linear? If it is linear, give values for a and b that result
when the equation is written in the form ax + b = 0.

9. 3x – 7 = 0 10. -2 x + 1 = 4

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Beginning and Intermediate Algebra with Applications & Visualization, Third edition, by Gary K. Rockswold and Terry A. Krieger. Published by Addison Wesley.
Copyright © 2013 by Pearson Education, Inc.

1092.2 LINEAR EQUATIONS

11. 12x = 0 12. –
3
4x = 0

13. 4 x2 – 6 = 11 14. -2 x2 + x = 4

15. 6x – 4 = 2 16. 22x – 1 = 0

17. 1.1x + 0.9 = 1.8 18. -5.7x – 3.4 = -6.8

19. 2(x – 3) = 0 20. 12(x + 4) = 0

21. �3x � + 2 = 1 22. 3x = 4 x3

SOLVING LINEAR EQUATIONS

Exercises 23–28: Complete the table for the given val-
ues of x. Then use the table to solve the given equation
numerically.

23. x – 3 = -1

24. -2 x = 0

25. -3x + 7 = 1

26. 5x – 2 = 3

27. 4 – 2 x = 6

28. 9 – (x + 3) = 4

x -2 -1 0 1 2

9 – (x + 3) 8

x -1 0 1 2 3

x – 3 -4

x -2 -1 0 1 2

-2 x 4

x 0 1 2 3 4

-3x + 7 7

x -1 0 1 2 3

5x – 2 -7

x -2 -1 0 1 2

4 – 2x 8

Exercises 29–58: Solve the given equation and check the
solution.

29. 11x = 3 30. -5x = 15

31. x – 18 = 5 32. 8 = 5 + 3x

33. 12 x – 1 = 13 34.
1
4 x + 3 = 9

35. -6 = 5x + 5 36. 31 = -7x – 4

37. 3� + 2 = � – 5 38. � – 5 = 5� – 3

39. 12y – 6 = 33 – y 40. -13y + 2 = 22 – 3y

41. 4(x – 1) = 5 42. -2(2 x + 7) = 1

43. 1 – (3x + 1) = 5 – x

44. 6 + 2(x – 7) = 10 – 3(x – 3)

45. (5t – 6) = 2(t + 1) + 2

46. -2(t – 7) – (t + 5) = 5

47. 3(4� – 1) – 2(� + 2) = 2(� + 1)

48. – (� + 4) + (3� + 1) = -2(� + 1)

49. 7.3x – 1.7 = 5.6 50. 5.5x + 3x = 51

51. -9.5x – 0.05 = 10.5x + 1.05

52. 0.04 x + 0.03 = 0.02 x – 0.1

53. 12x –
3
2 =

5
2 54. –

1
4x +

5
4 =

3
4

55. – 38 x +

1
4 =

1
8 56.

1
3 x +

1
4 =

1
6 – x

57. 4y – 2(y + 1) = 0

58. (15y + 20) – 5y = 5 – 10y

59. Thinking Generally A linear equation has exactly
one solution. Find the solution to the equation
ax + b = 0, where a � 0, by solving for x.

60. Thinking Generally Solve the linear equation given
by 1a x – b = 0 for x.

EQUATIONS WITH NO SOLUTIONS OR INFINITELY
MANY SOLUTIONS

Exercises 61–70: Determine whether the equation has no
solutions, one solution, or infinitely many solutions.

61. 5x = 5x + 1 62. 2(x – 3) = 2 x – 6

63. 8 x = 0 64. 9x = x + 1

65. 4 x = 5(x + 3) – x 66. 5x = 15 – 2(x + 7)

67. 5(2 x + 7) – (10 x + 5) = 30

68. 4(x + 2) – 2(2 x + 3) = 10

69. x – (3x + 2) = 15 – 2 x

70. 2 x – (x + 5) = x – 5

110 CHAPTER 2 LINEAR EQUATIONS AND INEQUALITIES

APPLICATIONS

71. Distance Traveled A bicyclist is 4 miles from home,
riding away from home at 8 miles per hour.
(a) Make a table that shows the bicyclist’s distance D

from home after 0, 1, 2, 3, and 4 hours.
(b) Write a formula that calculates D after x hours.
(c) Use your formula to determine D when x = 3

hours. Does your answer agree with the value
found in your table?

(d) Find x when D = 22 miles. Interpret the result.

x Ú 2002, can be approximated by the formula
N = 0.03x – 58.62. Determine the year in which there
were 1.5 million inmates. (Source: Bureau of Justice.)

76. Government Costs From 1960 to 2000 the cost C (in
billions of 1992 dollars) to regulate social and eco-
nomic programs could be approximated by the formula
C = 0.35x – 684 during year x. Estimate the year in
which the cost reached $6.6 billion. (Source: Center for the
Study of American Business.)

77. Hospitals The number of hospitals H with more than
100 beds during year x is estimated by the formula
H = -33x + 69,105, where x is any year from 2002
to 2008. In which year were there 2841 hospitals of
this type? (Source: AHA Hospital Statistics.)

78. Home Size The average size F in square feet of new
U.S. homes built during year x is estimated by the for-
mula F = 34x – 65,734, where x is any year from
2002 to 2008. In which year was the average home
size 2504 square feet? (Source: U.S. Census Bureau.)

WRITING ABOUT MATHEMATICS

79. A student says that the equation 4 x – 1 = 1 – x
is not a linear equation because it is not in the form
a x + b = 0. Is the student correct? Explain.

80. A student solves a linear equation as follows.

4(x + 3) = 5 – (x + 3)
4x + 3 � 5 – x + 3
4x + 3 � 8 – x

5x � 5

x � 1

Identify and explain the errors that the student made.
What is the correct answer?

Checking Basic ConceptsSECTIONS2.1 and 2.2
1. Determine whether the equation is linear.
(a) 4 x3 – 2 = 0 (b) 2(x + 1) = 4

2. Complete the table for each value of x. Then use
the table to solve 4 x – 3 = 13.

4. Determine whether each equation has no solu-
tions, one solution, or infinitely many solutions.

(a) x – 5 = 6x
(b) -2(x – 5) = 10 – 2x
(c) – (x – 1) = -x – 1

5. Distance Traveled A driver is 300 miles from
home and is traveling toward home on a freeway
at a constant speed of 75 miles per hour.

(a) Write a formula to calculate the distance D
that the driver is from home after x hours.

(b) Write an equation whose solution gives the
hours needed for the driver to reach home.

x 3 3.5 4 4.5 5

4 x – 3 9 17

3. Solve each equation and check your answer.
(a) x – 12 = 6
(b) 34 z =

1
8

72. Distance Traveled An athlete is 16 miles from home,
running toward home at 6 miles per hour.
(a) Write a formula that calculates the distance D that

the athlete is from home after x hours.
(b) Determine D when x = 1.5 hours.
(c) Find x when D = 5.5 miles. Interpret the result.

73. Internet Users (Refer to Example 4.) The number
of Internet users I in millions during year x, where
x Ú 2007, can be approximated by the formula

I = 241 x – 482,440.

Approximate the year in which there were 1730 mil-
lion (1.73 billion) Internet users.

74. HIV Infections The cumulative number of HIV
infections N in thousands for the United States in year
x can be approximated by the formula

N = 42 x – 83,197,

where x Ú 2000. Approximate the year when this
number reached 970 thousand. (Source: Centers for Disease
Control and Prevention.)

75. State and Federal Inmates The number N of state
and federal inmates in millions during year x, where

IS
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N
1-256-49082-2
Beginning and Intermediate Algebra with Applications & Visualization, Third edition, by Gary K. Rockswold and Terry A. Krieger. Published by Addison Wesley.
Copyright © 2013 by Pearson Education, Inc.

1112.3 INTRODUCTION TO PROBLEM SOLVING

Even if we understand the problem that we are trying to solve, we may not be able to
find a solution if we cannot write an appropriate equation. Table 2.7 lists several common
words that are associated with the math symbols needed to write equations.

2.3 Introduction to Problem Solving
Steps for Solving a Problem ● Percent Problems ● Distance Problems ●
Other Types of Problems

A LOOK INTO MATH N Throughout history people have found it quite difficult to predict human behavior. With the
development of the Internet and social networks, accurately predicting behavior is becom-
ing possible. For example, using mathematics, it is possible to analyze a person’s Facebook
friends to predict with “reasonable certainty” the person’s hobbies, interests, and even
health issues. Solving problems requires problem-solving skills, which we introduce in this
section.

Steps for Solving a Problem
Word problems can be challenging because formulas and equations are not usually given.
To solve such problems we need a strategy. The following steps are based on George
Polya’s (1888–1985) four-step process for problem solving.

NEW VOCABULARY

n Percent change
STEPS FOR SOLVING A PROBLEM

STEP 1: Read the problem carefully and be sure that you understand it. (You may
need to read the problem more than once.) Assign a variable to what you
are being asked to find. If necessary, write other quantities in terms of this
variable.

STEP 2: Write an equation that relates the quantities described in the problem. You
may need to sketch a diagram or refer to known formulas.

STEP 3: Solve the equation. Use the solution to determine the solution(s) to the origi-
nal problem. Include any necessary units.

STEP 4: Check your solution in the original problem. Does it seem reasonable?

READING CHECK

• Why doesn’t the problem-
solving strategy stop after
the equation is solved in
Step 3?

STUDY TIP

Step 4 in this problem-solving strategy provides good advice for working any math problem.
Checking your answer, especially when taking an exam, can lead to fewer errors and better scores.

TABLE 2.7 Math Symbols and Associated Words

Symbol Associated Words

+ add, plus, more, sum, total, increase

– subtract, minus, less, difference, fewer, decrease

# multiply, times, twice, double, triple, product
, divide, divided by, quotient, per

= equals, is, gives, results in, is the same as

112 CHAPTER 2 LINEAR EQUATIONS AND INEQUALITIES

EXAMPLE 1 Translating sentences into equations

Translate the sentence into an equation using the variable x. Then solve the resulting equation.
(a) Three times a number minus 6 is equal to 18.
(b) The sum of half a number and 5 is zero.
(c) Sixteen is 4 less than twice a number.

Solution
(a) The phrase “Three times a number” indicates that we multiply x by 3 to get 3x. The

word “minus” indicates that we then subtract 6 from 3x to get 3x – 6. This expression
“equals” 18, so the equation is 3x – 6 = 18. The solution is 8 as shown here.

3x – 6 = 18 Equation to be solved
3x = 24 Add 6 to each side.

3x
3
=

3
Divide each side by 3.

x = 8 Simplify the fractions.

(b) The word “sum” indicates that we add “half a number” and 5 to get 12 x + 5. The word
“is” implies equality, so the equation is 12 x + 5 = 0. The solution is -10 as shown
here.

2
x + 5 = 0 Equation to be solved

2
x = -5 Subtract 5 from each side.

x = -10 Multiply each side by 2.

(c) To translate “4 less than twice a number” into a mathematical expression, we write
2 x – 4. If this seems backwards, consider how you would calculate “4 less than your
age.” The equation is 16 = 2 x – 4 and the solution is 10 as shown here.

16 = 2 x – 4 Equation to be solved
20 = 2 x Add 4 to each side.

20
2
=
2 x
2
Divide each side by 2.

x = 10 Simplify and rewrite.

Now Try Exercises 13, 15, 19

In the next example we apply the four-step process to a word problem that involves
three unknown numbers.

EXAMPLE 2 Solving a number problem

The sum of three consecutive natural numbers is 81. Find the three numbers.

Solution
STEP 1: Start by assigning a variable n to an unknown quantity.

n: smallest of the three natural numbers

Next, write the other two natural numbers in terms of n.

n � 1: next consecutive natural number

n � 2: largest of the three consecutive natural numbers

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Copyright © 2013 by Pearson Education, Inc.

1132.3 INTRODUCTION TO PROBLEM SOLVING

STEP 2: Write an equation that relates these unknown quantities. The sum of the three con-
secutive natural numbers is 81, so the needed equation is

n + (n � 1) + (n � 2) = 81.

STEP 3: Solve the equation in Step 2.

n + (n + 1) + (n + 2) = 81 Equation to be solved
(n + n + n) + (1 + 2) = 81 Commutative and associative properties
3n + 3 = 81 Combine like terms.
3n = 78 Subtract 3 from each side.
n = 26 Divide each side by 3.

The smallest of the three numbers is 26, so the three numbers are 26, 27, and 28.
STEP 4: To check this solution we can add the three numbers to see if their sum is 81.

26 + 27 + 28 = 81

The solution checks.

Now Try Exercise 21

N REAL-WORLD CONNECTION The infant mortality rate for a country measures the number of
deaths of infants under one year of age per 1000 live births in a given year. A high infant
mortality rate may indicate a lack of good quality health care for infants. In the next exam-
ple, a number problem is solved to find the infant mortality rate in Iceland, which has one of
the lowest rates in the world.

EXAMPLE 3 Finding an infant mortality rate

Sierra Leone, in West Africa, has one of the highest infant mortality rates in the world at
160. This rate is 10 more than 50 times the rate in Iceland. Find the infant mortality rate in
Iceland. (Source: World Population Prospectus.)

Solution
STEP 1: Let r represent the unknown infant mortality rate in Iceland.
STEP 2: Since the rate 160 in Sierra Leone is 10 more than 50 times the rate r in Iceland, the

equation to solve is 160 = 10 + 50r.
STEP 3: Rewrite the equation in Step 2 and then solve it.

50r + 10 = 160 Equation to be solved (rewritten)
50r = 150 Subtract 10 from each side.
r = 3 Divide each side by 50.

The infant mortality rate in Iceland is 3 deaths per 1000 live births.
STEP 4: To check this solution, we verify that 10 more than 50 times the Iceland infant

mortality rate of 3 gives the Sierra Leone infant mortality rate of 160.

50(3) + 10 = 150 + 10 = 160

Now Try Exercise 33

Percent Problems
N REAL-WORLD CONNECTION Problems involving percentages occur in everyday life. Wage

increases, sales tax, and government data all make use of percentages. For example, it is
estimated that 10% of the world’s population will be older than 65 in 2025. (Source: World Health
Organization.)

114 CHAPTER 2 LINEAR EQUATIONS AND INEQUALITIES

Percent notation can be changed to either fraction or decimal notation.

PERCENT NOTATION

The expression x% represents the fraction x100 or the decimal number x # 0.01.

READING CHECK

• How do we write x% as a decimal?

NOTE: To write x% as a decimal number, move the decimal point in the number x two
places to the left and then remove the % symbol. For example, 13.6% can be written as a
decimal number as follows.

13.6% S 13.6% S 0.136

EXAMPLE 4 Converting percent notation

Convert each percentage to fraction and decimal notation.
(a) 23% (b) 5.2% (c) 0.3%

Solution
(a) Fraction Notation: To convert to fraction notation, divide 23 by 100. Thus 23% = 23100.
Decimal Notation: Move the decimal point two places to the left and then remove the %

symbol: 23% = 0.23.
(b) Fraction Notation: 5.2% = 5.2100 =

52
1000 =

13 # 4
250 # 4 = 13250

Decimal Notation: Move the decimal point two places to the left and then remove the %
symbol: 5.2% = 0.052.

1000
Decimal Notation: Move the decimal point two places to the left and then remove the %

symbol: 0.3% = 0.003.

Now Try Exercises 49, 53, 55

EXAMPLE 5 Converting to percent notation

Convert each real number to a percentage.
(a) 0.234 (b) 14 (c) 2.7

Solution
(a) Move the decimal point two places to the right and then insert the % symbol to obtain

0.234 = 23.4%.
(b) 14 = 0.25, so

1
4 = 25%.

(c) Move the decimal point two places to the right and then insert the % symbol to obtain
2.7 = 270%. Note that percentages can be greater than 100%.

Now Try Exercises 57, 59, 63

PERCENT CHANGE When prices increase (or decrease), the actual amount of the
increase (or decrease) is often not as significant as the percent change in the price. For
example, consider the $1000 increases shown in Figure 2.2.

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1152.3 INTRODUCTION TO PROBLEM SOLVING

Even though both prices increased by $1000, the percent increase in the price of the home
is four-tenths of 1% and the percent increase in the cost of tuition is more than 22%. The
increase in tuition is much more dramatic than the increase in the price of the home.

If a quantity changes from an old amount to a new amount, the percent change is given by

Percent Change =
new amount – old amount

old amount

# 100.

The reason for multiplying by 100 is to change the decimal representation to a percentage.

NOTE: A positive percent change corresponds to an increase and a negative percent change
corresponds to a decrease.

READING CHECK

• What does it mean when
a percent change is
negative?

Figure 2.2

Home Price:
$1000 Increase

Tuition:
$1000 Increase

$250,000 $251,000 $4500 $5500

EXAMPLE 6 Calculating percent increase in Skype subscriptions

From 2007 to 2010 the number of users paying for a Skype subscription increased from
4.6 million to 8.1 million. Calculate the percent increase in paying Skype subscribers from
2007 to 2010. (Source: Skype SEC filing.)

Solution
The old amount is 4.6 and the new amount is 8.1. The percent increase is

new amount – old amount
old amount

# 100 = 8.1 – 4.6
4.6

# 100 � 76%.
Now Try Exercise 69

As we previously indicated, percentages frequently occur in applications.

EXAMPLE 7 Solving an application involving a percent

In 2008, there were 1,766,695 surgical cosmetic procedures performed. In 2009, this num-
ber decreased by 16.7%. Find the number of surgical cosmetic procedures performed in
2009. (Source: American Society for Aesthetic Plastic Surgery.)

Solution
STEP 1: Let N represent the number of surgical cosmetic procedures performed in 2009.
STEP 2: N equals 1,766,695 minus 16.7% of 1,766,695, and 16.7% = 0.167, so

N = 1,766,695 – (0.167)1,766,695.

NOTE: The word “of” in percent problems often indicates multiplication.

STEP 3: To find N, evaluate the expression on the right side to get

N = 1,766,695 – (0.167)1,766,695 � 1,471,657.

There were about 1,471,657 surgical cosmetic procedures performed in 2009.

116 CHAPTER 2 LINEAR EQUATIONS AND INEQUALITIES

STEP 4: To check the answer, calculate the percent decrease.

1,471,657 – 1,766,695
1,766,695

# 100 � -16.7%
The answer checks.

Now Try Exercise

CRITICAL THINKING

If your salary increased
200%, by what factor did it
increase?

EXAMPLE 8 Solving an application involving a percent

In 2025, about 10% of the world’s population, or 800 million people, will be older than 65.
Find the estimated population of the world in 2025. (Source: World Population Prospectus.)

Solution
STEP 1: Let P represent the world’s population in millions in 2025.
STEP 2: Since 10% of P equals 800 million, this information is described by

0.10P = 800.

STEP 3: To solve the equation in Step 2, divide each side by 0.10.

0.10P = 800 Equation to be solved
P = 8000 Divide by 0.10.

In 2025 the estimated world population is 8000 million, or 8 billion.
STEP 4: To check the answer, determine whether 10% of 8 billion is 800 million.

(0.10)8,000,000,000 = 800,000,000

The answer checks.

Now Try Exercise 73

Distance Problems
If a person drives on an interstate highway at 70 miles per hour for 3 hours, then the total
distance traveled is 70 # 3 = 210 miles. In general, d = rt, where d is the distance trav-
eled, r is the rate (or speed), and t is time. In this example, the distance is in miles, the time
is in hours, and the rate is expressed in miles per hour. In general, the rate in a distance
problem is expressed in units of distance per unit of time.

Figure 2.3

180 mi
2 hr 30 min

EXAMPLE 9 Solving a distance problem

A person drives for 2 hours and 30 minutes at a constant speed and travels 180 miles. See
Figure 2.3. Find the speed of the car in miles per hour.

Solution
STEP 1: Let r represent the car’s rate, or speed, in miles per hour.
STEP 2: The rate is to be given in miles per hour, so change 2 hours and 30 minutes to 2.5

or 52 hours. Because d = 180 and t =
5
2, the equation d = rt becomes

180 = r # 5
2

.
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Beginning and Intermediate Algebra with Applications & Visualization, Third edition, by Gary K. Rockswold and Terry A. Krieger. Published by Addison Wesley.
Copyright © 2013 by Pearson Education, Inc.

1172.3 INTRODUCTION TO PROBLEM SOLVING

STEP 3: Solve the equation in Step 2 for r by multiplying each side of the equation by 25,
which is the reciprocal of 52.

180 =
5

2
# r Equation to be solved

2
5
# 180 = r # 5

2
# 2

5
Multiply each side by 25.

72 = r Simplify.

The speed of the car is 72 miles per hour.
STEP 4: Because 2 hours and 30 minutes is equivalent to 52 hours, traveling for 2 hours and

30 minutes at a constant rate of 72 miles per hour results in a distance of

d = rt = 72 # 5
2

= 180 miles.

The answer checks.

Now Try Exercise 83

EXAMPLE 10 Solving a distance problem

An athlete jogs at two speeds, covering a distance of 7 miles in 34 hour. If the athlete runs
1
4

hour at 8 miles per hour, find the second speed.

Solution
STEP 1: Let r represent the second speed of the jogger in miles per hour.
STEP 2: The total time spent jogging is 34 hour, so the time spent jogging at the second speed

must be 34 –
1
4 =

1
2 hour. The total distance of 7 miles is the result of jogging at

8 miles per hour for 14 hour and at r miles per hour for
1
2 hour. See Figure 2.4.

The distance traveled at 8 miles per hour for 14 hour is given by 8 # 14 and the
distance traveled at r miles per hour for 12 hour is given by r # 12. The sum of these
distances must equal 7 miles. Thus

8 # 1
4

+ r # 1
2

= 7.

STEP 3: Solve the equation in Step 2 for r.

8 # 1
4

+ r # 1
2

= 7 Equation to be solved

2 +
r

2
= 7 Simplify.

2
= 5 Subtract 2 from each side.

r = 10 Multiply each side by 2.

The athlete’s second speed is 10 miles per hour.
STEP 4: Jogging at a rate of 8 miles per hour for 14 hour results in a distance of 8 # 14 = 2

miles. Jogging at a rate of 10 miles per hour for 12 hour results in a distance of
10 # 12 = 5 miles. The total distance is 2 + 5 = 7 miles and the total time is

14 +
1
2 =

3
4 hour, so the answer checks.

Now Try Exercise 87

Figure 2.4

7 mi

hr at
8 mp

1
4 hr at

r mph

1
2
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118 CHAPTER 2 LINEAR EQUATIONS AND INEQUALITIES

Other Types of Problems
N REAL-WORLD CONNECTION Many applied problems involve linear equations. For example,

right after people have their wisdom teeth pulled, they may need to rinse their mouth with
salt water. In the next example, we use linear equations to determine how much water must
be added to dilute a concentrated saline solution.

EXAMPLE 11 Diluting a saline solution

A solution contains 4% salt. How much pure water should be added to 30 ounces of the
solution to dilute it to a 1.5% solution?

Solution
STEP 1: Assign a variable x as follows.

x: ounces of pure water (0% salt solution)

30: ounces of 4% salt solution

x � 30: ounces of 1.5% salt solution

In Figure 2.5 three beakers illustrate this situation.

Figure 2.5 Mixing a Saline Solution

x ounces 30 ounces x + 30 ounces

4% 1.5%=+

STEP 2: Note that the amount of salt in the first two beakers must equal the amount of salt
in the third beaker. We use Table 2.8 to organize our calculations. The amount of
salt in a solution equals the concentration times the solution amount, as shown in
the last column of the table.

TABLE 2.8 Mixing a Saline Solution

Concentration Solution Amount Salt
Solution Type (as a decimal) (ounces) (ounces)

Pure Water 0% = 0.00 x 0.00x

Initial Solution 4% = 0.04 30 0.04(30)

Diluted Solution 1.5% = 0.015 x + 30 0.015(x + 30)

The amount of salt in the first two beakers is

0.00 x + 0.04(30) = 0 + 1.2 = 1.2 ounces.

The amount of salt in the final beaker is

0.015(x � 30) ounces.

Because the amounts of salt in the solutions before and after mixing must be equal,
the following equation must hold.

0.015(x � 30) = 1.2

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Copyright © 2013 by Pearson Education, Inc.

1192.3 INTRODUCTION TO PROBLEM SOLVING

STEP 3: Solve the equation in Step 2.

0.015(x + 30) = 1.2 Equation to be solved
0.015x + 0.45 = 1.2 Distributive property
0.015x = 0.75 Subtract 0.45 from each side.

0.015x

0.015
=

0.75

0.015
Divide each side by 0.015.

x = 50 Simplify fractions.

Fifty ounces of water should be added.
STEP 4: Adding 50 ounces of water will yield 50 + 30 = 80 ounces of water containing

0.04(30) = 1.2 ounces of salt. The concentration is 1.280 = 0.015 or 1.5%, so the
answer checks.

Now Try Exercise 91

N REAL-WORLD CONNECTION Many times interest rates for student loans vary. In the next
example, we present a situation in which a student has to borrow money at two different
interest rates.

EXAMPLE 12 Calculating interest on college loans

A student takes out a loan for a limited amount of money at 5% interest and then must pay
7% for any additional money. If the student borrows $2000 more at 7% than at 5%, then the
total interest for one year is $440. How much does the student borrow at each rate?

Solution
STEP 1: Assign a variable x as follows.

x: loan amount at 5% interest

x + 2000: loan amount at 7% interest

STEP 2: The amount of interest paid for the 5% loan is 5% of x, or 0.05x. The amount of
interest paid for the 7% loan is 7% of x + 2000, or 0.07(x � 2000). The total
interest equals $440, so we solve the equation

0.05x + 0.07(x � 2000) = 440.

STEP 3: Solve the equation in Step 2 for x.

0.05x + 0.07(x + 2000) = 440 Equation to be solved
0.05x + 0.07x + 140 = 440 Distributive property
0.12 x + 140 = 440 Combine like terms.
0.12 x = 300 Subtract 140 from each side.
x = 2500 Divide each side by 0.12.

The student borrows $2500 at 5% and 2500 + 2000 = $4500 at 7%.
STEP 4: The amount of interest on $2500 borrowed at 5% is 0.05(2500) = $125 and the

amount of interest on $4500 borrowed at 7% is 0.07(4500) = $315. Thus the total
interest is 125 + 315 = $440. Furthermore, the amount borrowed at 7% is $2000
more than the amount borrowed at 5%. The answer checks.

Now Try Exercise 93

120 CHAPTER 2 LINEAR EQUATIONS AND INEQUALITIES

2.3 Putting It All Together

In this section on page 111 we presented a four-step approach to problem solving. How-
ever, because no approach works in every situation, solving mathematical problems takes
time, effort, and creativity.

CONCEPT COMMENTS EXAMPLES

Percent Notation x% represents either the fraction x100
or the decimal x # 0.01. 17% =

17
100 = 0.17

1.5% = 1.5100 =
15

1000 = 0.015

234% = 234100 = 2.34

Converting Fractions to
Percent Notation

Divide the numerator by the denomi-
nator, multiply by 100, and insert the
% symbol.

12 = 0.5 = 50%
2
3 = 0.6 � 66.7%

85 = 1.6 = 160%

Percent Change If a quantity changes from an old
amount to a new amount, then the
percent change is

new amount – old amount
old amount
# 100.

If a price changes from $1.50 to $1.35,
the percent change is

1.35 – 1.50
1.50

* 100 = -10%.

The price decreases by 10%.

Distance Problems Distance d, rate r, and time t are
related as expressed in the equation

d = rt.

If a car travels 60 mph for 3 hours, then
the distance d is

d = rt = 60 # 3 = 180 miles.

2.3 Exercises

CONCEPTS AND VOCABULARY

1. When you are solving a word problem, what is the
last step?

2. The words more, sum, and increase are associated
with the symbol .

3. The words is, gives, and results in are associated with
the symbol .

4. Given an integer n, what are the next two consecutive
integers?

5. The expression x% equals the fraction .

6. The expression x% equals the decimal x # .
7. To write 63.2% as a decimal, move the decimal point

2 places (left/right) and then remove the % symbol.

8. To write 0.349 as a percentage, move the decimal point
2 places (left/right) and then insert the % symbol.

9. If a price changes from A to B, then the percent
change equals _____.

10. A positive percent change corresponds to a(n) _____
and a negative percent change corresponds to a(n)
_____.

11. If a car travels at speed r for time t, then distance d is
given by d = .

12. In general, the rate in a distance problem is expressed
in units of _____ per unit of _____.

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Copyright © 2013 by Pearson Education, Inc.

1212.3 INTRODUCTION TO PROBLEM SOLVING

NUMBER PROBLEMS

Exercises 13–20: Using the variable x, translate the sen-
tence into an equation. Solve the resulting equation.

13. The sum of 2 and a number is 12.

14. Twice a number plus 7 equals 9.

15. A number divided by 5 equals the number decreased
by 24.

16. 25 times a number is 125.

17. If a number is increased by 5 and then divided by 2,
the result is 7.

18. A number subtracted from 8 is 5.

19. The quotient of a number and 2 is 17.

20. The product of 5 and a number equals 95.

Exercises 21–28: Find the number or numbers.

21. The sum of three consecutive natural numbers is 96.

22. The sum of three consecutive integers is -123.

23. Three times a number equals 102.

24. A number plus 18 equals twice the number.

25. Five times a number is 24 more than twice the number.

26. Three times a number is 18 less than the number.

27. Six times a number divided by 7 equals 18.

28. Two less than twice a number, divided by 5, equals 4.

29. Finding Age In 10 years, a child will be 3 years
older than twice her current age. What is the current
age of the child?

30. Finding Age A mother is 15 years older than twice
her daughter’s age. If the mother is 49 years old, how
old is the daughter?

31. Weight Loss After losing 30 pounds, an individual
weighs 110 pounds more than one-third his previous
weight. Find the previous weight of the individual.

32. Weight Gain After gaining 25 pounds, a person is 115
pounds lighter than double his previous weight. How
much did the person weigh before gaining 25 pounds?

33. Hazardous Waste In 2005, the number of federal
hazardous waste sites in California was 2 less than
twice the number of sites in Washington. How many
hazardous waste sites were there in Washington if
there were 24 such sites in California? (Source: Environ-
mental Protection Agency.)

34. Air Quality In 2007, there were 100 unhealthy air
quality days in Los Angeles. This is 1 more than 3
times the number of unhealthy air quality days in San
Diego. How many unhealthy air quality days were
there in San Diego? (Source: Environmental Protection Agency.)

35. Endangered Species There were 92 birds on the
endangered species list in 2010. This is 12 more than
twice the number of reptiles on the list. How many
reptiles were on the endangered species list in 2010?
(Source: U.S. Fish and Wildlife Service.)

36. Millionaires In 2009, there were three times as many
millionaires in Kentucky as there were in Rhode
Island. Together, a total of 84 thousand millionaires
lived in the two states. Find the number of millionaires
in each of these states. (Source: Internal Revenue Service.)

37. U.S. Oil Production There were 1562 million fewer
barrels of crude oil produced in Alaska in 2009 than
there were in the lower 48 states. Find the number of bar-
rels of crude oil (in millions) produced in the lower 48
states in 2009 if there were 250 million barrels produced
in Alaska that year. (Source: U.S. Energy Information Administration.)

38. Power Production In 1995, U.S. hydroelectric power
production reached 311 billion kilowatt hours. This is
189 billion kilowatt-hours less than twice the 2009
production level. Find the amount of hydroelectric
power produced in 2009. (Source: U.S. Department of Energy.)

39. Cosmetic Surgery If the number of cosmetic surger-
ies (in thousands) performed in 2009 on persons under
the age of 18 is multiplied by 2 and then decreased by
4, the result is the number (in thousands) of cosmetic
surgeries performed on persons over age 65 that year.
If there were 70 (thousand) such surgeries performed
on persons over age 65, find the number of cosmetic
surgeries performed on persons under age 18. (Source:
American Society for Aesthetic Plastic Surgery.)

40. Open Heart Surgery In 2002, open heart surgery
was performed 368 thousand times in the United
States on persons over the age of 65. This is 22 thou-
sand fewer than 13 times the number of this type of
surgery performed on persons under the age of 15.
How many times was open heart surgery performed
on persons under the age of 15 in 2002? (Source: U.S.
Department of Health and Human Services.)

41. Geometry If the perimeter of the following rectangle
is 106 inches, find the value of x.

x + 5
x
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Beginning and Intermediate Algebra with Applications & Visualization, Third edition, by Gary K. Rockswold and Terry A. Krieger. Published by Addison Wesley.
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122 CHAPTER 2 LINEAR EQUATIONS AND INEQUALITIES

42. Geometry If the perimeter of the following triangle
is 24 inches, find the value of x.

x + 2

x + 4
x

43. Rectangle Dimensions The length of a rectangle is
7 inches longer than the width. If the perimeter of the
rectangle is 62 inches, find the measures of the length
and width.

44. Triangle Dimensions The shortest side of a triangle
measures 15 feet less than the longest side. If the
third side is 6 feet shorter than the longest side and
the perimeter is 102 feet, find the measures of the
three sides of the triangle.

45. Social Networking In January 2008, the number
of active MySpace users was 10 million fewer than
double the number of active Facebook users. If the
two social networking sites had a total of 170 million
users, how many people used each of the Web sites?
(Source: comScore.)

46. Facebook Users In February 2010, Facebook had
225 million more active users than it had the same
month in 2009. If there were 400 million active users
in February 2010, how many were there in February
2009? (Source: Facebook.com)

47. Troops in Iraq In 2010, there were 52,000 fewer
U.S. troops in Iraq than there were in 2003. If the
total number of U.S. troops in Iraq for the two years
was 248,000, find the number of troops for each year.
(Source: USA Today.)

48. Farm Land In 2009, there were 12 million fewer
acres of farm land in Arizona than there were in 1980.
If the sum of acreage for these two years is 64 million
acres, find the number of farm land acres in Arizona
for each of these years. (Source: U.S. Department of Agriculture,
2010.)

PERCENT PROBLEMS

Exercises 49–56: Convert the percentage to fraction and
decimal notation.

49. 37% 50. 52%

51. 148% 52. 252%

53. 6.9% 54. 8.1%

55. 0.05% 56. 0.12%

Exercises 57–68: Convert the number to a percentage.

57. 0.45 58. 0.08

59. 1.8 60. 2.97

61. 0.006 62. 0.0001

63. 25 64.
1
3

65. 34 66.
7

67. 56 68.
53
50

69. Voter Turnout In the 1980 election for president
there were 86.5 million voters, whereas in 2008 there
were 132.6 million voters. Find the percent change in
the number of voters.

70. College Degrees In 2005, about 594,000 people
received a master’s degree, and by 2010 this number
had increased to 659,000. Find the percent change
in the number of master’s degrees received over this
time period. (Source: The College Board.)

71. Wages A part-time instructor is receiving $950 per
credit taught. If the instructor receives a 4% increase,
how much will the new per credit compensation be?

72. Tuition Increase Tuition is currently $125 per credit.
There are plans to raise tuition by 8% for next year.
What will the new tuition be per credit?

73. Rural Forestland There are about 13.6 million acres
of rural forestland in Wisconsin. This is about 38%
of the state’s total area. Approximate the total area of
Wisconsin in millions of acres. (Source: U.S. Department of
Agriculture.)

74. Women in Uniform In 2009 there were 203,375
female active-duty military personal, up 0.38% from
2008. How many female military personnel were there
in 2008? (Source: U.S. Department of Defense.)

75. Thinking Generally Calculate the percent change if
the price of an item increases from $1.20 to $1.50.
Now calculate the percent change if the price of the
item decreases from $1.50 to $1.20.

76. Thinking Generally Refer to the previous exercise.
Suppose an amount increases from A1 to A2 and the
percent change is calculated to be 30%. If that amount
now decreases from A2 back to A1, will the percent
change be -30%?

DISTANCE PROBLEMS

Exercises 77–82: Use the formula d = rt to find the
value of the missing variable.

77. r = 4 mph, t = 2 hours

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Beginning and Intermediate Algebra with Applications & Visualization, Third edition, by Gary K. Rockswold and Terry A. Krieger. Published by Addison Wesley.
Copyright © 2013 by Pearson Education, Inc.

1232.3 INTRODUCTION TO PROBLEM SOLVING

78. r = 70 mph, t = 2.5 hours

79. d = 1000 feet, t = 50 seconds

80. d = 1250 miles, t = 5 days

81. d = 200 miles, r = 40 mph

82. d = 1700 feet, r = 10 feet per second

83. Driving a Car A person drives a car at a constant
speed for 4 hours and 15 minutes, traveling 255 miles.
Find the speed of the car in miles per hour.

255 mi
4 hr 15 min

84. Flying an Airplane A pilot flies a plane at a constant
speed for 5 hours and 30 minutes, traveling 715 miles.
Find the speed of the plane in miles per hour.

85. Jogging Speeds One runner passes another runner
traveling in the same direction on a hiking trail. The
faster runner is jogging 2 miles per hour faster than
the slower runner. Determine how long it will be
before the faster runner is 34 mile ahead of the slower
runner.

86. Distance Running An athlete runs 8 miles, first at a
slower speed and then at a faster speed. The total time
spent running is 1 hour. If the athlete runs 13 hour at 6
miles per hour, find the second speed.

87. Jogging Speeds At first an athlete jogs at 5 miles
per hour and then jogs at 8 miles per hour, traveling
7 miles in 1.1 hours. How long does the athlete jog at
each speed? (Hint: Let t represent the amount of time
the athlete jogs at 5 mph. Then 1.1 – t represents the
amount of time the athlete jogs at 8 mph.)

7 mi

t hr at
5 mph

1.1 – t hr
at 8 mph

88. Distance and Time A bus is 160 miles east of the
North Dakota–Montana border and is traveling west
at 70 miles per hour. How long will it take for the bus
to be 295 miles west of the border?

89. Distance and Time A plane is 300 miles west of
Chicago, Illinois, and is flying west at 500 miles per
hour. How long will it take for the plane to be 2175
miles west of Chicago?

2175 mi

500 mph 300 mi
Chicago

90. Finding Speeds Two cars pass on a straight high-
way while traveling in opposite directions. One car
is traveling 6 miles per hour faster than the other
car. After 1.5 hours the two cars are 171 miles apart.
Find the speed of each car.

OTHER TYPES OF PROBLEMS

91. Saline Solution (Refer to Example 11.) A solution
contains 3% salt. How much water should be added
to 20 ounces of this solution to make a 1.2% solution?

92. Acid Solution A solution contains 15% hydrochlo-

ric acid. How much water should be added to 50 mil-
liliters of this solution to dilute it to a 2% solution?

93. College Loans (Refer to Example 12.) A student

takes out two loans, one at 5% interest and the other
at 6% interest. The 5% loan is $1000 more than the
6% loan, and the total interest for 1 year is $215.
How much is each loan?

94. Bank Loans Two bank loans, one for $5000 and
the other for $3000, cost a total of $550 in interest
for one year. The $5000 loan has an interest rate 3%
lower than the interest rate for the $3000 loan. Find
the interest rate for each loan.

95. Mixing Antifreeze How many gallons of 70%

antifreeze should be mixed with 10 gallons of 30%
antifreeze to obtain a 45% antifreeze mixture?

96. Mixing Antifreeze How many gallons of 65%
antifreeze and how many gallons of 20% antifreeze
should be mixed to obtain 50 gallons of a 56% mix-
ture of antifreeze? (Hint: Let x represent the number
of gallons of 65% antifreeze. Then 50 – x repre-
sents the amount of 20% antifreeze.)

97. Hydrocortisone Cream A pharmacist needs to

make a 1% hydrocortisone cream. How many grams
of 2.5% hydrocortisone cream should be added to
15 grams of cream base (0% hydrocortisone) to
make the 1% cream?

98. Credit Card Debt A person carries a balance on two
credit cards, one with a monthly interest rate of 1.5%
and the other with a monthly rate of 1.75%. The bal-
ance on the 1.5% card is $600 less than the balance
on the 1.75% card. If the total interest for the month
is $49.50, what is the balance on each card?

WRITING ABOUT MATHEMATICS

99. State the four steps for solving a word problem.

100. The cost of living has increased about 600% during
the past 50 years. Does this percent change corre-
spond to a cost of living increase of 6 times? Explain.

124 CHAPTER 2 LINEAR EQUATIONS AND INEQUALITIES

Working with Real DataGroup Activity

Exercises 1–5: In this set of exercises you are to use your
mathematical problem-solving skills to find the thickness
of a piece of aluminum foil without measuring it directly.

1. Area of a Rectangle The area of a rectangle equals
length times width. Find the area of a rectangle with
length 12 centimeters and width 11 centimeters.

2. Volume of a Box The volume of a box equals
length times width times height. Find the volume
of the box shown, which is 12 centimeters long,
11 centimeters wide, and 5 centimeters high.

11 cm
12 cm

5 cm

3. Height of a Box Suppose that a box has a volume
of 100 cubic centimeters and that the area of the
bottom of the box is 50 square centimeters. Find
the height of the box.

4. Volume of Aluminum Foil One cubic centimeter
of aluminum weighs 2.7 grams. If a piece of alumi-
num foil weighs 5.4 grams, find the volume of the
aluminum foil.

5. Thickness of Aluminum Foil A rectangular sheet
of aluminum foil is 50 centimeters long and
20 centimeters wide, and weighs 5.4 grams. Find
the thickness of the aluminum foil in centimeters.

Directions: Form a group of 2 to 4 people. Select someone to record the group’s responses for
this activity. All members of the group should work cooperatively to answer the questions. If
your instructor asks for your results, each member of the group should be prepared to respond.

2.4 Formulas
Formulas from Geometry ● Solving for a Variable ● Other Formulas

A LOOK INTO MATH N Have you ever wondered how your grade point average (GPA) is calculated? A formula is
used that involves the number of credits earned at each possible grade. Once a formula
has been derived, it can be used over and over by any number of people. For this reason,
formulas provide a convenient and easy way to solve certain types of recurring problems. In
this section we discuss several types of formulas.

NEW VOCABULARY

n Degree
n Circumference

STUDY TIP

If you are studying with classmates, make sure that they do not “do the work for you.” A class-
mate with the best intentions may give too many verbal hints while helping you work through a
problem. Remember that members of your study group will not be giving hints during an exam.

Formulas from Geometry
Formulas from geometry are frequently used in various fields, including surveying and con-
struction. In this subsection we discuss several important formulas from geometry.

RECTANGLES If a rectangle has length l and width w, then its perimeter P and its area A
are given by the formulas

P = 2l + 2w and A = lw.

For example, the rectangle in Figure 2.6 has length 12 centimeters and width 7 centimeters.
Its perimeter is

P = 2(12) + 2(7) = 24 + 14 = 38 centimeters,Figure 2.6 Rectangle
12 cm

7 cm ISB
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Beginning and Intermediate Algebra with Applications & Visualization, Third edition, by Gary K. Rockswold and Terry A. Krieger. Published by Addison Wesley.
Copyright © 2013 by Pearson Education, Inc.

1252.4 FORMULAS

and its area is

A = 12 # 7 = 84 square centimeters.
NOTE: All measurements used in a geometry formula must have the same units.

TRIANGLES If a triangle has base b and height h, as shown in Figure 2.7(a), then its area
A is given by the formula

A =
1

2
bh.

For example, to find the area of a triangle with base 4 feet and height 18 inches as illustrated
in Figure 2.7(b), we begin by making sure that all measurements have the same units. By
converting 18 inches to 1.5 feet, the area of the triangle is

A =
1

2
bh =

2
(4)(1.5) = 3 square feet.

If the base had been converted to 48 inches, the area is

A =
1
2
bh =
1

2
(48)(18) = 432 square inches.

READING CHECK

• What do l and w stand for
in formulas for rectangles?

Figure 2.7 Triangles

b
(a)

4 ft

(b)

8 in.

READING CHECK

• What do b and h stand for in the triangle area formula?

EXAMPLE 1 Calculating area of a region

A residential lot is shown in Figure 2.8. It comprises a rectangular region and an adjacent
triangular region.
(a) Find the area of this lot.
(b) An acre contains 43,560 square feet. How many acres are there in this lot?

Solution
(a) The rectangular portion of the lot has length 405 feet and width 235 feet. Its area AR is

AR = lw = 405 # 235 = 95,175 square feet.
The triangular region has base 124 feet and height 235 feet. Its area AT is

AT =
1

2
bh =
1

2
# 124 # 235 = 14,570 square feet.

The total area A of the lot equals the sum of AR and AT.

A = 95,175 + 14,570 = 109,745 square feet

(b) Each acre equals 43,560 square feet, so divide 109,745 by 43,560 to calculate the num-
ber of acres.

109,745

43,560
� 2.5 acres

Now Try Exercise 27

Figure 2.8

5 ft

405 ft 124 ft

126 CHAPTER 2 LINEAR EQUATIONS AND INEQUALITIES

ANGLES Angles are often measured in degrees. A degree (�) is 1360 of a revolution, so
there are 360� in one complete revolution. In any triangle, the sum of the measures of the
angles equals 180�. In Figure 2.9, triangle ABC has angles with measures x, y, and z. Therefore

x + y + z = 180�.Figure 2.9

A
B

x
y

EXAMPLE 2 Finding angles in a triangle

In a triangle the two smaller angles are equal in measure and are each half the measure of
the largest angle. Find the measure of each angle.

Solution
Let x represent the measure of each of the two smaller angles, as illustrated in Figure 2.10. Then
the measure of the largest angle is 2x, and the sum of the measures of the three angles is given by

x + x + 2 x = 180�.
This equation can be solved as follows.

x + x + 2 x = 180� Equation to be solved
4 x = 180� Combine like terms.

4 x

4
=

180�

4
Divide each side by 4.

x = 45� Divide the real numbers.

The measure of the largest angle is 2 x = 2 # 45� = 90�. Thus the measures of the three
angles are 45�, 45�, and 90�.

Now Try Exercise 33

Figure 2.10

x x

CIRCLES The radius of a circle is the distance from its center to the perimeter of the circle.
The perimeter of a circle is called its circumference C and is given by C = 2pr, where r
is the radius of the circle. The area A of a circle is given by A = pr2. See Figure 2.11. The
distance across a circle, through its center, is called the diameter. Note that a circle’s radius
is half of its diameter. (Recall that p � 3.14.)

Figure 2.11 Circle

EXAMPLE 3 Finding the circumference and area of a circle

A circle has a diameter of 25 inches. Find its circumference and area.

Solution
The radius is half the diameter, or 12.5 inches.

Circumference: C = 2pr = 2p(12.5) = 25p � 78.5 inches.
Area: A = pr2 = p(12.5)2 = 156.25p � 491 square inches.

Now Try Exercise 37

CRITICAL THINKING

Write formulas for the circumference and area of a circle having diameter d.

READING CHECK

• What is the approximate value of p when rounded to the nearest hundredth?

CALCULATOR HELP
To evaluate p on a calculator, see
Appendix A (page AP-1).

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Beginning and Intermediate Algebra with Applications & Visualization, Third edition, by Gary K. Rockswold and Terry A. Krieger. Published by Addison Wesley.
Copyright © 2013 by Pearson Education, Inc.

1272.4 FORMULAS

TRAPEZOIDS A trapezoid such as the one shown in Figure 2.12 has bases a and b, and
height h. The area A of a trapezoid is given by the formula

A =
1

2
(a + b)h.

Figure 2.12 Trapezoid

b
a
h

EXAMPLE 4 Finding the area of a trapezoid

Find the area of the trapezoid shown in Figure 2.13.

Figure 2.13

2 mm

6 mm

5 mm

Solution
The bases of the trapezoid in Figure 2.13 are 126 millimeters and 182 millimeters, and its
height is 95 millimeters. Substituting these values in the area formula gives an area of

A =
1

2
(a + b)h =

2
(126 + 182) # 95 = 1

2
(308) # 95 = 14,630 square millimeters.

Now Try Exercise 19

BOXES The box in Figure 2.14 has length l, width w, and height h. Its volume V is given by

V = lwh.

The surface of the box comprises six rectangular regions: top and bottom, front and back,
and left and right sides. The total surface area S of the box is given by

S = lw + lw + wh + wh + lh + lh.
(top + bottom + front + back + left side + right side)

When we combine like terms, this expression simplifies to

S = 2lw + 2wh + 2lh.

Figure 2.14 Rectangular Box

w
l

EXAMPLE 5 Finding the volume and surface area of a box

Find the volume and surface area of the box shown in Figure 2.15.

Figure 2.15

8 in.

10 in.

5 in.

Solution
Figure 2.15 shows that the box has length l = 10 inches, width w = 8 inches, and height
h = 5 inches. The volume of the box is

V = lwh = 10 # 8 # 5 = 400 cubic inches.

128 CHAPTER 2 LINEAR EQUATIONS AND INEQUALITIES

Since l = 10, w = 8, and h = 5, the surface area of the box is

S = 2lw + 2wh + 2lh
= 2(10)(8) + 2(8)(5) + 2(10)(5)
= 160 + 80 + 100
= 340 square inches.

Now Try Exercise 41

CYLINDERS A soup can is usually made in the shape of a cylinder. The volume of a
cylinder having radius r and height h is V = pr2h. See Figure 2.16.

Figure 2.16 Cylinder

h
r

EXAMPLE 6 Calculating the volume of a soda can

A cylindrical soda can has a radius of 114 inches and a height of 4
3
8 inches.

(a) Find the volume of the can.
(b) If 1 cubic inch equals 0.554 fluid ounce, find the number of fluid ounces in the can.

Solution
(a) Changing mixed numbers to improper fractions gives the radius as r = 54 inches and

the height as h = 358 inches.

V = pr2h Volume of the soda can

= pa 5
4
b2a 35

8
b Substitute.

= pa 875
128
b Multiply the fractions.

� 21.48 cubic inches Approximate.

(b) To calculate the number of fluid ounces in 21.48 cubic inches, multiply by 0.554.

21.48(0.554) � 11.9 fluid ounces

Note that a typical aluminum soda can holds 12 fluid ounces.

Now Try Exercise 49

Solving for a Variable
A formula establishes a relationship between two or more variables (or quantities). Some-
times a formula must be rewritten to solve for the needed variable. For example, if the area
A and the width w of a rectangular region are given, then its length l can be found by solv-
ing the formula A = lw for l.

A = lw Area formula

A
w

=
lw
w

Divide each side by w.

A
w

= l Simplify the fraction.

l =
A
w

Rewrite the equation.
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Beginning and Intermediate Algebra with Applications & Visualization, Third edition, by Gary K. Rockswold and Terry A. Krieger. Published by Addison Wesley.
Copyright © 2013 by Pearson Education, Inc.

1292.4 FORMULAS

If the area A of a rectangle is 400 square inches and its width w is 16 inches, then the
rectangle’s length l is

l =
A
w

=
400

16
= 25 inches.

Once the area formula has been solved for l, the resulting formula can be used to find the
length of any rectangle whose area and width are known.

EXAMPLE 7 Finding the base of a trapezoid

The area of a trapezoid is given by

A =
1

2
(a + b)h,

where a and b are the bases of the trapezoid and h is the height.
(a) Solve the formula for b.
(b) A trapezoid has area A = 36 square inches, height h = 4 inches, and base a = 8 inches.

Find b.

Solution
(a) To clear the equation of the fraction, multiply each side by 2.

A =
1

2
(a + b)h Area formula

2A = (a + b)h Multiply each side by 2.

h
= a + b Divide each side by h.

h
– a = b Subtract a from each side.

b =
2A

h
– a Rewrite the formula.

(b) Let A = 36, h = 4, and a = 8 in b = 2Ah – a. Then

b =
2(36)

4
– 8 = 18 – 8 = 10 inches.

Now Try Exercise 59

EXAMPLE 8 Solving for a variable

Solve each equation for the indicated variable.
(a) c = a + b2 for b (b) ab – bc = ac for c

Solution
(a) To clear the equation of the fraction, multiply each side by 2.

c =
a + b

2
Given formula

2c = a + b Multiply each side by 2.
2c – a = b Subtract a.

The formula solved for b is b = 2c – a.

130 CHAPTER 2 LINEAR EQUATIONS AND INEQUALITIES

(b) In ab – bc = ac, the variable c appears in two terms. We will combine the terms con-
taining c by using the distributive property. Begin by moving the term on the left side
containing c to the right side of the equation.

ab – bc = ac Given formula
ab – bc + bc = ac + bc Add bc to each side.
ab = (a + b)c Combine terms; distributive property.

a + b
=

(a + b)c
(a + b)

Divide each side by (a + b).

a + b
= c Simplify the fraction.

The formula solved for c is c = aba + b.

Now Try Exercises 63, 67

CRITICAL THINKING

Are the formulas c = 1a – b and c =
-1

b – a equivalent? Why?

Other Formulas
To calculate a student’s GPA, the number of credits earned with a grade of A, B, C, D, and
F must be known. If a, b, c, d, and ƒ represent these credit counts respectively, then

GPA =

4a + 3b + 2c + d
a + b + c + d + f

This formula is based on the assumption that a 4.0 GPA is an A, a 3.0 GPA is a B, and so on.

EXAMPLE 9 Calculating a student’s GPA

A student has earned 16 credits of A, 32 credits of B, 12 credits of C, 2 credits of D, and 5
credits of F. Calculate the student’s GPA to the nearest hundredth.

Solution
Let a = 16, b = 32, c = 12, d = 2, and ƒ = 5. Then

GPA =
4 # 16 + 3 # 32 + 2 # 12 + 2

16 + 32 + 12 + 2 + 5
=

186

67
� 2.78.

The student’s GPA is 2.78.

Now Try Exercise 71

EXAMPLE 10 Converting temperature scales

In the United States, temperature is measured with either the Fahrenheit or the Celsius
temperature scale. To convert Fahrenheit degrees F to Celsius degrees C, the formula
C = 59 (F – 32) can be used.
(a) Solve the formula for F to find a formula that converts Celsius degrees to Fahrenheit

degrees.
(b) If the outside temperature is 20� C, find the equivalent Fahrenheit temperature.

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Beginning and Intermediate Algebra with Applications & Visualization, Third edition, by Gary K. Rockswold and Terry A. Krieger. Published by Addison Wesley.
Copyright © 2013 by Pearson Education, Inc.

1312.4 FORMULAS

Solution
(a) The reciprocal of 59 is

9
5, so multiply each side by

9
5.

C =
5

9
(F – 32) Given equation

C =
9
5
# 5

9
(F – 32) Multiply each side by 95.

5
C = F – 32 Multiplicative inverses

5
C + 32 = F Add 32 to each side.

The required formula is F = 95 C + 32.
(b) If C = 20� C, then F = 95 (20) + 32 = 36 + 32 = 68� F.

Now Try Exercise 75

2.4 Putting It All Together

In this section we discussed formulas and how to solve for a variable. The following table
summarizes some of these formulas.

CONCEPT FORMULA EXAMPLES

Area and Perimeter of a
Rectangle

A = lw and P = 2l + 2w,

where l is the length and w is the
width.

If l = 10 feet and w = 5 feet, then

A = 10 # 5 = 50 square feet

and

P = 2(10) + 2(5) = 30 feet.

5 ft

10 ft

Area of a Triangle A = 12 bh,

where b is the base and h is the height.

If b = 5 inches and h = 6 inches, then
the area is

A =
1

2
(5)(6) = 15 square inches.

5 in.

6 in.

Angle Measure in a
Triangle

x + y + � = 180�,

where x, y, and � are the angle
measures.

If x = 40� and y = 60�, then � = 80�
because

40� + 60� + 80� = 180�

40° 60°

continued on next page

CHAPTER 2 LINEAR EQUATIONS AND INEQUALITIES132

CONCEPT FORMULA EXAMPLES

Volume and Surface
Area of a Box

If a box has length l, width w, and
height h, then its volume is

V = lwh

and its surface area is

S = 2lw + 2wh + 2lh.

If a box has dimensions l = 4 feet,
w = 3 feet, and h = 2 feet, then

V = 4(3)(2) = 24 cubic feet

and

S = 2(4)(3) + 2(3)(2) + 2(4)(2)
= 52 square feet.

4 ft

3 ft

2 ft

Circumference and Area
of a Circle

If a circle has radius r, then its
circumference is

C = 2pr

and its area is

A = pr2.

If r = 6 inches,
C = 2p(6) = 12p � 37.7 inches and
A = p(6)2 = 36p � 113.1 square
inches.

6 in.

Area of a Trapezoid A = 12 (a + b)h,

where a and b are the bases and h is
the height.

If a = 4, b = 6, and h = 3, then the
area is

A =
1

2
(4 + 6)(3) = 15 square units.

4
6
3

Volume of a Cylinder V = pr2h,

where r is the radius and h is the
height.

If r = 5 inches and h = 20 inches, then
the volume is

V = p(52)(20) = 500p cubic inches.

20 in.

5 in.

Calculating Grade
Point Average (GPA)

GPA is calculated by

4a + 3b + 2c + d
a + b + c + d + f

where a, b, c, d, and ƒ represent the
number of credits earned with grades
of A, B, C, D, and F, respectively.

10 credits of A, 8 credits of B, 6 credits
of C, 12 credits of D, and 8 credits of F
results in a GPA of

4(10) + 3(8) + 2(6) + 12
10 + 8 + 6 + 12 + 8

= 2.0.

continued from previous page

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Beginning and Intermediate Algebra with Applications & Visualization, Third edition, by Gary K. Rockswold and Terry A. Krieger. Published by Addison Wesley.
Copyright © 2013 by Pearson Education, Inc.

CONCEPT FORMULA EXAMPLES

Converting Between
Fahrenheit and Celsius
Degrees

C =
5

9
(F – 32)

F =
9

5
C + 32

212� F

is equivalent to

C =
5

9
(212 – 32) = 100� C.

100� C is equivalent to

F =
9

5
(100) + 32 = 212� F.

1332.4 FORMULAS

2.4 Exercises

CONCEPTS AND VOCABULARY

1. A(n) _____ can be used to calculate one quantity by
using known values of other quantities.

2. The area A of a rectangle with length l and width w is
A = .

3. The area A of a triangle with base b and height h is
A = _____.

4. One degree equals of a revolution.

5. There are degrees in one revolution.

6. The sum of the measures of the angles in a triangle
equals degrees.

7. The volume V of a box with length l, width w, and
height h is V = .

8. The surface area S of a box with length l, width w,
and height h is S = _____.

9. The circumference C of a circle with radius r is
C = .

10. The area A of a circle with radius r is A = .

11. The volume V of a cylinder with radius r and height h
is V = _____.

15. 16.

3 in.

6 in.

1 mi

3 mi

17. 18.

4 cm

6 km

19. 20.

2 mm
6 mm
5 mm

4 ft
3 ft
2 ft

21. Find the area of a rectangle having a 7-inch width and
a 13-inch length.

22. Find the area of a triangle having a 9-centimeter base
and a 72-centimeter height.

23. Find the area of a rectangle having a 5-foot width and
a 7-yard length.

24. Find the area of a triangle having a 12-millimeter
base and a 6-millimeter height.

25. Find the circumference of a circle having an 8-inch
diameter.

26. Find the area of a circle having a 9-foot radius.

12. The area A of a trapezoid with height h and bases a
and b is A = _____.

FORMULAS FROM GEOMETRY

Exercises 13–20: Find the area of the region shown.

13. 14.

6 ft

3 ft

4 yd

2.5 yd

134 CHAPTER 2 LINEAR EQUATIONS AND INEQUALITIES

27. Area of a Lot Find the area of the lot shown, which
consists of a square and a triangle.

52 ft

73 ft

28. Area of a Lot Find the area of the lot shown, which
consists of a rectangle and two triangles.

20 yd

10 yd

60 yd 30 yd

Exercises 29 and 30: Angle Measure Find the measure
of the third angle in the triangle.

29. 30.

75°

40°

30°

25°

31. A triangle contains two angles having measures of
23� and 76�. Find the measure of the third angle.

32. The measures of the angles in an equilateral triangle
are equal. Find their measure.

33. The measures of the angles in a triangle are x, 2x, and
2x. Find the value of x.

34. The measures of the angles in a triangle are 3x, 4x,
and 11x. Find the value of x.

35. In a triangle the two smaller angles are equal in mea-
sure and are each one third of the measure of the
larger angle. Find the measure of each angle.

36. In a triangle the two larger angles differ by 10�. The
smaller angle is 50� less than the largest angle. Find
the measure of each angle.

37. The diameter of a circle is 12 inches. Find its circumfer-
ence and area.

38. The radius of a circle is 54 feet. Find its circumference
and area.

39. The circumference of a circle is 2p inches. Find its
radius and area.

40. The circumference of a circle is 13p feet. Find its
radius and area.

Exercises 41–44: A box with a top has length l, width w,
and height h. Find the volume and surface area of the box.

41. l = 22 inches, w = 12 inches, h = 10 inches

42. l = 5 feet, w = 3 feet, h = 6 feet

43. l = 23 yard, w =
2
3 foot, h =

3
2 feet

44. l = 1.2 meters, w = 0.8 meter, h = 0.6 meter

Exercises 45–48: Use the formula V = pr2h to find the
volume of a cylindrical container for the given r and h.
Leave your answer in terms of p.

45. r = 2 inches, h = 5 inches

46. r = 12 inch, h =
3
2 inches

47. r = 5 inches, h = 2 feet

48. r = 2.5 feet, h = 1.5 yards

49. Volume of a Can (Refer to Example 6.)
(a) Find the volume of a can with a radius of 34 inch

and a height of 2 12 inches.
(b) Find the number of fluid ounces in the can if one

cubic inch equals 0.554 fluid ounces.

50. Volume of a Barrel (Refer to Example 6.) A cylin-
drical barrel has a diameter of 1 34 feet and a height of
3 feet. Find the volume of the barrel.

SOLVING FOR A VARIABLE

Exercises 51–68: Solve the formula for the given variable.

51. 9x + 3y = 6 for y 52. -2x – 2y = 10 for y

53. 4x + 3y = 12 for y 54. 5x – 2y = 22 for y

55. A = lw for w 56. A = 12 bh for b

57. V = pr2h for h 58. V = 13 pr
2h for h

59. A = 12 (a + b)h for a 60. C = 2pr for r

61. V = lwh for w 62. P = 2l + 2w for w

63. s =
a + b + c

2
for b 64. t =

x – y
3

for x

65.
a

b
–

b
= 1 for b 66.

x
y

+
�
y

= 5 for �

67. ab = cd + ad for a

68. S = 2lw + 2lh + 2wh for w

69. Perimeter of a Rectangle If the width of a rectangle
is 5 inches and its perimeter is 40 inches, find the
length of the rectangle.

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Beginning and Intermediate Algebra with Applications & Visualization, Third edition, by Gary K. Rockswold and Terry A. Krieger. Published by Addison Wesley.
Copyright © 2013 by Pearson Education, Inc.

1352.4 FORMULAS

70. Perimeter of a Triangle Two sides of a triangle have
lengths of 5 feet and 7 feet. If the triangle’s perimeter
is 21 feet, what is the length of the third side?

OTHER FORMULAS AND APPLICATIONS

Exercises 71–74: (Refer to Example 9.) Let the variable a
represent the number of credits with a grade of A, b the
number of credits with a grade of B, and so on. Calculate
the corresponding grade point average (GPA). Round
your answer to the nearest hundredth.

71. a = 30, b = 45, c = 12, d = 4, ƒ = 4

72. a = 70, b = 35, c = 5, d = 0, ƒ = 0

73. a = 0, b = 60, c = 80, d = 10, ƒ = 6

74. a = 3, b = 5, c = 8, d = 0, ƒ = 22

Exercises 75–78: (Refer to Example 10.) Convert the Cel-
sius temperature to an equivalent Fahrenheit temperature.

75. 25� C 76. 100� C 77. -40� C 78. 0� C

Exercises 79–82: (Refer to Example 10.) Convert the Fahr-
enheit temperature to an equivalent Celsius temperature.

79. 23� F 80. 98.6� F 81. -4� F 82. -31� F

83. Gas Mileage The formula M = DG can be used to

calculate a car’s gas mileage M after it has traveled D

miles on G gallons of gasoline. Suppose that a truck
driver leaves a gas station with a full tank of gas and the
odometer showing 87,625 miles. At the next gas stop, it
takes 38 gallons to fill the tank and the odometer reads
88,043 miles. Find the gas mileage for the truck.

84. Gas Mileage (Refer to Exercise 83.) A car that gets
34 miles per gallon is driven 578 miles. How many
gallons of gasoline are used on this trip?

85. Lightning If there is an x-second delay between
seeing a flash of lightning and hearing the thunder, then
the lightning is D = x5 miles away. Suppose that the
time delay between a flash of lightning and the sound of
thunder is 12 seconds. How far away is the lightning?

86. Lightning (Refer to Exercise 85.) Doppler radar
shows an electrical storm 2.5 miles away. If you see
the lightning from this storm, how long will it be
before you hear the thunder?

WRITING ABOUT MATHEMATICS

87. A student solves the formula A = 12 bh for h and
obtains the formula h = 12 bA. Explain the error that
the student is making. What is the correct answer?

88. Give an example of a formula that you have used and
explain how you used it.

Checking Basic ConceptsSECTIONS2.3 and 2.4
1. Translate the sentence into an equation containing

the variable x. Then solve the resulting equation.
(a) The product of 3 and a number is 36.
(b) A number subtracted from 35 is 43.

2. When three consecutive integers are added, the
sum is -93. Find the three integers.

3. Convert 9.5% to a decimal.

4. Convert 54 to a percentage.

5. Violent Crime In 2009, New York City experi-
enced 46,357 violent crimes. This figure repre-
sented a 38.8% decrease from the number in 2000.
Find the number of violent crimes in 2000. (Source:
New York State Crime Update.)

6. Driving a Car How many hours does it take the
driver of a car to travel 390 miles at 60 miles per hour?

7. College Loans A student takes out two loans, one
at 6% and the other at 7%. The 6% loan is $2000
more than the 7% loan, and the total interest for
one year is $510. Find the amount of each loan.

8. Gas Mileage A car that gets 28 miles per gallon
is driven 504 miles. How many gallons of gaso-
line are used on this trip?

9. Height of a Triangle The area of a triangle hav-
ing a base of 6 inches is 36 square inches. Find the
height of the triangle.

10. Area and Circumference Find the area and cir-
cumference of the circle shown.

3 ft

11. Angles Find the value of x in the triangle shown.

x
3x

12. Solving a Formula Solve A = pr2 + prl for l.

136 CHAPTER 2 LINEAR EQUATIONS AND INEQUALITIES

2.5 Linear Inequalities
Solutions and Number Line Graphs ● The Addition Property of Inequalities ●
The Multiplication Property of Inequalities ● Applications

A LOOK INTO MATH N At an amusement park, a particular ride might be restricted to people at least 48 inches
tall. A child who is x inches tall may go on the ride if x Ú 48 but may not go on the ride if
x 6 48. A height of 48 inches represents the boundary between being allowed on the ride
and being denied access to the ride.

Solving linear inequalities is closely related to solving linear equations because equality
is the boundary between greater than and less than. In this section we discuss techniques
used to solve linear inequalities.

Solutions and Number Line Graphs
A linear inequality results whenever the equals sign in a linear equation is replaced with
any one of the symbols 6 , … , 7 , or Ú . Examples of linear equations include

x � 5, 2 x + 1 � 0, 1 – x � 6, and 5x + 1 � 3 – 2 x.

Therefore examples of linear inequalities include

x + 5, 2 x + 1 * 0, 1 – x # 6, and 5x + 1 ” 3 – 2 x.

Table 2.9 shows how each of the inequality symbols is read.
NEW VOCABULARY

n Linear inequality
n Solution
n Solution set
n Interval notation
n Set-builder notation

READING CHECK

• How many solutions do
inequalities have?

TABLE 2.9 Inequality Symbols

Symbol How the Symbol Is Read

7 greater than

6 less than

Ú greater than or equal to

… less than or equal to

A solution to an inequality is a value of the variable that makes the statement true. The
set of all solutions is called the solution set. Two inequalities are equivalent if they have
the same solution set. Inequalities often have infinitely many solutions. For example, the
solution set for the inequality x 7 5 includes all real numbers greater than 5.

A number line can be used to graph the solution set for an inequality. The graph of all
real numbers satisfying x 6 2 is shown in Figure 2.17(a), and the graph of all real num-
bers satisfying x … 2 is shown in Figure 2.17(b). A parenthesis “)” is used to show that
2 is not included in Figure 2.17(a), and a bracket “]” is used to show that 2 is included in
Figure 2.17(b).

Figure 2.17

–2 –1 0 1 2 3

(a)

x * 2

–2 –1 0 1 2 3
(b)

x ” 2 ISB
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1372.5 LINEAR INEQUALITIES

EXAMPLE 1 Graphing inequalities on a number line

Use a number line to graph the solution set to each inequality.
(a) x 7 0 (b) x Ú 0 (c) x … -1 (d) x 6 3

Solution
(a) First locate x = 0 (or the origin) on a number line. Numbers greater than 0 are located

to the right of the origin, so shade the number line to the right of the origin. Because
the inequality is x 7 0, the number 0 is not included, so place a parenthesis “(” at 0, as
shown in Figure 2.18(a).

(b) Figure 2.18(b) is similar to the graph in part (a) except that a bracket “[” is placed at the
origin because the inequality symbol is Ú and 0 is included in the solution set.

(c) First locate x = -1 on the number line. Numbers less than -1 are located to the left of
-1. Because -1 is included, a bracket “]” is placed at -1, as shown in Figure 2.18(c).

(d) Real numbers less than 3 are graphed in Figure 2.18(d).

Figure 2.18

–2 –1 0 1 2

(a) x > 0

–2 –1 0 1 2

(b) x ≥ 0

–2–3 –1 0 1

10 2 3 4

(d) x < 3

Now Try Exercises 11, 13, 15, 17

INTERVAL NOTATION (OPTIONAL) The solution sets graphed in Figure 2.18 can
also be represented in a convenient notation called interval notation. Rather than draw
the entire number line, we can use brackets or parentheses to indicate the interval of values
that represent the solution set. For example, the solution set shown in Figure 2.18(a) can be
represented by the interval (0, �), and the solution set shown in Figure 2.18(b) can be rep-
resented by the interval [0, �). The symbol � refers to infinity and is used to indicate that
the values increase without bound. Similarly, – � can be used when the values decrease
without bound. The solution sets shown in Figure 2.18(c) and (d) can be represented by
(- � , -1] and (- � , 3), respectively.

EXAMPLE 2 Writing solution sets in interval notation

Write the solution set to each inequality in interval notation.
(a) x 7 4 (b) y … -3 (c) � Ú -1

Solution
(a) Real numbers greater than 4 are represented by the interval (4, �).
(b) Real numbers less than or equal to -3 are represented by the interval (- � , -3].
(c) The solution set is represented by the interval [-1, �).

Now Try Exercises 25, 27, 29

CHECKING SOLUTIONS We can check possible solutions to an inequality in the same
way that we checked possible solutions to an equation. For example, to check whether 5 is a
solution to the equation 2 x + 3 = 13, we substitute 5 for x in the equation.

2(5) + 3 � 13 Replace x with 5.
13 = 13 A true statement

138 CHAPTER 2 LINEAR EQUATIONS AND INEQUALITIES

Thus 5 is a solution to this equation. Similarly, to check whether 7 is a solution to the
inequality 2 x + 3 7 13, we substitute 7 for x in the inequality.

2(7) + 3 7
?
13 Replace x with 7.

17 7 13 ✓ A true statement

Thus 7 is a solution to the inequality.

EXAMPLE 3 Checking possible solutions

Determine whether the given value of x is a solution to the inequality.
(a) 3x – 4 6 10, x = 6 (b) 4 – 2 x … 8, x = -2

Solution
(a) Substitute 6 for x and simplify.

3(6) – 4 6
?
10 Replace x with 6.

14 6 10 ✗ A false statement
Thus 6 is not a solution to the inequality.
(b) Substitute �2 for x and simplify.

4 – 2(�2) …
?
8 Replace x with -2.

8 … 8 ✓ A true statement

Thus -2 is a solution to the inequality.

Now Try Exercises 33, 35

Just as solving a linear equation means finding the value of the variable that makes the
equation true, solving a linear inequality means finding the values of the variable that make
the inequality true.

Making a table is an organized way of checking possible values of the variable to see
if there are values that make an inequality true. In the next example, we use a table to find
solutions to an equation and related inequalities.

EXAMPLE 4 Finding solutions to equations and inequalities

In Table 2.10 the expression 2 x – 6 has been evaluated for several values of x. Use the
table to determine any solutions to each equation or inequality.
(a) 2 x – 6 = 0 (b) 2 x – 6 7 0 (c) 2 x – 6 Ú 0 (d) 2 x – 6 6 0

TABLE 2.10

x 0 1 2 3 4 5 6

2 x – 6 -6 -4 -2 0 2 4 6

Solution
(a) From Table 2.10, 2 x – 6 equals 0 when x = 3.
(b) The values of x in the table that make the expression 2 x – 6 greater than 0 are 4, 5, and

6. These values are all greater than 3, which is the solution found in part (a). It follows
that 2 x – 6 7 0 when x 7 3.

(c) The values of x in the table that make the expression 2 x – 6 greater than or equal to 0
are 3, 4, 5, and 6. It follows that 2 x – 6 Ú 0 when x Ú 3.

(d) The expression 2 x – 6 is less than 0 when x 6 3.

Now Try Exercise 39

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Beginning and Intermediate Algebra with Applications & Visualization, Third edition, by Gary K. Rockswold and Terry A. Krieger. Published by Addison Wesley.
Copyright © 2013 by Pearson Education, Inc.

1392.5 LINEAR INEQUALITIES

The Addition Property of Inequalities
N REAL-WORLD CONNECTION Suppose that the speed limit on a country road is 55 miles per

hour, and this is 25 miles per hour faster than the speed limit in town. If x represents lawful
speeds in town, then x satisfies the inequality

x + 25 … 55.

To solve this inequality we add �25 to (or subtract 25 from) each side of the inequality.

x + 25 � (�25) … 55 � (�25) Add -25 to each side.
x … 30 Add the real numbers.

Thus drivers are obeying the speed limit in town when they travel at 30 miles per hour or
less. To solve this inequality the addition property of inequalities was used.

ADDITION PROPERTY OF INEQUALITIES

Let a, b, and c be expressions that represent real numbers. The inequalities

a 6 b and a + c 6 b + c

are equivalent. That is, the same number may be added to (or subtracted from) each
side of an inequality. Similar properties exist for 7 , … , and Ú .

To solve some inequalities, we apply the addition property of inequalities to obtain a
simpler, equivalent inequality.

EXAMPLE 5 Applying the addition property of inequalities

Solve each inequality. Then graph the solution set.
(a) x – 1 7 4 (b) 3 + 2 x … 5 + x

Solution
(a) Begin by adding 1 to each side of the inequality.

x – 1 7 4 Given inequality
x – 1 � 1 7 4 � 1 Add 1 to each side.
x 7 5 Add the real numbers.

The solution set is given by x 7 5 and is graphed as follows.

–2 0 2 4 6 8

(b) Begin by subtracting x from (or adding -x to) each side of the inequality.

3 + 2 x … 5 + x Given inequality
3 + 2 x � x … 5 + x � x Subtract x from each side.
3 + x … 5 Combine like terms.
3 + x � 3 … 5 � 3 Subtract 3 from each side.
x … 2 Subtract the real numbers.

The solution set is given by x … 2 and is graphed as follows.

–2 –1 0 1 2 3
Now Try Exercise 47
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Beginning and Intermediate Algebra with Applications & Visualization, Third edition, by Gary K. Rockswold and Terry A. Krieger. Published by Addison Wesley.
Copyright © 2013 by Pearson Education, Inc.

140 CHAPTER 2 LINEAR EQUATIONS AND INEQUALITIES

MAKING CONNECTIONS

The addition property of inequalities can be illustrated by an old-fashioned pan balance. In the
figure on the left, the weight on the right pan is heavier than the weight on the left because the
right pan rests lower than the left pan. If we add an equal amount of weight (blue in the figure)
to both pans, or if we subtract an equal amount of weight from both pans, the pan on the right
still weighs more than the pan on the left by the same amount that it previously did.

The Multiplication Property of Inequalities
The multiplication property of inequalities differs from the multiplication property of equality.
When we multiply each side of an inequality by the same nonzero number, we may need
to reverse the inequality symbol to make sure that the resulting inequality remains true.
Table 2.11 shows various results that occur when each side of a true inequality is multiplied
by the same nonzero number.

EXAMPLE 6 Applying the addition property of inequalities

Solve 5 + 12 x … 3 –
1
2 x. Then graph the solution set.

Solution
Begin by subtracting 5 from each side of the inequality.

5 +
1

2
x … 3 –

2
x Given inequality

5 +
1

2
x � 5 … 3 –

2
x � 5 Subtract 5 from each side.

2
x … –

2
x – 2 Subtract real numbers.

2
x +

1
2

x … –
1

2
x +
1
2

x – 2 Add 12 x to each side.

x … -2 Combine like terms.

The solution set is given by x … -2 and is graphed as follows.

–3

–2 –1 0 1 2

Now Try Exercise 53

STUDY TIP

If you miss something in class, section video lectures provide a short lecture for each section
in this text. These lectures, presented by actual math instructors, offer you the opportunity to
review topics that you may not have fully understood before. These videos are available on the
Video Resources on DVD*, and as streaming videos and video podcasts within MyMathLab.

*www.mypearsonstore.com

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Beginning and Intermediate Algebra with Applications & Visualization, Third edition, by Gary K. Rockswold and Terry A. Krieger. Published by Addison Wesley.
Copyright © 2013 by Pearson Education, Inc.

1412.5 LINEAR INEQUALITIES

Table 2.11 indicates that the inequality symbol must be reversed when each side of the
given inequality is multiplied by a negative number. This result is summarized below.

TABLE 2.11 Determining When the Inequality Symbol Should Be Reversed

True Multiply Each Resulting Is the Result Reverse the
Statement Side By Inequality True or False? Inequality Symbol

-3 6 5 4 -12 6
?

20 True Not needed

7 7 -1 -2 -14 +
?

2 False -14 * 2

-2 7 -5 3 -6 7
?
-15 True Not needed

4 6 9 -11 -44 *
?

-99 False -44 + -99

MULTIPLICATION PROPERTY OF INEQUALITIES

Let a, b, and c be expressions that represent real numbers with c � 0.

1. If c 7 0, then the inequalities a 6 b and ac 6 bc are equivalent. That is, each
side of an inequality may be multiplied (or divided) by the same positive number.

2. If c 6 0, then the inequalities a 6 b and ac 7 bc are equivalent. That is, each
side of an inequality may be multiplied (or divided) by the same negative number,
provided the inequality symbol is reversed.

Note that similar properties exist for … and Ú .

NOTE: Remember to reverse the inequality symbol when either multiplying or dividing by
a negative number.

READING CHECK

• When solving an inequality, when does it become necessary to reverse the inequality
symbol?

EXAMPLE 7 Applying the multiplication property of inequalities

Solve each inequality. Then graph the solution set.
(a) 3x 6 18 (b) -7 … – 12 x

Solution
(a) To solve for x, divide each side by 3.

3x 6 18 Given inequality

3
6

18
3
Divide each side by 3.

x 6 6 Simplify fractions.

The solution set is given by x 6 6 and is graphed as follows.

–2 0 2 4 6 8
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Beginning and Intermediate Algebra with Applications & Visualization, Third edition, by Gary K. Rockswold and Terry A. Krieger. Published by Addison Wesley.
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142 CHAPTER 2 LINEAR EQUATIONS AND INEQUALITIES

(b) To isolate x in -7 … – 12 x, multiply each side by -2 and reverse the

inequality symbol.

-7 ” –
1

2
x Given inequality

�2(-7) # �2a – 1
2
b x Multiply by -2; reverse the inequality.

14 Ú 1 # x Multiply the real numbers.
x … 14 Rewrite the inequality.

The solution set is given by x … 14 and is graphed as follows.

–10 –5 0 5 10 15

Now Try Exercises 55, 57

SET-BUILDER NOTATION Because x … 14 is an inequality with infinitely many
solutions and is not itself a set of solutions, a notation called set-builder notation
has been devised for writing the solutions to an inequality as a set. For example, the
solution set consisting of “all real numbers x such that x is less than or equal to 14” can be
written as {x 0 x … 14}. The vertical line segment “ 0 ” is read “such that.”

In the next example, the solution sets are expressed in set-builder notation. However,
this notation is not widely used throughout this text.

EXAMPLE 8 Applying both properties of inequalities

Solve each inequality. Write the solution set in set-builder notation.
(a) 4x – 7 Ú -6 (b) -8 + 4x … 5x + 3 (c) 0.4(2 x – 5) 6 1.1x + 2

Solution
(a) Start by adding 7 to each side.

4x – 7 Ú -6 Given inequality
4x – 7 � 7 Ú – 6 � 7 Add 7 to each side.
4x Ú 1 Add real numbers.

4
Ú

4
Divide each side by 4.

x Ú
1

4
Simplify.

In set-builder notation, the solution set is 5x 0 x Ú 146 .
(b) Begin by adding 8 to each side.

-8 + 4x … 5x + 3 Given inequality
-8 + 4x � 8 … 5x + 3 � 8 Add 8 to each side.
4x … 5x + 11 Add real numbers.
4x � 5x … 5x + 11 � 5x Subtract 5x from each side.
-x ” 11 Combine like terms.
�1 # (-x) # �1 # 11 Multiply by -1; reverse the inequality.
x Ú -11 Simplify.

The solution set is {x 0 x Ú -11}.

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Beginning and Intermediate Algebra with Applications & Visualization, Third edition, by Gary K. Rockswold and Terry A. Krieger. Published by Addison Wesley.
Copyright © 2013 by Pearson Education, Inc.

1432.5 LINEAR INEQUALITIES

0.4(2 x – 5) 6 1.1x + 2 Given inequality
0.8x – 2 6 1.1x + 2 Distributive property
0.8x 6 1.1x + 4 Add 2 to each side.
-0.3x * 4 Subtract 1.1x from each side.

x + –
4

0.3
Divide by -0.3; reverse the inequality.

Since – 40.3 = –
40
3 , the solution set is 5x 0 x 7 – 403 6 .

Now Try Exercises 67, 77, 85

CRITICAL THINKING

Solve
-5 – 3x 7 -2x + 7 with-
out having to reverse the
inequality symbol.

READING CHECK

• Give two phrases associ-
ated with each inequality
symbol.

Applications
To solve applications involving inequalities, we often have to translate words or phrases to
mathematical statements. Table 2.12 lists words and phrases that are associated with each
inequality symbol.

TABLE 2.12 Words and Phrases Associated with Inequality Symbols

Symbol Associated Words and Phrases

7 greater than, more than, exceeds, above, over

6 less than, fewer than, below, under

Ú greater than or equal to, at least, is not less than

… less than or equal to, at most, does not exceed

EXAMPLE 9 Translating words to inequalities

Translate each phrase to an inequality. Let the variable be x.
(a) A number that is more than 30
(b) An age that is at least 18
(c) A grade point average that is at most 3.25

Solution
(a) The inequality x 7 30 represents a number x that is more than 30.
(b) The inequality x Ú 18 represents an age x that is at least 18.
(c) The inequality x … 3.25 represents a grade point average x that is at most 3.25.

Now Try Exercises 95, 101

N REAL-WORLD CONNECTION In the atmosphere the air temperature generally becomes colder
as the altitude increases. One mile above Earth’s surface the temperature is about 19� F
colder than the ground-level temperature. As the air cools, there is an increased chance
of clouds forming. In the next example we estimate the altitudes where clouds may form.
(Source: A. Miller and R. Anthes, Meteorology.)

EXAMPLE 10 Finding the altitude of clouds

If the ground temperature is 79� F, then the temperature T above Earth’s surface is given by
the formula T = 79 – 19x, where x is the altitude in miles. Suppose that clouds form only
where the temperature is 3� F or colder. Determine the heights at which clouds may form.

144 CHAPTER 2 LINEAR EQUATIONS AND INEQUALITIES

Solution
Clouds may form at altitudes at which T … 3 degrees. Since T = 79 – 19x, we can substi-
tute the expression 79 – 19x for T to obtain the inequality 79 – 19x … 3.

79 – 19x … 3 Inequality to be solved
-19x ” -76 Subtract 79 from each side.

-19x
-19

#
-76
-19

Divide by -19; reverse the inequality.

x Ú 4 Simplify the fractions.

Clouds may form at 4 miles or higher.

Now Try Exercise 115

EXAMPLE 11 Calculating revenue, cost, and profit

For a computer company, the cost to produce one laptop computer (variable cost) is $1320
plus a one-time cost (fixed cost) of $200,000 for research and development. The revenue
received from selling one laptop computer is $1850.
(a) Write a formula that gives the cost C of producing x laptop computers.
(b) Write a formula that gives the revenue R from selling x laptop computers.
(c) Profit equals revenue minus cost. Write a formula that calculates the profit P from sell-

ing x laptop computers.
(d) How many computers need to be sold to yield a positive profit?

Solution
(a) The cost of producing the first laptop is

1320 * 1 + 200,000 = $201,320.

The cost of producing two laptops is

1320 * 2 + 200,000 = $202,640.

And, in general, the cost of producing x laptops is

1320 * x + 200,000 = 1320x + 200,000.

Thus C = 1320x + 200,000.
(b) Because the company receives $1850 for each laptop, the revenue for x laptops is given

by R = 1850x.
(c) Profit equals revenue minus cost, so

P = R – C
= 1850x – (1320x � 200,000)
= 530x – 200,000.

Thus P = 530x – 200,000.
(d) To determine how many laptops need to be sold to yield a positive profit, we must

solve the inequality P 7 0.

530x – 200,000 7 0 Inequality to be solved
530x 7 200,000 Add 200,000 to each side.

x 7
200,000

530
Divide each side by 530.

Because 200,000530 � 377.4, the company must sell at least 378 laptops. Note that the
company cannot sell a fraction of a laptop.

Now Try Exercise 111

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Beginning and Intermediate Algebra with Applications & Visualization, Third edition, by Gary K. Rockswold and Terry A. Krieger. Published by Addison Wesley.
Copyright © 2013 by Pearson Education, Inc.

1452.5 LINEAR INEQUALITIES

2.5 Putting It All Together

Linear Inequality If the equals sign in a linear equation
is replaced with 6 , 7 , … , or Ú , a
linear inequality results.

Linear Equation Linear Inequality

4x – 1 = 0 4x – 1 7 0
2 – x = 3x 2 – x … 3x

4(x + 3) = 1 – x 4(x + 3) 6 1 – x
-6 x + 3 = 5 -6 x + 3 Ú 5

Solution to an Inequality A value for the variable that makes
the inequality a true statement

6 is a solution to 2 x 7 5 because
2(6) 7 5 is a true statement.

Set-Builder Notation A notation that can be used to identify
the solution set to an inequality

The solution set for x – 2 6 5 can be
written as {x 0 x 6 7} and is read “the
set of real numbers x such that x is less
than 7.”

CONCEPT COMMENTS EXAMPLES

Solution Set to an
Inequality

The set of all solutions to an inequality The solution set to x + 1 7 5 is given
by x 7 4 and can be written in set-
builder notation as {x 0 x 7 4}.

Number Line Graphs The solutions to an inequality can be
graphed on a number line.

x 6 2 is graphed as follows.

–2 –1 0 1 2 3

x Ú -1 is graphed as follows.

–2 –1 0 1 2

Addition Property of
Inequalities

a 6 b

is equivalent to

a + c 6 b + c,

where a, b, and c represent real num-
ber expressions.

x – 5 Ú 6 Given inequality
x Ú 11 Add 5.

3x 7 5 + 2 x Given inequality
x 7 5 Subtract 2x.

Multiplication Property
of Inequalities

a 6 b
is equivalent to

ac 6 bc

when c 7 0, and is equivalent to

ac 7 bc

when c 6 0.

2
x Ú 6 Given inequality

x Ú 12 Multiply by 2.

-3x 7 5 Given inequality

x 6 –
5

3
Divide by -3; reverse the

inequality symbol.
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Beginning and Intermediate Algebra with Applications & Visualization, Third edition, by Gary K. Rockswold and Terry A. Krieger. Published by Addison Wesley.
Copyright © 2013 by Pearson Education, Inc.

146 CHAPTER 2 LINEAR EQUATIONS AND INEQUALITIES

2.5 Exercises

CONCEPTS AND VOCABULARY

1. A linear inequality results whenever the equals sign
in a linear equation is replaced by any one of the sym-
bols , , , or .

2. Equality is the boundary between _____ and _____.

3. A(n) _____ is a value of the variable that makes an
inequality statement true.

4. Two linear inequalities are _____ if they have the
same solution set.

5. (True or False?) When a linear equation is solved,
the solution set contains one solution.

6. (True or False?) When a linear inequality is solved,
the solution set contains infinitely many solutions.

7. The solution set to a linear inequality can be graphed
by using a(n) _____.

8. The addition property of inequalities states that if
a 7 b, then a + c b + c.

9. The multiplication property of inequalities states that
if a 6 b and c 7 0, then ac bc.

10. The multiplication property of inequalities states that
if a 6 b and c 6 0, then ac bc.

SOLUTIONS AND NUMBER LINE GRAPHS

Exercises 11–18: Use a number line to graph the solution
set to the inequality.

11. x 6 0 12. x 7 -2

13. x 7 1 14. x 6 – 52
15. x … 1.5 16. x Ú -3

17. � Ú -2 18. � … -p

Exercises 19–24: Express the set of real numbers graphed
on the number line as an inequality.

19. 20.

–2 –1 0 1 2

–2–3 –1 0 1

21. 22.

10 2 3 4

–15 –10 –5 0 5 10 15

23. 24.

–10–20 0 10 20

–50 –40 –30 –20 –10 0 10

Exercises 25–30: Write the solution set to the inequality
in interval notation.

25. x Ú 6 26. x 6 3

27. y 7 -2 28. y Ú 1

29. � … 7 30. � 6 -5

Exercises 31–38: Determine whether the given value of
the variable is a solution to the inequality.

31. x + 5 7 5 x = 4

32. -3x … -8 x = -2

33. 5x Ú 25 x = 3

34. 4y – 3 … 5 y = -3

35. 3y + 5 Ú -8 y = -3

36. – (� + 7) 7 3(6 – �) � = 2

37. 5(� + 1) 6 3� – 7 � = -4

38. 32 t –
1
2 Ú 1 – t t =

3
5

TABLES AND LINEAR INEQUALITIES

Exercises 39–42: Use the table to solve the inequality.

39. 3x + 6 7 0

40. 6 – 3x … 0

41. -2 x + 7 7 5

42. 5(x – 3) … 4

x -4 -3 -2 -1 0

3x + 6 -6 -3 0 3 6

x 1 2 3 4 5

6 – 3x 3 0 -3 -6 -9

x -1 0 1 2 3

-2 x + 7 9 7 5 3 1

x 3.2 3.4 3.6 3.8 4

5(x – 3) 1 2 3 4 5

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Beginning and Intermediate Algebra with Applications & Visualization, Third edition, by Gary K. Rockswold and Terry A. Krieger. Published by Addison Wesley.
Copyright © 2013 by Pearson Education, Inc.

1472.5 LINEAR INEQUALITIES

Exercises 43–46: Complete the table. Then use the table
to solve the inequality.

43. -2x + 6 … 0

44. 3x – 1 6 8

45. 5 – x 7 x + 7

46. 2(3 – x) Ú -3(x – 2)

SOLVING LINEAR INEQUALITIES

Exercises 47–54: Use the addition property of inequali-
ties to solve the inequality. Graph the solution set.

47. x – 3 7 0 48. x + 6 6 3

49. 3 – y … 5 50. 8 – y Ú 10

51. 12 6 4 + � 52. 2� … � + 17

53. 5 – 2t Ú 10 – t 54. -2t 7 -3t + 1

Exercises 55–62: Use the multiplication property of inequali-
ties to solve the inequality. Graph the solution set.

55. 2 x 6 10 56. 3x 7 9

57. – 12 t Ú 1 58. -5t … -6

59. 34 7 -5y 60. 10 Ú –
1
7 y

61. – 23 …
1
7 � 62. –

3
10 � 6 11

Exercises 63–68: Solve the linear inequality and write the
solution in set-builder notation.

63. x + 6 7 7 64. x + 4 6 1

65. -3x … 21 66. 4x Ú -20

67. 2x – 3 6 9 68. -5x + 4 6 44

x -2 -1 0 1 2

2(3 – x)

-3(x – 2)

x 1 2 3 4 5

-2 x + 6 4 -4

x 0 1 2 3 4

3x – 1 -1

x -3 -2 -1 0 1

5 – x 8 4

x + 7 4 8

Exercises 69–94: Solve the linear inequality.

69. 3x + 1 6 22 70. 4 + 5x … 9

71. 5 – 34 x Ú 6 72. 10 –
2
5 x 7 0

73. 45 7 6 – 2 x 74. 69 Ú 3 – 11x

75. 5x – 2 … 3x + 1 76. 12 x + 1 6 25 – 3x

77. -x + 24 6 x + 23 78. 6 – 4x … x + 1

79. – (x + 1) Ú 3(x – 2)

80. 5(x + 2) 7 -2(x – 3)

81. 3(2 x + 1) 7 – (5 – 3x)

82. 4x Ú -3(7 – 2 x) + 1

83. 1.6 x + 0.4 … 0.4x

84. -5.1x + 1.1 6 0.1 – 0.1x

85. 0.8x – 0.5 6 x + 1 – 0.5x

86. 0.1(x + 1) – 0.1 … 0.2 x – 0.5

87. – 12123 x + 42 Ú x
88. -5x 7 451103 x + 102
89. 37 x +

2
7 7 –

1
7 x –

5
14

90. 56 –
1
3 x Ú –

1
3156 x – 12

91.
x

3
+

6
…

3
92.

4
–

2
6 1

93.
6 x

7
6

3
x + 1 94.

8
–

4
… 8

TRANSLATING PHRASES TO INEQUALITIES

Exercises 95–102: Translate each phrase to an inequal-
ity. Let x be the variable.

95. A speed that is greater than 60 miles per hour

96. A speed that is at most 60 miles per hour

97. An age that is at least 21 years old

98. An age that is less than 21 years old

99. A salary that is more than $40,000

100. A salary that is less than or equal to $40,000

101. A speed that does not exceed 70 miles per hour

102. A speed that is not less than 70 miles per hour

148 CHAPTER 2 LINEAR EQUATIONS AND INEQUALITIES

APPLICATIONS

103. Geometry Find all values for x so that the perimeter
of the rectangle is less than 50 feet.

x
x + 5

104. Geometry A triangle with height 12 inches is to
have area less than 120 square inches. What must be
true about the base of the triangle?

105. Grade Average A student scores 74 out of 100
on a test. If the maximum score on the next test is
also 100 points, what score does the student need to
maintain at least an average of 80?

106. Grade Average A student scores 65 and 82 on two
different 100-point tests. If the maximum score on
the next test is also 100 points, what score does the
student need to maintain at least an average of 70?

107. Parking Rates Parking in a student lot costs $2 for

the first half hour and $1.25 for each hour thereafter.
A partial hour is charged the same as a full hour.
What is the longest time that a student can park in
this lot for $8?

108. Parking Rates Parking in a student lot costs $2.50
for the first hour and $1 for each hour thereafter. A
nearby lot costs $1.25 for each hour. In both lots a
partial hour is charged as a full hour. In which lot
can a student park the longest for $5? For $11?

109. Car Rental A rental car costs $25 per day plus
$0.20 per mile. If someone has $200 to spend and
needs to drive the car 90 miles each day, for how
many days can that person rent the car? Assume that
the car cannot be rented for part of a day.

110. Car Rental One car rental agency charges $20 per
day plus $0.25 per mile. A different agency charges
$37 per day with unlimited mileage. For what mile-
ages is the second rental agency a better deal?

111. Revenue and Cost (Refer to Example 11.) The
cost to produce one compact disc is $1.50 plus a
one-time fixed cost of $2000. The revenue received
from selling one compact disc is $12.

(a) Write a formula that gives the cost C of produc-
ing x compact discs. Be sure to include the fixed
cost.

(b) Write a formula that gives the revenue R from
selling x compact discs.

(d) How many compact discs need to be sold to
yield a positive profit?

112. Revenue and Cost The cost to produce one lap-
top computer is $890 plus a one-time fixed cost of
$100,000 for research and development. The revenue
received from selling one laptop computer is $1520.

(a) Write a formula that gives the cost C of produc-
ing x laptop computers.

(b) Write a formula that gives the revenue R from
selling x laptop computers.

(d) How many computers need to be sold to yield a
positive profit?

113. Distance and Time Two athletes are jogging in
the same direction along an exercise path. After x
minutes the first athlete’s distance in miles from a
parking lot is given by 16 x and the second athlete’s
distance is given by 18 x + 2.

(a) When are the athletes the same distance from
the parking lot?

(b) When is the first athlete farther from the park-
ing lot than the second?

114. Altitude and Dew Point If the dew point on the
ground is 65� F, then the dew point x miles high is
given by D = 65 – 5.8 x. Determine the altitudes
at which the dew point is greater than 36� F. (Source:
A. Miller.)

115. Altitude and Temperature (Refer to Example 10.)
If the temperature on the ground is 90°F, then the air
temperature x miles high is given by T = 90 – 19x.
Determine the altitudes at which the air temperature is
less than 4.5°F. (Source: A. Miller.)

116. Size and Weight of a Fish If the length of a bass is
between 20 and 25 inches, its weight W in pounds can
be estimated by the formula W = 0.96 x – 14.4,
where x is the length of the fish. (Source: Minnesota Depart-
ment of Natural Resources.)

(a) What length of bass is likely to weigh 7.2
pounds?

(b) What lengths of bass are likely to weigh less
than 7.2 pounds?

WRITING ABOUT MATHEMATICS

117. Explain each of the terms and give an example.
(a) Linear equation (b) Linear inequality

118. Suppose that a student says that a linear equation
and a linear inequality can be solved the same way.
How would you respond?

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Beginning and Intermediate Algebra with Applications & Visualization, Third edition, by Gary K. Rockswold and Terry A. Krieger. Published by Addison Wesley.
Copyright © 2013 by Pearson Education, Inc.

149CHAPTER 2 SUMMARY

Checking Basic ConceptsSECTION2.5
1. Use a number line to graph the solution set to the

inequality x + 1 Ú -1.

2. Express the set of real numbers graphed on the
number line by using an inequality.

–2 –1 0 1 2

3. Determine whether -3 is a solution to the in-
equality 4x – 5 … -15.

4. Complete the table. Then use the table to solve the
inequality 5 – 2 x … 7.

5. Solve each inequality.
(a) x + 5 7 8
(b) – 57 x … 25
(c) 3x Ú -2(1 – 2 x) + 3

6. Translate the phrase “a price that is not more than
$12” to an inequality using the variable x.

7. Geometry The length of a rectangle is 5 inches
longer than twice its width. If the perimeter of the
rectangle is more than 88 inches, find all possible
widths for the rectangle.

x -2 -1 0 1 2

5 – 2 x 1

CHAPTER 2 Summary
SECTION 2.1 . INTRODUCTION TO EQUATIONS

Equations An equation is a mathematical statement that two expressions are equal.
Every equation contains an equals sign. An equation can be either true or false.

Important Terms
Solution A value for a variable that makes the equation true

Solution Set The set of all solutions

Equivalent Equations Equations that have the same solution set

Checking a Solution Substituting the solution in the given equation to verify that the solution makes
the equation true

Example: 3 is the solution to 4x – 2 = 10 because 4(3) – 2 = 10 is a true
statement.

Properties of Equality
Addition Property a = b is equivalent to a + c = b + c.

Example: x – 3 = 0 and x – 3 + 3 = 0 + 3 are equivalent equations.

Multiplication Property a = b is equivalent to ac = bc, provided c � 0.

Example: 2 x = 5 and 2 x # 12 = 5 # 12 are equivalent equations.

SECTION 2.2 . LINEAR EQUATIONS

Linear Equation Can be written in the form ax + b = 0, where a � 0

Examples: 3x – 5 = 0 is linear, whereas 5×2 + 2 x = 0 is not linear.

150 CHAPTER 2 LINEAR EQUATIONS AND INEQUALITIES

Solving Linear Equations Numerically To solve a linear equation numerically, complete a
table for various values of the variable and then select the solution from the table.

Example: The solution to 3x – 4 = -1 is 1.

Solving Linear Equations Symbolically The following steps can be used as a guide for solving
linear equations symbolically.

STEP 1: Use the distributive property to clear any parentheses on each side of the equation. Com-
bine any like terms on each side.

STEP 2: Use the addition property of equality to get all of the terms containing the variable on
one side of the equation and all other terms on the other side of the equation. Combine
any like terms on each side.

STEP 3: Use the multiplication property of equality to isolate the variable by multiplying each
side of the equation by the reciprocal of the number in front of the variable (or divide
each side by that number).

STEP 4: Check the solution by substituting it in the given equation.

Distributive Properties a(b + c) = ab + ac or a(b – c) = ab – ac

Examples: 5(2 x + 3) = 10x + 15 and 5(2 x – 3) = 10x – 15

Clearing Fractions and Decimals When fractions or decimals appear in an equation,
multiplying each side by the least common denominator can be helpful.

Examples: Multiply each side of 13 x –
1
6 =

2
3 by 6 to obtain 2 x – 1 = 4.

Multiply each side of 0.04x + 0.1 = 0.07 by 100 to obtain 4x + 10 = 7.

Number of Solutions Equations that can be written in the form ax + b = 0, where a and b are
any real number, can have no solutions, one solution, or infinitely many solutions.

Examples: x + 3 = x is equivalent to 3 = 0. (No solutions)

2y + 1 = 9 is equivalent to y = 4. (One solution)

� + � = 2� is equivalent to 0 = 0. (Infinitely many solutions)

SECTION 2.3 . INTRODUCTION TO PROBLEM SOLVING

Steps for Solving a Problem The following steps can be used as a guide for solving word
problems.

STEP 1: Read the problem carefully and be sure that you understand it. (You may need to read the
problem more than once.) Assign a variable to what you are being asked to find. If neces-
sary, write other quantities in terms of this variable.

STEP 2: Write an equation that relates the quantities described in the problem. You may need to
sketch a diagram or refer to known formulas.

STEP 3: Solve the equation. Use the solution to determine the solution(s) to the original problem.
Include any necessary units.

STEP 4: Check your solution in the original problem. Does it seem reasonable?

x -1 0 1

3x – 4 -7 -4 -1

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Beginning and Intermediate Algebra with Applications & Visualization, Third edition, by Gary K. Rockswold and Terry A. Krieger. Published by Addison Wesley.
Copyright © 2013 by Pearson Education, Inc.

151CHAPTER 2 SUMMARY

Percent Problems
The expression x% Represents the fraction x100 or the decimal number given by x # 0.01.
Examples: 45% =

100
= 0.45

7.1% =
7.1

100
=

1000
= 0.071

Changing Fractions to Percents Divide the numerator by the denominator, multiply by 100, and insert the % symbol.

Example: Since 3 , 4 = 0.75, the fraction 34 is written in percent notation as
0.75 # 100 = 75%.

Percent Change If a quantity changes from an old amount to a new amount, then the percent

change equals
(new amount ) – (old amount)

(old amount)
# 100.

Example: If a price increases from $2 to $3, then the percent change equals
3 – 22 # 100 = 50%.
Distance Problems If an object travels at speed (rate) r for time t, then the distance d traveled
is calculated by d = rt.

Example: A car moving at 65 mph for 2 hours travels

d = rt = 65 # 2 = 130 miles.

SECTION 2.4 . FORMULAS

Formula A formula is an equation that can be used to calculate a quantity by using known values
of other quantities.

Formulas from Geometry
Area of a Rectangle A = lw, where l is the length and w is the width.

Perimeter of a Rectangle P = 2l + 2w, where l is the length and w is the width.

Area of a Triangle A = 12 bh, where b is the base and h is the height.

Degree Measure There are 360� in one complete revolution.

Angle Measure The sum of the measures of the angles in a triangle equals 180�.

Circumference C = 2pr, where r is the radius.

Area of a Circle A = pr2, where r is the radius.

Area of a Trapezoid A = 12 (a + b)h, where h is the height and a and b are the bases of the trapezoid.

Volume of a Box V = lwh, where l is the length, w is the width, and h is the height.

Surface Area of a Box S = 2lw + 2wh + 2lh, where l is the length, w is the width, and h is the height.

Volume of a Cylinder V = pr2h, where r is the radius and h is the height.

Other Formulas
Grade Point Average (GPA) GPA = 4a + 3b + 2c + da + b + c + d + ƒ , where a represents the number of A credits

earned, b the number of B credits earned, and so on.

Temperature Scales F = 95 C + 32 and C =
5
9 (F – 32), where F is the Fahrenheit temperature

and C is the Celsius temperature.

See Putting It All Together in Section 2.4 for examples.

152 CHAPTER 2 LINEAR EQUATIONS AND INEQUALITIES

SECTION 2.5 . LINEAR INEQUALITIES

Linear Inequality When the equals sign in a linear equation is replaced with any one of the sym-
bols 6 , … , 7 , or Ú , a linear inequality results.

Examples: x 7 0, 6 – 23 x … 7, and 4(x – 1) 6 3x – 1

Solution to an Inequality Any value for the variable that makes the inequality a true statement

Example: 3 is a solution to 2 x 6 9 because 2(3) 6 9 is a true statement.

Number Line Graphs A number line can be used to graph the solution set to a linear inequality.

Example: The graph of x … 1 is shown in the figure.

–2 –1 0 1 2

Properties of Inequality
Addition Property a 6 b is equivalent to a + c 6 b + c.

Example: x – 3 6 0 and x – 3 + 3 6 0 + 3 are equivalent inequalities.

Multiplication Property When c 7 0, a 6 b is equivalent to ac 6 bc.
When c 6 0, a 6 b is equivalent to ac 7 bc.

Examples: 2 x 6 6 is equivalent to 2 x1122 6 61122 or x 6 3.
-2 x 6 6 is equivalent to -2 x1 – 122 7 61 – 122 or x 7 -3.

CHAPTER 2 Review Exercises
SECTION 2.1

Exercises 1–8: Solve the equation. Check your solution.

1. x + 9 = 3 2. x – 4 = -2

3. x – 34 =
3
2 4. x + 0.5 = 0

5. 4x = 12 6. 3x = -7

7. -0.5x = 1.25 8. – 13 x =
7
6

SECTION 2.2

Exercises 9 and 10: Decide whether the equation is
linear. If the equation is linear, give values for a and b so
that it can be written in the form ax + b = 0.
9. -4x + 3 = 2 10. 38 x

2 – x = 14

Exercises 11–20: Solve the equation. Check the solution.

11. 4x – 5 = 3 12. 7 – 12 x = -4

13. 5(x – 3) = 12 14. 3 + x = 2 x – 4

15. 2(x – 1) = 4(x + 3)

16. 1 – (x – 3) = 6 + 2 x

17. 3.4x – 4 = 5 – 0.6 x

18. – 13 (3 – 6 x) = – (x + 2) + 1

19. 23 x –
1
6 =

5
12

20. 2y – 3(2 – y) = 5 + y

Exercises 21–24: Determine whether the equation has no
solutions, one solution, or infinitely many solutions.

21. 4(3x – 2) = 2(6 x + 5)

22. 5(3x – 1) = 15x – 5

23. 8x = 5x + 3x

24. 9x – 2 = 8x – 2

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Beginning and Intermediate Algebra with Applications & Visualization, Third edition, by Gary K. Rockswold and Terry A. Krieger. Published by Addison Wesley.
Copyright © 2013 by Pearson Education, Inc.

153CHAPTER 2 REVIEW EXERCISES

Exercises 25 and 26: Complete the table. Then use the
table to solve the given equation.

25. -2 x + 3 = 0

26. – (x + 1) + 3 = 2

SECTION 2.3

Exercises 27–30: Using the variable x, translate the sen-
tence into an equation. Solve the resulting equation.

27. The product of a number and 6 is 72.

28. The sum of a number and 18 is -23.

29. Twice a number minus 5 equals the number plus 4.

30. The sum of a number and 4 equals the product of the
number and 3.

Exercises 31 and 32: Find the number or numbers.

31. Five times a number divided by 3 equals 15.

32. The sum of three consecutive integers is -153.

Exercises 33–36: Convert the percentage to fraction and
decimal notation.

33. 85% 34. 5.6%

35. 0.03% 36. 342%

Exercises 37–40: Convert the number to a percentage.

37. 0.89 38. 0.005

39. 2.3 40. 1

Exercises 41–44: Use the formula d = rt to find the
value of the missing variable.

41. r = 8 miles per hour, t = 3 hours

42. r = 70 feet per second, t = 55 seconds

43. d = 500 yards, t = 20 seconds

44. d = 125 miles, r = 15 miles per hour

SECTION 2.4

Exercises 45 and 46: Find the area of the region shown.

45. 46.

3 m

5 m

6 ft

47. Find the area of a rectangle having a 24-inch width
and a 3-foot length.

48. Find the perimeter of a rectangle having a 13-inch
length and a 7-inch width.

49. Find the circumference of a circle having an 18-foot
diameter.

50. Find the area of a circle having a 5-inch radius.

51. Find the measure of the third angle in the triangle.

40°

90°

52. The angles in a triangle have measures x, 3x, and 4x.
Find the value of x.

53. If a cylinder has radius 5 inches and height 25 inches,
find its volume. (Hint: V = pr2h.)

54. Find the area of a trapezoid with height 5 feet and
bases 3 feet and 18 inches. (Hint: A = 12 (a + b)h.)

Exercises 55 and 56: Find the area of the figure shown.

55.

25 in.

6 in.
8 in.

56.

12 ft

4 ft

Exercises 57–62: Solve the given formula for the specified
variable.

57. 3x = 5 + y for y 58. 16 = 2 x + 2y for y

59. � = 2 xy for y 60. S =
a + b + c

3
for b

61. T =
a

3
+
b

4
for b 62. cd = ab + bc for c

x 0.5 1.0 1.5 2.0 2.5

-2 x + 3 2

x -2 -1 0 1 2

– (x + 1) + 3

154 CHAPTER 2 LINEAR EQUATIONS AND INEQUALITIES

Exercises 63 and 64: Let the variable a represent the num-
ber of credits with a grade of A, b the number of credits with
a grade of B, and so on. Calculate the grade point average
(GPA). Round your answer to the nearest hundredth.

63. a = 20, b = 25, c = 12, d = 4, ƒ = 4

64. a = 64, b = 32, c = 20, d = 10, ƒ = 3

65. Convert 15�C to an equivalent Fahrenheit temperature.

66. Convert 113�F to an equivalent Celsius temperature.

SECTION 2.5

Exercises 67–70: Use a number line to graph the solution
set to the inequality.

67. x 6 2 68. x 7 -1

69. y Ú – 32 70. y … 2.5

Exercises 71 and 72: Express the set of real numbers
graphed on the number line with an inequality.

71. 72.

10 2 3 4

–2 –1 0 1 2

Exercises 73 and 74: Determine whether the given value
of x is a solution to the inequality.

73. 1 – (x + 3) Ú x x = -2

74. 4(x + 1) 6 – (5 – x) x = -1

Exercises 75 and 76: Complete the table and then use the
table to solve the inequality.

75. 5 – x 7 3

76. 2 x – 5 … 0

Exercises 77–82: Solve the inequality.

77. x – 3 7 0 78. -2 x … 10

79. 5 – 2 x Ú 7

80. 3(x – 1) 6 20

81. 5x … 3 – (4x + 2)

82. 3x – 2(4 – x) Ú x + 1

x 0 1 2 3 4

5 – x 5

x 1 1.5 2 2.5 3

2 x – 5 -3

Exercises 83–86: Translate the phrase to an inequality.
Let x be the variable.

83. A speed that is less than 50 miles per hour

84. A salary that is at most $45,000

85. An age that is at least 16 years old

86. A year before 1995

APPLICATIONS

87. Rainfall On a stormy day 2 inches of rain fall before
noon and 34 inch per hour fall thereafter until 5 P.M.
(a) Make a table that shows the total rainfall at each

hour starting at noon and ending at 5 P.M.
(b) Write a formula that calculates the rainfall R in

inches, x hours past noon.
(c) Use your formula to calculate the total rainfall at

5 P.M. Does your answer agree with the value in
your table from part (a)?

(d) How much rain had fallen at 3:45 P.M.?

88. Cost of a Laptop A 5% sales tax on a laptop com-
puter amounted to $106.25. Find the cost of the
laptop.

89. Distance Traveled At noon a bicyclist is 50 miles
from home, riding toward home at 10 miles per hour.
(a) Make a table that shows the bicyclist’s distance

D from home after 1, 2, 3, 4, and 5 hours.
(b) Write a formula that calculates the distance D

from home after x hours.
(c) Use your formula to determine D when x = 3

hours. Does your answer agree with the value
shown in your table?

(d) For what times was the bicyclist at least 20 miles
from home? Assume that 0 … x … 5.

90. Master’s Degree In 2001, about 468,500 people
received a master’s degree, and in 2008, about
625,000 did. Find the percent change in the number
of master’s degrees received between 2001 and 2008.

91. Car Speeds One car passes another car on a freeway.
The faster car is traveling 12 miles per hour faster
than the slower car. Determine how long it will be
before the faster car is 2 miles ahead of the slower
car.

92. Dimensions of a Rectangle The width of a rectangle
is 10 inches less than its length. If the perimeter is 112
inches, find the dimensions of the rectangle.

93. Saline Solution A saline solution contains 3% salt.
How much water should be added to 100 milliliters of
this solution to dilute it to a 2% solution?

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Beginning and Intermediate Algebra with Applications & Visualization, Third edition, by Gary K. Rockswold and Terry A. Krieger. Published by Addison Wesley.
Copyright © 2013 by Pearson Education, Inc.

155CHAPTER 2 TEST

94. Investment Money A student invests two sums of
money, $500 and $800, at different interest rates,
receiving a total of $55 in interest after one year. The
$500 investment receives an interest rate 2% lower
than the interest rate for the $800 investment. Find the
interest rate for each investment.

95. Geometry A triangle with height 8 inches is to have
an area that is not more than 100 square inches. What
lengths are possible for the base of the triangle?

96. Grade Average A student scores 75 and 91 on two
different tests of 100 points. If the maximum score on
the next test is also 100 points, what score does the
student need to maintain an average of at least 80?

97. Parking Rates Parking in a lot costs $2.25 for the
first hour and $1.25 for each hour thereafter. A partial

hour is charged the same as a full hour. What is the
longest time that someone can park for $9?

98. Profit The cost to produce one DVD player is $85
plus a one-time fixed cost of $150,000. The revenue
received from selling one DVD player is $225.
(a) Write a formula that gives the cost C of produc-

ing x DVD players.
(b) Write a formula that gives the revenue R from

selling x DVD players.
(c) Profit equals revenue minus cost. Write a for-

mula that calculates the profit P from selling x
DVD players.

(d) What numbers of DVD players sold will result
in a loss? (Hint: A loss corresponds to a negative
profit.)

Exercises 1–4: Solve the equation. Check your solution.

1. 9 = 3 – x 2. 4x – 3 = 7

3. 4x – (2 – x) = -3(2 x + 6)

4. 112 x –
2
3 =

1
2134 – 13 x2

Exercises 5 and 6: Determine the number of solutions to
the given equation.

5. 6(2 x – 1) = -4(3 – 3x)

6. 4(2 x – 1) = 8x – 4

7. Complete the table. Then use the table to solve the
equation 6 – 2 x = 0.

Exercises 8 and 9: Translate the sentence into an equation,
using the variable x. Then solve the resulting equation.

8. The sum of a number and -7 is 6.

9. Twice a number plus 6 equals the number minus 7.

10. The sum of three consecutive natural numbers is 336.
Find the three numbers.

11. Convert 3.2% to fraction and decimal notation.

12. Convert 0.345 to a percentage.

13. Use the formula d = rt to find r when d = 200 feet
and t = 4 seconds.

14. Area Find the area of the triangle shown.

3 in.
5 in.

15. Find the circumference and area of a circle with a
30-inch diameter.

16. The measures of the angles in a triangle are x, 2x, and
3x. Find the value of x.

Exercises 17 and 18: Solve the formula for x.

17. � = y – 3xy 18. R =
x

4
+
y
5

19. Use a number line to graph the solution set to x … 3.

20. Express the set of real numbers graphed on the num-
ber line with an inequality.

–2 –1 0 1 2

Exercises 21 and 22: Solve the inequality.

21. -3x + 9 Ú x – 15

22. 3(6 – 5x) 6 20 – x

x 0 1 2 3 4

6 – 2 x 6

CHAPTER 2 Test Step-by-step test solutions are found on the Chapter Test Prep Videos available via the Video Resources on DVD, in , and on (search “RockswoldBeginAlg” and click on “Channels”).

156 CHAPTER 2 LINEAR EQUATIONS AND INEQUALITIES

23. Snowfall Suppose 5 inches of snow fall before noon
and 2 inches per hour fall thereafter until 10 p.m.
(a) Write a formula that calculates the snowfall S in

inches, x hours past noon.
(b) Use your formula to calculate the total snowfall

at 8 p.m.
(c) How much snow had fallen at 6:15 p.m.?

24. Mixing Acid A solution is 45% hydrochloric acid.
How much water should be added to 1000 milliliters
of this solution to dilute it to a 15% solution?

25. Cable TV In 2010, the average cable TV subscriber
was paying $75 per month. By 2015, this cost is
expected to rise to $95 per month. Find the percent
increase in cable TV costs over this time period.
(Source: CNN Money.)

Exercises 1–4: Average Speed If someone travels a dis-
tance d in time t, then the person’s average speed is dt .
Use this fact to solve the problem.

1. A driver travels at 50 mph for the first hour and then
travels at 70 mph for the second hour. What is the
average speed of the car?

2. A bicyclist rides 1 mile uphill at 5 mph and then rides
1 mile downhill at 10 mph. Find the average speed of
the bicyclist. Does your answer agree with what you
expected?

3. At a 3-mile cross-country race an athlete runs 2 miles
at 8 mph and 1 mile at 10 mph. What is the athlete’s
average speed?

4. A pilot flies an airplane between two cities and trav-
els half the distance at 200 mph and the other half at
100 mph. Find the average speed of the airplane.

5. A Puzzle About Coins Suppose that seven coins look

exactly alike but that one coin weighs less than any
of the other six coins. If you have only a balance with
two pans, devise a plan to find the lighter coin. What
is the minimum number of weighings necessary?
Explain your answer.

6. Global Warming If the global climate were to warm
significantly as a result of the greenhouse effect or
other climatic change, the Arctic ice cap would start
to melt. This ice cap contains the equivalent of some
680,000 cubic miles of water. More than 200 million
people live on land that is less than 3 feet above sea
level. In the United States several large cities have
low average elevations. Three examples are Boston
(14 feet), New Orleans (4 feet), and San Diego (13
feet). In this exercise you are to estimate the rise in
sea level if the Arctic ice cap were to melt and to
determine whether this event would have a signifi-
cant impact on people living in coastal areas.
(a) The surface area of a sphere is given by the for-

mula 4pr2, where r is its radius. Although the
shape of Earth is not exactly spherical, it has an
average radius of 3960 miles. Estimate the sur-
face area of Earth.

(b) Oceans cover approximately 71% of the total
surface area of Earth. How many square miles of
Earth’s surface are covered by oceans?

(c) Approximate the potential rise in sea level by
dividing the total volume of the water from the
ice cap by the surface area of the oceans. Convert
your answer from miles to feet.

(d) Discuss the implications of your calculation.
How would cities such as Boston, New Orleans,
and San Diego be affected?

(e) The Antarctic ice cap contains some 6,300,000
cubic miles of water. Estimate how much the
sea level would rise if this ice cap melted. (Source:
Department of the Interior, Geological Survey.)

CHAPTER 2 Extended and Discovery Exercises

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1-256-49082-2
Beginning and Intermediate Algebra with Applications & Visualization, Third edition, by Gary K. Rockswold and Terry A. Krieger. Published by Addison Wesley.
Copyright © 2013 by Pearson Education, Inc.

157CHAPTERS 1–2 CUMULATIVE REVIEW EXERCISES

CHAPTERS 1–2 Cumulative Review Exercises

Exercises 1 and 2: Classify the number as prime or com-
posite. If the number is composite, write it as a product
of prime numbers.

1. 45 2. 37

Exercises 3 and 4: Multiply or divide and then simplify to
lowest terms when appropriate.

3. 43 # 38 4. 23 , 6
Exercises 5 and 6: Add or subtract and then simplify to
lowest terms when appropriate.

5. 1112 –
3
8 6.

2
3 +

1
5

Exercises 7 and 8: Classify the number as one or more
of the following: natural number, whole number, integer,
rational number, or irrational number.

7. -1 8. 23

Exercises 9–12: Evaluate the expression by hand.

9. 15 – 4 # 3 10. 30 , 6 # 2
11. 23 – 42 , 2 12. 11 – 3 + 16 – 4

Exercises 13 and 14: Simplify the expression.

13. 5×3 – x3 14. 4 + 2 x – 1 + 3x

Exercises 15–18: Solve the equation.

15. x – 3 = 11 16. 4x – 6 = -22

17. 5(6y + 2) = 25

18. 11 – (y + 2) = 3y + 5

Exercises 19 and 20: Determine whether the equation has
no solutions, one solution, or infinitely many solutions.

19. 6x + 2 = 2(3x + 1)

20. 2(3x – 4) = 6(x – 1)

21. Find three consecutive integers whose sum is 90.

22. Convert 4.7% to decimal notation.

23. Convert 0.17 to a percentage.

24. Find the average speed of a car that travels 325 miles
in 5 hours.

25. The angles in a triangle have measures 2x, 3x, and
4x. Find the value of x.

26. Find the area of a circle having a 10-inch diameter.

Exercises 27 and 28: Solve the formula for x.

27. a = 3xy – 4 28. A =
x + y + �

Exercises 29 and 30: Solve the inequality and graph the
solution set.

29. 7 – 3x 7 4 30. 6 x … 5 – (x – 9)

31. Yards to Inches There are 36 inches in 1 yard. Write
a formula that converts Y yards to I inches.

32. Checking Account The initial balance in a check-
ing account is $468. Find the final balance result-
ing from the following sequence of withdrawals and
deposits: -$14, $200, -$73, -$21, and $58.

33. Acid Solution How much of a 4% acid solution
should be added to 150 milliliters of a 10% acid
solution to dilute it to a 6% acid solution?

34. Energy Production In 1997, U.S. hydroelectric
power production hit an all-time high of 356 bil-
lion kilowatt-hours. This is 188 billion kilowatt-
hours less than twice the 2007 production level. Find
the production level in 2007. (Source: U.S. Department of
Energy.)

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