Explain in your own words what the meaning of domain is.

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min 250 words Algebra discussion

Explain in your own words what the meaning of domain is. Also, explain why a denominator cannot be zero. Find the domain for each of your two rational expressions. Write the domain of each rational expression .

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Rational Expressions
Advanced technical developments have made sports equipment faster, lighter,

and more responsive to the human body. Behind the more flexible skis, lighte

bats, and comfortable athletic shoes lies the science of biomechanics, which is the

study of human movement and the factors that influence it

Designing and testing an athletic shoe go hand in hand. While a shoe is being

designed, it is tested in a multitude of ways, including long-term wear, rear foo

stability, and strength of materials. Testing basketball shoes usually includes a

evaluation of the force applied to the ground by the foot during running, jumping,

and landing. Many biome-

chanics laboratories have

special platform that can mea-

sure the force exerted when a

player cuts from side to side, a

well as the force against the

bottom of the shoe. Force

exerted in landing from a lay-

up shot can be as high as

14 times the weight of the

body. Side-to-side force is usu-

ally about 1 to 2 body weights

in a cutting movement.

a
p

t
e

In Exercises 53 and 54 of
Section 6.7 you will see ho

designers of athletic shoes use
proportions to find the amount
of force on the foot and soles of

shoes for activities such as
running and jumping.

Reducing Rational
Expressions

Multiplication an

Division

Finding the Least
Common Denominat

6.4 Addition and Subtraction

6.5 Complex Fractions

6.6
Solving Equations with
Rational Expressions

Applications of Ratios
and Proportions

Applications of Rational
Expressions

2
1

50 100 150 200 250 300

Weight (pounds

)

Fo
r

ho
us

an
ds

o
f

po
un

ds
)

dug84356_ch06a.qxd 9/14/10 12:39 PM Page 381

382 Chapter 6 Rational Expressions 6-2

E X A M P L E 1

U Calculator Close-Up V

To evaluate the rational expression
in Example 1(a) with a calculator, first
use Y � to define the rational expres-
sion. Be sure to enclose both numera-
tor and denominator in parentheses.

Then find y1(�3).

Evaluating a rational expression

a) Find the value of �4

�

2
1

� for x � �3. b) If R(x)

� �

3
2

�
�
2
1

�, find R(4).

Solution

a) To find the value of �4

x
x

�

�
2
1

� for x � �3, replace x by �3 in the rational expression:

�
4

(

�
�

3)
�

�
2
1

� �

�
�

�

1
3

� � 13

So the value of the rational expression is 13. The Calculator Close-Up shows how to
evaluate the expression with a graphing calculator using a variable. With a scientific or
graphing calculator you could also evaluate the expression by entering (4(�3) � 1)�
(�3 � 2). Be sure to enclose the numerator and denominator in parentheses.

b) R(4) is the value of the rational expression when x � 4. To find R(4), replace x by 4
in R(x) � �

2
x

x
�
�
2
1

�:

R(4) �

�
3

2
(
(

4
4

)
)

�
�
2
1
�

R(4) � �

1
7

� � 2

So the value of the rational expression is 2 when x � 4, or R(4) � 2 (read “R of 4
is 2”).

Now do Exercises 1–6

In This Section

U1V Rational Expressions and
Functions

U2V Reducing to Lowest Terms

U3V Reducing with the Quotient
Rule for Exponents

U4V Dividing a � b by b � a

U5V Factoring Out the Opposite
of a Common Factor

U6V Writing Rational Expressions

6.1

Reducing Rational Expressions

Rational expressions in algebra are similar to the rational numbers in arithmetic. In
this section, you will learn the basic ideas of

rational expressions.

U1V Rational Expressions and Functions

A rational number is the ratio of two integers with the denominator not equal to 0. For
example,

�

3
4

�, �
�

�

6
�, �

�

1
7
�, and �

0
2

�

are rational numbers. Of course, we usually write the last three of these rational num-
bers in their simpler forms �3

2
�, �7, and 0. A rational expression is the ratio of two poly-

nomials with the denominator not equal to 0. Because an integer is a monomial, a
rational number is also a rational expression. As with rational numbers, if the denom-
inator is 1, it can be omitted. Some examples of rational expressions are

�
x

x
2
�
�

8
1

�, �
3

a
�
�

5a
9

� 3

�, �

3
7

� , and 9x.

A rational expression involving a variable has no value unless we assign a value
to the variable. If the value of a rational expression is used to determine the value of
a second variable, then we have a rational function. For example,

y �

�
x
x

2
�

–
8
1

� and w � �
3a2

a
�

�
5a
9

–

3
�

are rational functions. We can evaluate a rational expression with or without function
notation as we did for polynomials in Chapter 5.

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6-3 6.1 Reducing Rational Expressions 383

An expression such as

�
5

� is undefined because the definition of rational numbers
does not allow zero in the denominator. When a variable occurs in a denominator, an

real number can be used for the variable except numbers that make the expression
undefined.

E X A M P L E 2 Ruling out values for x
Which numbers cannot be used in place of x in each rational expression

a) �
x
x

2
�
�
8
1

� b)

�
2

x
�

�
2
1

� c) �
x

x
2

�
�

5
4

�

Solution
a) The denominator is 0 if x � 8 � 0, or x � �8. So �8 cannot be used in place of x.

(All real numbers except �8 can be used in place of x.)

b) The denominator is zero if 2x � 1 � 0, or x � ��

1
2

�. So we cannot use �

�1

� in place
of x. �All real numbers except ��12� can be used in place of x.�

c) The denominator is zero if x2 � 4 � 0. Solve this equation:

x2 � 4 � 0

(x � 2)(x � 2) � 0

Factor.

x � 2 � 0 or x � 2 � 0

Zero factor property

x � 2 or x � �2

So 2 and �2 cannot be used in place of x. (All real numbers except 2 and �2 can
be used in place of x.)

Now do Exercises 7–14

E X A M P L E 3 Domain
Find the domain of each expression.

a) �
x
x
2
�
�

3
9

� b) �
x

2 �

x
x � 6
� c) �

x �

5
�

Solution
a) The denominator is 0 if x � 3 � 0, or x � �3. So �3 can’t be used for x. The

domain is the set of all real numbers except �3, which is written in set notation
as {x � x � �3}.

b) The denominator is 0 if x2 � x � 6 � 0:

x2 � x � 6 � 0

(x � 3)(x � 2) � 0

x � 3 � 0 or x � 2 � 0

x � 3 or x � �2

So �2 and 3 can’t be used in place of x. The domain is the set of all real numbers
except �2 and 3, which is written as {x � x � �2 and x � 3}.

In Example 2 we determined the real numbers that could not be used in place of
the variable in a rational expression. The domain of any algebraic expression in one
variable is the set of all real numbers that can be used in place of the variable. For
rational expressions, the domain must exclude any real numbers that cause the denom-
inator to be zero.

dug84356_ch06a.qxd 9/14/10 12:39 PM Page 383

384 Chapter 6 Rational Expressions 6-4

c) Since the denominator is 4, the denominator can’t be 0 no matter what number is
used for x. The domain is the set of all real numbers, R.

Now do Exercises 15–22

Note that if a rational expression is used to define a function, then the domain of
the rational expression is also called the domain of the function. For example, the
domain of the function y �

�x

x
2

�
–

3
9

� is the set of all real numbers except �3 or {x � x � �3}.
When dealing with rational expressions in this book, we will generally assume

that the variables represent numbers for which the denominator is not zero.

U2V Reducing to Lowest Terms
Rational expressions are a generalization of rational numbers. The operations that we
perform on rational numbers can be performed on rational expressions in exactly the
same manner.

Each rational number can be written in infinitely many equivalent forms. For
example,

� � � � � � �

� .

Each equivalent form of �

3
5

� is obtained from �3
5

� by multiplying both numerator and

denominator by the same nonzero number. This is equivalent to multiplying the frac-
tion by 1, which does not change its value. For example,

� � 1 � � � and � � .

If we start with

�
1

0
� and convert it into �3

5
�, we say that we are reducing �

1
6

� to lowest terms.

We reduce by dividing the numerator and denominator by the common factor 2:

� �

A rational number is expressed in lowest terms when the numerator and the denomi-
nator have no common factors other than 1.

We can reduce fractions only by dividing the numerator and the denom-
inator by a common factor. Although it is true that

� ,

we cannot eliminate the 2’s, because they are not factors. Removing
them from the sums in the numerator and denominator would not result
in �3

�.

Reducing Fractions

If a � 0 and c � 0, then

�
a

� � �

b
c

�.

2 �

4
�

2 � 8

6
�

CAUTION

3
�
5

� � 3

�
2

� � 5

�
10

9
�

3 � 3
�
5 � 3

3
�
5
6
�
10

2
�

3
�
5
3
�
5
3
�
5

15
�

�
20

�
15

6
�
10
3
�
5

U Helpful Hint V

How would you fill in the blank i

n
�

3
5� � 10

—? Most students learn to divide
5 into 10 to get 2, and then multiply 3
by 2 to get 6. In algebra, it is better to
multiply the numerator and denomi-
nator of �35� by 2, as shown here.

dug84356_ch06a.qxd 9/14/10 12:39 PM Page 384

6-5 6.1 Reducing Rational Expressions 385

To reduce rational expressions to lowest terms, we use exactly the same procedure
as with fractions:

Dividing the numerator and denominator by the GCF is often referred to as dividing
out or canceling the GCF.

Reducing Rational Expressions

1. Factor the numerator and denominator completely.

2. Divide the numerator and denominator by the greatest common factor.

E X A M P L E 4 Reducing

Reduce to lowest terms.

�
3
4

2
� b) �

6
x

2
�
�

1
9

�

c) �
3

2
�

2
9

�

x
8
� 6

�
Solution

a) �
3
4
0

2
� � �

2�

�
�

3�

�
�

5
7

� Factor.

� �
5
7

� Divide out the GCF: 2 � 3 or 6.

b) Since

�
1
9

8
� � �

9
9

�
�
1
2

�

� �
1

� , it is tempting to apply that fact here. However, 9 is not a common

factor of the numerator and denominator of

�
6
x

2
�
�
1
9
8

�, as it is i

n �

1
9
8
�. You must factor

the numerator and denominator completely before reducing.

�
6
x
x
2
�
�
1
9
8

�

��

(x �

x
3)

�

(

x
3

�

)
3)

� Factor.

�

�
x �

6
3

� Divide out the GCF: x � 3.

This reduction is valid for all real numbers except �3, because that is the domain of the
original expression. If x � �3, then x � 3 � 0 and we would be dividing out 0 from the
numerator and denominator, which is prohibited in the rule for reducing

fractions.

c) �
3×2
2
�
x2
9
�
x
8
� 6

� ��
3
2
(
(
x
x

�
�

2
2

)
)
(
(
x
x

�
�

1
2
)
)

� Factor completely.

�

2
3

x �

�

� Divide out the GCF: x � 2.

This reduction is valid for all real numbers except �2 and 2, because that is the
domain of the original expression.

Now do Exercises 23–46

In reducing, you can divide out or cancel common factors only. Yo

cannot cancel x from �xx
�
�

3
2�, because it is not a factor of either x � 3 or

x � 2. But x is a common factor in �32
x
x�, and �

3
2
x
x� � �

3
2�.

Note that there are four ways to write the answer to Example 3(c) depending on
whether the numerator and denominator are factored. Since

�
2
3

(
x
x

�
�
3
2)

� � �
3

2
(
(
x
x

�
�
1
2
)
)

� � �
3
2

(x
x

�
�

1
4

)

� � �

3
2
x
x

�
�
3
4

�,

CAUTION

dug84356_ch06a.qxd 9/14/10 12:39 PM Page 385

E X A M P L E 6 Reducing expressions involving large integers
Reduce �4

6
2

1
0
6

� to lowest terms.

Solution
Use the method of Section 5.1 to get a prime factorization of 420 and 616:

2 420 2 6

2 210 2 308

3 105 2 154

5 35 7 77
7 11

�

�
��

��
��

E X A M P L E 5 Using the quotient rule in reducing
Reduce to lowest terms.

a) �
3
6
a
a

1
7
5

� b) �

6
4
x
x

y
5

2
�
Solution
a) �
3
6
a
a
1
7
5

� � �
3�

3�
�

a
2
15

a7
� Factor. b) �

6
4
x
x
4

y
y

5
2

� � �
2�
2�

�
�
3
2
x
x
4
y
y
5
2
� Factor.

� �
a1

5�7

�
Quotient rule

� �
3×4�

1y2�5
�

Quotient rule
for exponents for exponents

� �

a
2

� � �
3×3

2
y�

3
� � �

3
2

x
y

3
3�

Now do Exercises 47–58

any of these four rational expressions is correct. We usually give such answers with the
denominator factored and the numerator not factored. With the denominator factored
you can easily spot the values for x that will cause an undefined expression.

U3V Reducing with the

Quotient Rule for Exponents

To reduce rational expressions involving exponential expressions, we use the quotient
rule for exponents from Chapter 4. We restate it here for reference.

386 Chapter 6 Rational Expressions 6-6

The essential part of reducing is getting a complete factorization for the numerator
and denominator. To get a complete factorization, you must use the techniques for fac-
toring from Chapter 5. If there are large integers in the numerator and denominator, you
can use the technique shown in Section 5.1 to get a prime factorization of each integer.

Quotient Rule for Exponents

If a � 0, and m and n are any integers, then

�
a
a

n� � a
m�n.

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6-7 6.1 Reducing Rational Expressions 387

E X A M P L E 7

The complete factorization for 420 is 22 � 3 � 5 � 7, and the complete factorization for 616
is 23 � 7 � 11. To reduce the fraction, we divide out the common factors:

�

4
6

2
1
0
6

�

� �
2

2
3
�
�

7
�

�

1
�

1
7
�
� �
2
3
�
�

1
5

�

�
1
2

5
2

�

Now do Exercises 59–66

U4V Dividing a � b by b � a
In Section 4.5 you learned that a � b � �(b � a) � �1(b � a). So if a � b is divided
by b � a, the quotient is �1:

�

a
b

�
�

b
a

� � �
�1

b
(

b
�

�
a

a)
�

� �1

We will use this fact in Example 7.

Expressions with a � b and b � a
Reduce to lowest terms.

�
5
4
x

y
�
�

4
y

x
� b) �

m
n
2
�
�
m

n2
�

Solution
a) Factor out 5 from the numerator and 4 from the denominator and use �x

y
�

�

y
x

�

� �1:

�
5
4
x
y
�
�
5
4
y

x
� � �

5
4
(
(
x
y

�
�

y
x
)
)

� � �
5
4

�(�1) � ��
5
4

�

Another way is to factor out �5 from the numerator and 4 from the denominator
and then use �y

y
�
�
x
x
� �1:
�
5
4
x
y
�
�
5
4
y
x
� � �
�

4
5

(y
(y

�
�

x
x
)
)

�

� �
�

4
5
�(1) � ��

5
4
�
�1

b) �
m

n
2
�
�
m

n2
� ��

(

m �

n
�

)(m
m

� n)
� Factor.

� �1(m � n) �
m

n �
�
m

� � �1

� �m � n

Now do Exercises 67–74

We can reduce �a
b

�
�
b
a

� to �1, but we cannot reduce �a
a

�
�

b
b

�. There is no factor

that is common to the numerator and denominator of �a
a

�
�
b
b

� or �a
a

�
�
b
b
�.

U5V Factoring Out the Opposite of a Common Factor
If we can factor out a common factor, we can also factor out the opposite of that com-
mon factor. For example, from �3x � 6y we can factor out the common factor 3 or
the common factor �3:

�3x � 6y � 3(�x � 2y) or �3x � 6y � �3(x � 2y)
To reduce an expression, it is sometimes necessary to factor out the opposite of a
common factor.

CAUTION

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E X A M P L E 9 Writing rational expressions
Answer each question with a rational expression.

a) If a trucker drives 500 miles in x � 1 hours, then what is his average speed?

b) If a wholesaler buys 100 pounds of shrimp for x dollars, then what is the price per pound?

c) If a painter completes an entire house in 2x hours, then at what rate is she painting?

Solution
a) Because R � �D

T
�, he is averaging �x

5
�

0
1

� mph.

b) At x dollars for 100 pounds, the wholesaler is paying �
10

x
0

� dollars per pound
or �

0
x

� dollars/pound.

c) By completing 1 house in 2x hours, her rate is �

1
x

� house/hour.

Now do Exercises 107–112

The main points to remember for reducing rational expressions are summarized in
the following reducing strategy.

388 Chapter 6 Rational Expressions 6-8

E X A M P L E 8 Factoring out the opposite of a common factor
Reduce ��3

w
w

�
�

3
1

� to lowest terms.

Solution
We can factor 3w or �3w from the numerator. If we factor out �3w, we get a common
factor in the numerator and denominator:

�
�3

w
w
2
�
�
3
1
w2

� ��
(w

�
�

3w
1

(
)
1
(w
�

�

� Factor.

� �
w
�

�
3w
1

� Since 1 � w � w � 1, we divide out w � 1.

� �
1
3
�
w
w

� Multiply numerator and denominator by �1.

The last step is not absolutely necessary, but we usually perform it to express the answer
with one less negative sign.

Now do Exercises 75–

Strategy for Reducing Rational Expressions

1. Factor the numerator and denominator completely. Factor out a common fac-
tor with a negative sign if necessary.

2. Divide out all common factors. Use the quotient rule if the common factors
are powers.

U6V Writing Rational Expressions
Rational expressions occur in applications involving rates. For uniform motion, rate is
distance divided by time, R � �DT�. For example, if you drive 500 miles in 10 hours, your
rate is �51

0
0
0

� or 50 mph. If you drive 500 miles in x hours, your rate is �5

� mph. In work prob-
lems, rate is work divided by time, R � �W

T
�. For example, if you lay 400 tiles in 4 hours,

your rate is �404
0

� or 100 tiles/hour. If you lay 400 tiles in x hours, your rate is �40x
0

� tiles/hour.

dug84356_ch06a.qxd 9/14/10 12:39 PM Page 388

6-9 6.1 Reducing Rational Expressions 389

Warm-Ups ▼

Fill in the blank.

1. A rational number is a ratio of two with the

denominator not 0.
2. A rational expression is a ratio of two with

the denominator not 0.
3. A rational expression is reduced to lowest terms by

the numerator and denominator by the GCF.
4. The rule is used in reducing a ratio of

monomials.
5. The expressions a � b and b � a are .
6. If a rational expression is used to determine y from x,

then y is a function of x.

True or false?

7. A complete factorization of 3003 is 2 � 3 � 7 � 11 � 13.

8. A complete factorization of 120 is 23 � 3 � 5.

9. We can’t replace x by �1 or 3 in �
x
x �

�
1
3
�.

10. For any real number x, �

2
2
x

�

� x.

11. Reducing �
a
a

2 �
�
b
b

2
� to lowest terms yields a � b.

U1V Rational Expressions and Functions

Evaluate each rational expression. See Example 1.

1. Evaluate �

3
x

x
�
�
5

3
� for x � �2.

2. Evaluate �
3

4
x
x
�
�
1

4
� for x � 5.

3. If R(x) � �

� 9
�, find R(3).

4. If R(x) � ��2
x

0
�

x �
8

2
�, find R(�1).

5. If R(x)

� �
x

x
�
�
5

3
�, find R(2), R(�4), R(�3.02), and

R(�2.96). Note how a small difference in x (�3.02 to
�2.96) can make a big difference in R(x).

6. If R(x) � �

x2 �

x �
2x
2

� 3
�, find R(3), R(5), R(2.05),

and R(1.999).

Which numbers cannot be used in place of the variable in each
rational expression? See Example 2.

7. �

x �

1
� 8. �

x
3
�
x
7
�

9. �
3a

7
�
a

5
� 10. �

3 �

�

11. �
x

2
2
x
�
�
1
3

6
� 12. �

�

y �

y �
1
6
�

13. �
p �

1
� 14. �

m �
5

31
�

Find the domain of each rational expression. See Example 3.

15. �
x

2
�
�
2
x
�

16. �
x
x

�
�
4
5
�

Exercises

U Study Tips V
• If you must miss class, let your instructor know. Be sure to get notes from a reliable classmate.
• Take good notes in class for yourself and your classmates. You never know when a classmate will ask to see your notes.

6
.1

dug84356_ch06a.qxd 9/17/10 8:06 PM Page 389

390 Chapter 6 Rational Expressions 6-10

17. �
x2 � 5

x
x

� 6
�

18.

�
x2 �

x2
x
�
�

2
12

�

19. �
x2 �

2
4

�

20. �
x2 �

�

21. �
x �

x
5

�

22. �
x
x

2
�
�

9
3

�
U2V Reducing to Lowest Terms

Reduce each rational expression to lowest terms. Assume that the
variables represent only numbers for which the denominators are
nonzero. See Example 4.

23. �
2

7
� 24. �

1
2
4
1
�

25. �
4

9
2

0
� 26. �

4
5
2
4
�

27. �
3

9
6
0

a
� 28. �

5
4
6
0
y
�

29. �
3

7
0
8

w
� 30. �

4
6
4
8
y
�

31. �

� 2
� 32. �

� 2
�

33. �
2

6
x
y
�
�
4
3
y

x
� 34. �

1
5
0

x
x
�

�

1
2
0
0
a
a

�

35. �
6

3
b

b
�

�

1
9
5

� 36. �

3
3

m
m

�
�

9
6

w
w

�

37. �
w

w
2 �
�
4
7

9
� 38. �

a
a
2 �
�
b
b
2
�

39. �

a2 �

a2
2
�
a
1

� 1
� 40. �

x2 �
x2
2
�

y
�
2

�

41. �

2×2

4
�

x2
4

�
x
4

� 2
� 42. �

2×2
3
�

x2
1

�
0x
2
�
7

12
�

43. �
3×2

2
�

1
�

6
�
3

27
� 44. �

x3 �

x2
3

�
x2
4
�
x

�

45. �
2

4
a

3 �
�

1
8

� 46. �
w
w

2
3
�
�

3
2
w
7

�

U3V Reducing with the Quotient Rule
for Exponents

Reduce each expression to lowest terms. Assume that all
variables represent nonzero real numbers, and use only positive
exponents in your answers. See Example 5.

47. �
x
x

1
7
0

� 48. �
y
y

5� 49. �

z
z

8�

50. �
w
w

1
9

2� 51. ��

4
2

x
x

5� 52. �
�

3
6
y
y
9

3
�

53. �
�

8
1
m
2m

9
1

n
6

� 54. �
�

6
9
u
u
9

9v
14

� 55. �
�

6
8
b
b

10
1

c
0c

7�

56. �
�

9
6
x
x

2
2
0
5

y
y3

� 57.

�
1
3

8
0
a
a

7
3
b
b
1

c
7� 58. �2

1
4
5
m
m

1
1
2
0

n
n

3
�

Reduce each expression to lowest terms. Assume that all
variables represent nonzero real numbers, and use only positive
exponents in your answers. See Example 6.

59. �
2

2
1
6
0

4
� 60. �

6
6
1
6
6
0
�

61. �
2

1
3
6
1

8
�

62. �

9
6
3
2
6
4
�

63. �

3
3
0
0
0
x
x

9� 64. �1
9
0
6
8
y
y

5�

65. �

9
4

2
4

4
8

a
a

2
1
3

9� 66. �
2
1

7
6

0
5

b
b

7
1
5
2�
U4V Dividing a � b by b � a

Reduce each expression to lowest terms. See Example 7.

67. �
3

2
a

b
�
�

2
3
b

a
� 68. �

5
6

n �
�
5
6
m
n
�

69. �
h

t
2
�
�
h

t2
� 70. �

r
s
2 �
�
s
r
2
�

71. �

2
h

g
2

�
�

6
g
h
2� 72. �

5
4
a

b2
�

�
10
a

b
2�

73.

�
x2

9
�
�
x
x
�
2

6
� 74. �

a2
1

�
�
a
a
�
2
2
�

dug84356_ch06a.qxd 9/14/10 12:39 PM Page 390

6-11 6.1 Reducing Rational Expressions 391

U5V Factoring Out the Opposite
of a Common Factor

Reduce each expression to lowest terms. See Example 8.

75. �

�

x
x
�
�
6

6
�

76. �

�
3
5
x
x
�
�
1
2
2
0
�

77. �

�

�
�
9
6
y

y2
�

78. �

�
y2
8
�
�
1
2
6
y
�

79. �
�

3
3
x
x
�
�
6

6
� 80. �

�
8
8x
�
�
4x

16
�

81. �
2a

�
2
1
�
2a

�
�
6

3
� 82. �

�2b2

b2
�
�

� 4
�

83.

�
2b

a
2
3
�
�

2
b

b
� 84. �

x
x
3
�
�

x
1

2�

Reduce each expression to lowest terms. See the Strategy for
Reducing Rational Expressions box on page 388.

85. �
2
4
x
x

1
8
2

� 86. �
4
2
x
x

9�

87.

�
2x

4
�
x

4
� 88. �

2x �

x2
�

89. �
a

4
�
�
4

a
� 90. �

2
2
b
b
�
�
4
4
�

91. �
2

4
c
�
�
c

4
2� 92. �

�
4
2
�

t �

t2
4

�

3. �
x2 �

x2
4
�
x
4

� 4
� 94. �

x2 �
3x
4
�
x �
6
4
�

95. �
x2

�
�
2x

�
�
4

6
� 96. �

x2
�
�
2x
2x
�
�
8

8
�

97. �
2

q
8
6
�
�
q
q
7

5� 98. �12s6
8

�

s12

16s5
�

99. �
u

u
2

2
�
�
1
6
6
u
u
�
�
1
6
6

4
� 100. �

v
2

2
�
�
1
3
2
v
v
�
�
1
3
8
6
�

101. �
2

a
a
3 �
�
8

4
� 102. �

4w2

2
�

w3
1

�
2w
5
�
4

36
�

103. �
y3 �

y2
2

�
y2
4
�
y �

� 8
� 104.

105.

106.

U6V Writing Rational Expressions

Answer each question with a rational expression. Be sure to
include the units. See Example 9.

107. If Sergio drove 300 miles at x � 10 miles per hour, then
how many hours did he drive?

108. If Carrie walked 40 miles in x hours, then how fast did
she walk?

109. If x � 4 pounds of peaches cost $4.50, then what is the
cost per pound?

110. If nine pounds of pears cost x dollars, then what is the
price per pound?

111. If Ayesha can clean the entire swimming pool in x hours,
then how much of the pool does she clean per hour?

112. If Ramon can mow the entire lawn in x � 3 hours, then
how much of the lawn does he mow per hour?

Applications

Solve each problem.

113. Annual reports. The Crest Meat Company found that the
cost per report for printing x annual reports at Peppy
Printing is

given by the formula

C(x) � �
150 �

0.60x
�,

where C(x) is in dollars.

x2 � ax � 4x � 4a
��

x2 � 16

2x � 2w � ax � aw
��

x3 � xw2

mx � 3x � my � 3y

��

m2 � 3m � 18

dug84356_ch06a.qxd 9/14/10 12:39 PM Page 391

392 Chapter 6 Rational Expressions 6-12

a) Use the accompanying graph to estimate the cost per
report for printing 1000 reports.

b) Use the formula to find C(1000), C(5000), and
C(10,000).

c) What happens to the cost per report as the number of
reports gets very large?

given by the formula,

C(p) � �
1

5
0
0
0

�

000

p
�.

a) Use the accompanying graph to estimate the cost for
removing 90% and 95% of the toxic chemicals.

b) Use the formula to find C(99.5) and C(99.9).
c) What happens to the cost as the percentage of

pollutants removed approaches 100%?

Number of reports (thousands)

C
os

t p
er

r
ep

or
t (

do
ll

ar
s)

0.60

0.70

0.80

0.50

0.40

2 3 4 51

Figure for Exercise 113
Percentage of chemicals removed

ua
l c

os
t

(h
un

dr
ed

th
ou

sa
nd

d
ol

la
rs

)
0
2
3
4
5
1

91 92 93 9594 96 97 98 9990

Figure for Exercise 114

In This Section

U1V Multiplication of Rational
Numbers

U2V Multiplication of Rational
Expressions

U3V Division of Rational
Numbers

U4V Division of Rational
Expressions

U5V Applications

6.2 Multiplication and Division

In Section 6.1, you learned to reduce rational expressions in the same way that we
reduce rational numbers. In this section, we will multiply and divide rational
expressions using the same procedures that we use for rational numbers.

U1V

Multiplication of Rational Numbers

Two rational numbers are multiplied by multiplying their numerators and multiplying
their denominators.

Multiplication of Rational Numbers

If b � 0 and d � 0, then

�
a

b
� � �

c
� � �

b
a
d

c
�.

114. Toxic pollutants. The annual cost in dollars for removing
p% of the toxic chemicals from a town’s water supply is

dug84356_ch06a.qxd 9/14/10 12:39 PM Page 392

6-13 6.2 Multiplication and Division 393

E X A M P L E 1 Multiplying rational numbers
Find the product �6

7
� � �

1
1

4
5
�.

Solution
The product is found by multiplying the numerators and multiplying the denominators:

�

6
7

� � �
1

1
4

5
� � �

1
8

0
4

�

� �

2
2
1
1

�
�
4
5

� Factor the numerator and denominator

� �
4
5

� Divide out the GCF 21.

The reducing that we did after multiplying is easier to do before multiplying. First factor
all terms, reduce, and then multiply:

�
6
7
� � �
1
1
4
5
� � �

2
7�
� 3�
� � �

�
2
3
�
�

7�
5

�
� �
4
5
�

Now do Exercises 1–8

U2V Multiplication of Rational Expressions

Rational expressions are multiplied just like rational numbers: factor, reduce, and then
multiply. A rational number cannot have zero in its denominator and neither can a
rational expression. Since a rational expression can have variables in its denominator,
the results obtained in Examples 2 and 3 are valid only for values of the variable(s)
that would not cause a denominator to be 0.

E X A M P L E 2 Multiplying rational expressions
Find the indicated products.

a) �
9

5
x

y
� � �

1
3
0
xy
y

�

b) �
�

3
8
z
x
3

4
� � �

2
1
x
5
5y
z

3�
Solution
a) �
9
5
x
y
� � �
1
3
0
xy
y
� � �
3

5�
�

y�
3�x�
� � �

2
3�
�

x�
5�
y
y�

� Factor.

� �
6
y

�
b) �
�
3
8
z
x
3
y4
� � �
2
1
x
5
5y
z

3� � �
�2 �

3�
2
z3
� 2�xy4
� � �

3�
2�x

�
5

5
y3
z

� Factor.
� �
�

2
3x

x
y
y
3

� Reduce.

� �
�

z2
2
x
0
4

y
� Quotient rule

Now do Exercises 9–18

U Helpful Hint V

Did you know that the line separating
the numerator and denominator in a
fraction is called the vinculum?

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394 Chapter 6 Rational Expressions 6-14

E X A M P L E 3 Multiplying rational expressions
Find the indicated products.

a) �
2x �

4
2y

�

� �
x2

2
�

x
y2

�

b) �
x2 �

2x
7
�

x �
6

12
� � �

x2 �
x

16
� c) �

a
6
�

a
b

� � �
a2 � 2

8
a
a
b

� b2
�

Solution
a) �
2x �
4
2y

� � �
x2

2
�
x
y2

� � �
2�(

2�
x �

�

2�
y)

�

� �
(x �

2�
y)
�
(x
x

� y)
� Factor.

� �
x �
x
y
� Reduce.
b) �
x2 �
2x
7
�
x �
6
12
� � �
x2 �
x

6
� � �

(x �
2(

3
x
)
�

(

x
3
�

)
4)

� � �
(x � 4)

x
(x � 4)
� Factor.

� �
2(x

x
� 4)
� Reduce.

c) �
a

6
�
a
b

� �

�
a2 �

8
2

a
a

b � b2
� � �

a �

� 3
b
a

� � �
(
2
a

�
�

4a
b

)2
� Factor.

� �
3(a

4
�

a
b)

� Reduce.

Now do Exercises 19–26

U3V

Division of Rational Numbers

By the definition of division, a quotient is found by multiplying the dividend by the re-
ciprocal of the divisor. If the divisor is a rational number �

d
c

�, its reciprocal is simply �d
c

�.
Division of Rational Numbers

If b � 0, c � 0, and d � 0, then

�
a
b
� � �
d
c
� � �
a
b
� � �
d
c
�.

E X A M P L E 4 Dividing rational numbers
Find each quotient.

a) 5 � �
1
2

� b) �
6
7

� � �
1
3

4
�

Solution
a) 5 � �
1
2

� � 5 � 2 � 10 b) �
6
7

� � �
1
3
4
� � �

6
7
� � �
1
3
4
� � �
2
7�
� 3�
� � �

2
3�
� 7�

� � 4

Now do Exercises 27–34

U4V Division of Rational Expressions

We divide rational expressions in the same way we divide rational numbers: Invert the
divisor and multiply.

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6-15 6.2 Multiplication and Division 395

E X A M P L E 5 Dividing rational expressions
Find each quotient.

a) �

3
5
x
� � �

6
5

x
� b) �

x
2
7

� � (2×2) c) �4
x2

�
�
x
x
2

�

� �
x
x

�
�
2
1
�
Solution
a) �
3
5
x
� � �

6
5
x
� � �

3
5
x
� � �

6
5
x
� Invert the divisor and multiply.

� �
3�
5�
x�
� � �

2 �
5�
3�x�
� Factor.

� 2 Divide out the common factors.

b) �
x

� � (2×2) � �x
2

� � �
2
1

� Invert and multiply.

� �

x
4

� Quotient rule

c) �
4
x2

�
�
x
x
2

� � �
x

2
�
�
2
1

� � �
4
x2

�
�
x
x
2

� � �
x
x

2
�
�
2
1
� Invert and multiply.
�
�1

�
(2 �

x(x
x)

�

(2
1
�

)
x)

� � �
(x �

x
1
�

)(x

2
� 1)

� Factor.

��
�1(2 �

x
x)(x � 1)
� �

2
x �

�
2
x
� � �1

� Simplify.

� �
�x2 �

x
x � 2
�

Now do Exercises 35–48

We sometimes write division of rational expressions using the fraction bar. For
example, we can write

�
a �

3
b

� �

�
1
6

� as .

No matter how division is expressed, we invert the divisor and multiply.

�
a �
3
b
�

—

�
1
6

�

�1(x2 � x � 2)
��x

U Helpful Hint V

A doctor told a nurse to give a patient
half of the usual 500-mg dose of a
drug. The nurse stated in court,
“dividing in half means dividing by
1/2, which means multiply by 2.” The
nurse was in court because the
patient got 1000 mg instead of
250 mg and died (true story).
Dividing a quantity in half and divid-
ing by one-half are not the same.

E X A M P L E 6 Division expressed with a fraction bar
Find each quotient.

a) b) c)

�
a2

� 5
�

—
2

�
x2 �
2
1
�

—
�
x �

3
1
�
�
a �
3
b
�
—

�

6�

dug84356_ch06a.qxd 9/14/10 12:39 PM Page 395

396 Chapter 6 Rational Expressions 6-16

E X A M P L E 7 Using rational expressions with uniform motion
Shasta drove 200 miles on I-10 in x hours before lunch.

a) Write a rational expression for her average speed before lunch.
b) She drives for 3 hours after lunch at the same average speed. Write a rational

expression for her distance after lunch.

Solution
a) Because R � �DT�, her rate before lunch is �

20
x
0
ho

m
u
i
r
l
s
es

� or �
20

x
0
� mph.
Solution

� �
a �

3
b

� � �
1
6

� Rewrite as division.

� �
a �
3
b

� � �

6
1

� Invert and multiply.
� �
a �

3�
b

� � �
2

1
� 3�
� Factor.

� (a � b)2 Reduce.

� 2a � 2b

� �
x2 �

2
1

� � �
x �

3
1
� Rewrite as division.
� �
x2 �
2
1
� � �
x �
3
1
� Invert and multiply.

� �
(x � 1)

2
(x � 1)
��

x �
3

1
� Factor.

�

�
3x

2
� 3
� Reduce.

� �
a2

3
� 5
� � 2 Rewrite as division.

� �
a2

3
� 5
� � �

1
2

� � �
a2

6
� 5
�

Now do Exercises 49–56

U5V Applications
We saw in Section 6.1 that rational expressions can be used to represent rates. Note
that there are several ways to write rates. For example, miles per hour is written mph,
mi/hr, or �m

hr
i

�. The last way is best when doing operations with rates because it helps us
reconcile our answers. Notice how hours “cancels” when we multiply miles per hour
and hours in Example 7, giving an answer in miles, as it should be.

�
a2
3
� 5
�
—
2
�
x2 �
2
1
�
—
�
x �
3
1
�
�
a �
3
b
�
—
�
1
6
�
U Helpful Hint V

In Section 6.5 you will see another
technique for finding the quotients in
Example 6.

dug84356_ch06a.qxd 9/20/10 11:45 AM Page 396

6-17 6.2 Multiplication and Division 397

b) Because D � R � T, her distance after lunch is the product of �20x
0

� mph (her rate) and
3 hours (her time):

D � �
20

x
0

� �
m

i
� � 3 hr � �

60
x
0

� mi

Now do Exercises 77–78

The amount of work completed is the product of rate and time, W � R � T. So if a
machine washes cars at the rate of 12 per hour and it works for 3 hours, the amount of

work completed is 36 cars washed. Note that the rate is given by R � �
W
T

�.
Warm-Ups ▼

Fill in the blank.
1. expressions are multiplied by multiplying their

numerators and multiplying their denominators.

2. can be done before multiplying rational
expressions.

3. To rational expressions, invert the divisor and
multiply.

True or false?
4. One-half of one-fourth is one-sixth.

5. �
2
3

� � �
5
7

� � �
1
2

0
1
�

E X A M P L E 8 Using rational expressions with work
It takes x minutes to fill a bathtub.

a) Write a rational expression for the rate at which the tub is filling.
b) Write a rational expression for the portion of the tub that is filled in 10 minutes.

Solution
a) The work completed in this situation is 1 tub being filled. Because R � �WT�, the rate

at which the tub is filling is �x
1

m
tu

i
b
n

� or

�
1
x

� tub/min.

b) Because W � R � T, the work completed in 10 minutes or the portion of the tub that

is filled in 10 minutes is the product of �
1
x

� tub/min (the rate) and 10 minutes (the time):

W � �
1
x

� �
m
tu

i
b
n

� � 10 min � �
1
x
0
� tub

Now do Exercises 79–80

6. The product of �
x �

3
7

� and �
7 �

6
x

� is �2.

7. Dividing by 2 is equivalent to multiplying b

y �
1

�.

8. For any real number a, �

a
3

� � 3 � �
a
9

�.

9. �
2
3

� � �
1
2

� � �

4
3

�

dug84356_ch06a.qxd 9/14/10 12:39 PM Page 397

U1V Multiplication of Rational Numbers

Perform the indicated operation. See Example 1.

1. �
2
3

� �

�
5
6

� 2. �
3
4

� � �

2
5

� 3. �
1

8
5
� � �
3
2
5
4
�

4. �
3

4
� � �
2
8

1
� 5. �

1
1
2
7
� � �
5
1
1

0
� 6. �

2
4
5
8
� � �
5
3
6
5
�

7. 24 � �
2

0
� 8. �

1
3

0
� � 35

U2V Multiplication of Rational Expressions

Perform the indicated operation. See Example 2.

9. �
2
3
x
� � �

4
5
x
� 10. �

3
7
y
� � �

2
2

1
y

�

11. �
5
6
x2
� � �

3
x

� 12. �
1
9
0
x
� � �

x
5
2�

13. �
1

5
2
a
b
� � �
3
5
a
5
b

a
� 14. �

3
7
m

p
� � �

6
3
m

p
�

15. �
�

7
2

a
x
5

6
� � �
2
6
1
x

a2
� 16. �

�

5z
9

y
w

� � �
�

20
6
z
y
9

5
�

17. �
1

2
5
0
t
w

3y
7

� � 24t5w3y2 18. 22x2y3z � �
33

6
y
x
3

z4
�

Perform the indicated operation. See Example 3.

19. �
2x �

7
2y

�

� �
6x

1
�

5
6y

�

20. �
a2

3
�

a
� � �

2a
6
� 2
�

21. �
3a

1
�
5

3b
� � �

a2
1
�

b2
�

22. �
b3 �

5
b
� � �

b2
1

�
0
b
�

23. (x2 � 6x � 9) � �
x �

3
3
�

24. �
4x

1
�
2

10
� � (4×2 � 20x � 25)

5. �
1

5
6
a

a
2 �

�
5
8
� � �

a
4

�
2 �

a �
1

1
�

26. �
2×2

6x
�

�

5x
18

� 3
� � �

4×2

6
�

x �
4x

3
� 1

�

U3V Division of Rational Numbers

Perform the indicated operation. See Example 4.

27.

�
1
4

� � �
1
2

� 28. �
1
6

� � �
1
2

� 29. 12 � �
2

5
�

30. 32 � �
1

4
� 31. �

5
7
� � �
1
1
5

4
� 32. �

3
4
� � �
1
2
5
�

33. �
4

0
� � 12 34. � 9

U4V Division of Rational Expressions

Perform the indicated operation. See Example 5.

35. �
x
4

2
� � �
2
x

� 36. �
2
3
a2
� � �

2
6
a
�

37. � �
1

2
0
1

x
� 38. �

4
3
u
v
2
� � �
1
1
5
4
v

u
6�

39. �
8
n
m

4
3

� � (12mn2) 40. �2
3
p
q

3� � (4pq5)

41.

�
y �

2
6
� � �

6 �

y
� 42. �

4 �

5
a
� � �
a2 �
3
16
�

43. �
x2 � 4

� 4
� � �

(x �
16

2)3
�

44. �
a2 � 2

a �

1
� � �

a2 �
a
1
�

45. �

t2 �

3t �

10
� � (4t � 8)

46. �
w2

w
�
2
7
�
w
4
�
w

12
�� (w2 � 9)

5×2
�
3

22
�
9

Exercises

U Study Tips V
• Personal issues can have a tremendous effect on your progress in any course. If you need help, get it.
• Most schools have counseling centers that can help you to overcome personal issues that are affecting your studies.

6
.2

dug84356_ch06a.qxd 9/17/10 8:07 PM Page 398

6-19 6.2 Multiplication and Division 399

47. (2×2 � 3x � 5) � �2
x
x
�

�
1
5
�

48. (6y2 � y � 2) � �2
3
y
y

�
�
1
2
�

Perform the indicated operation. See Example 6.

49. 50.

51. 52.

53. 54.

55. 56.

Miscellaneous

Perform the indicated operation.

57. �
x �

3
1
� � �

1 �

x
� 58. �

2x �
3

2y
� � �

y �
1
x
�

59. �
3a �

a
3b

� � �
1

3
� 60. �

61. 62.

63. �
6
3

y
� � (2x) 64. �

8
9

x
� � (18x)

65. �
�

a
2
3

a
b
b

� � �
a

5b
b

� 66. �
�

3
2
a
a
2

� � �
1
2
5
0
a
a
3�

�
2
3
g
h
�

�

�
1
h

�

�
b
a

�
�
�
1
2
�

2
�
5

a � b
�
2b � 2a

x2 � 6x � 8

——

�
x
x
�
�
2
1
�

x2 � y2
—
�
x �

9
y

�
�
a �
1
3
�

—
4

�
x2 �
3
9
�

—
5

�
6a2

5
� 6
�

—

�

5
� 6
�
�
x2
1
�
2
4
�
—
�
x �
6
2
�

�
3m �

6n
�

——
�
3
4
�
�
x �
5

2y
�

—
�
1
1
0
�

67. �
2

6
m
m

n
n
4

2� � �
3
m
m

2
5

n
n
4

7
�

68. �
r
r

t
t

2� � �r
r
3

t
t
2
2�

69. �
3×2 �

x � 5
� � �

9×2
x

�
2
1
�

70. �
x2 � 6

x
x � 5
� � �
3x
x
�
4
3
�
71. �

72. �
(

w
w
2
�
�
1
1

)2
� � �

w2 �

w
2
�
w
1

� 1
�

73. �
2×2

x
�
2 �
19
1
x
0
�
0

10
��

2×2 �

4×2
19
�
x
1

� 10
�

74. �
x

x
3
2
�
�
1

1
� �

75. �
76. �
U5V Applications

Solve each problem. Answers could be rational expressions.
Be sure to give your answers with appropriate units.
See Examples 7 and 8.

77. Marathon run. Florence ran 26.2 miles in x hours in the
Boston Marathon.
a) Write a rational expression for her average speed.

b) She runs at the same average speed for �12� hour in the
Cripple Creek Fun Run. Write a rational expression for
her distance at Cripple Creek.

78. Driving marathon. Felix drove 800 miles in x hours on
Monday.
a) Write a rational expression for his average speed.

b) On Tuesday he drove for 6 hours at the same average
speed. Write a rational expression for his distance on
Tuesday.

6 � 2b
�
9 � b2

3x � 3w � bx � bw
��

x2 � w2

m2 � 9
��
m2 � mk � 3m � 3k

9 � 6m � m2
��
9 � 6m � m2

9×2 � 9x � 9
��

2 � x

(a � 2)3
�
2a � 4

a2 � 2a � 4
��

a2 � 4

dug84356_ch06a.qxd 9/14/10 12:39 PM Page 399

400 Chapter 6 Rational Expressions 6-20

79. Filling the tank. Chantal filled her empty gas tank in x
minutes.
a) Write a rational expression for the rate at which she

filled her tank.

b) Write a rational expression for the portion of the tank
that is filled in 2 minutes.

80. Magazine sales. Henry sold 120 magazine subscriptions in
x days.
a) Write a rational expression for the rate at which he sold

the subscriptions.

b) Suppose that he continues to sell at the same rate for
5 more days. Write a rational expression for the number
of magazines sold in those 5 days.

81. Area of a rectangle. If the length of a rectangular flag is
x meters and its width is �5

x
� meters, then what is the area of

the rectangle?

— m
x
5

x m

Figure for Exercise 81

1—
x � 2

8x � 16 yd

Figure for Exercise 82

In This Section

U1V Building Up the Denominator

U2V Finding the Least Common
Denominator

U3V Converting to the LCD

6.3 Finding the Least Common Denominator

Every rational expression can be written in infinitely many equivalent forms.
Because we can add or subtract only fractions with identical denominators, we
must be able to change the denominator of a fraction. You have already learned
how to change the denominator of a fraction by reducing. In this section, you will
learn the opposite of reducing, which is called building up the denominator.

U1V Building Up the Denominator
To convert the fraction �2

3
� into an equivalent fraction with a denominator of 21,

we factor 21 as 21 � 3 � 7. Because �2
3

� already has a 3 in the denominator, multiply

Getting More Involved

83. Discussion

Evaluate each expression.

a) One-half of �1
4

� b) One-third of 4

c) One-half of �4
3
x
� d) One-half of �3

2
x
�

84. Exploration

Let R � �6
2
x
4

2
x2
�
�

23
2
x
9x

�
�

20
4

� and H � �
8
2
x
x

�

5
1

�.

a) Find R when x � 2 and x � 3. Find H when x � 2 and
x � 3.

b) How are these values of R and H related and why?

82. Area of a triangle. If the base of a triangle is 8x � 16 yards
and its height is �

x �
1

2
� yards, then what is the area of the

triangle?

dug84356_ch06a.qxd 9/14/10 12:39 PM Page 400

the numerator and denominator of �2
3

� by the missing factor 7 to get a denominator
of 21:

�
2
3

� � �
2
3

� � �
7
7

� � �
1
2

4
1

�

For rational expressions the process is the same. To convert the rational
expression

�
x �

5
3

�

into an equivalent rational expression with a denominator of x2 � x � 12, first factor
x2 � x � 12:

x2 � x � 12 � (x � 3)(x � 4)

From the factorization we can see that the denominator x � 3 needs only a factor of
x � 4 to have the required denominator. So multiply the numerator and denominator
by the missing factor x � 4:

�
x �
5
3

� ��
(x �

5(x
3)

�

(x
4
�

)
4)

��
x2

5
�

x �
x �

20
12

�

6-21 6.3 Finding the Least Common Denominator 401

E X A M P L E 1

E X A M P L E 2

Building up the denominator
Build each rational expression into an equivalent rational expression with the indicated
denominator.

�
3x �

7
3y

� � �
6y �

?
6x

� b) �
x

�
�
2
2

� � �
x2 � 8

?
x � 12
�

Building up the denominator
Build each rational expression into an equivalent rational expression with the indicated
denominator.

a) 3 � �
1

?
2
� b) �
w
3
� � �
w

?
x
� c) �

3
2
y3

� � �

12
?
y8
�

Solution
a) Because 3 � �3

1
�, we get a denominator of 12 by multiplying the numerator and

denominator by 12:

3 � �
3
1

� � �
3
1

�
�

1
1
2

� � �
3
1
6
2
�

b) Multiply the numerator and denominator by x :

�
w
3

� � �
w

3 �
�

x
x
� � �
w

3x
x
�

c) Note that 12y8 � 3y3 � 4y5. So to build 3y3 up to 12y8 multiply by 4y5:

�
3

2
y3
� � �

3
2
y3
�
�

4
4
y
y

5� � �1
8

2
y

5
8�

Now do Exercises 1–20

In Example 2 we must factor the original denominator before building up the
denominator.

dug84356_ch06a.qxd 9/14/10 12:39 PM Page 401

402 Chapter 6 Rational Expressions 6-22

U Helpful Hint V

Notice that reducing and building up
are exactly the opposite of each
other. In reducing you remove a
factor that is common to the numera-
tor and denominator, and in building
up you put a common factor into the
numerator and denominator.

Solution
a) Because 3x � 3y � 3(x � y), we factor �6 out of 6y � 6x. This will give a factor

of x � y in each denominator:

3x � 3y � 3(x � y)
6y � 6x � �6(x � y) � �2 � 3(x � y)

To get the required denominator, we multiply the numerator and denominator
by �2 only:

�
3x �
7
3y

� � �
(3x �

7(�
3y

2
)
)
(�2)

�

� �
6y

�
�

14
6x

�

b) Because x2 � 8x � 12 � (x � 2)(x � 6), we multiply the numerator and
denominator by x � 6, the missing factor:

�
x
x
�
�
2
2

�

� �
(
(
x
x

�
�

2
2
)
)
(
(
x
x

�
�

6
6

)
)

�
� �
x
x
2
2
�
�

4
8
x
x

�
�
1
1
2
2
�

Now do Exercises 21–32

When building up a denominator, both the numerator and the denomina-
tor must be multiplied by the appropriate expression.

U2V Finding the Least Common Denominator

We can use the idea of building up the denominator to convert two fractions with
different denominators into fractions with identical denominators. For example,

�
5
6

� and �
1
4

�

can both be converted into fractions with a denominator of 12, since 12 � 2 � 6
and 12 � 3 � 4:

�
5
6

� � �
5
6

�
�
2
2

� � �
1
1
0

2
� �

1
4

� � �
1
4

�
�
3
3

� � �
1
3
2
�

The smallest number that is a multiple of all of the denominators is called the least
common denominator (LCD). The LCD for the denominators 6 and 4 is

12.

To find the LCD in a systematic way, we look at a complete factorization of each
denominator. Consider the denominators 24 and 30:

24 � 2 � 2 � 2 � 3 � 23 � 3

30 � 2 � 3 � 5

Any multiple of 24 must have three 2’s in its factorization, and any multiple of 30
must have one 2 as a factor. So a number with three 2’s in its factorization will have
enough to be a multiple of both 24 and 30. The LCD must also have one 3 and one 5
in its factorization. We use each factor the maximum number of times it appears in
either factorization. So the LCD is 23 � 3 � 5:

23 � 3 � 5 � 2 � 2 � 2 � 3 � 5 � 120
30

CAUTION
� �

dug84356_ch06a.qxd 9/14/10 12:39 PM Page 402

If we omitted any one of the factors in 2 � 2 � 2 � 3 � 5, we would not have a
multiple of both 24 and 30. That is what makes 120 the least common denominator.
To find the LCD for two polynomials, we use the same strategy.

6-23 6.3 Finding the Least Common Denominator 403

E X A M P L E 4 Converting to the LCD
Find the LCD for the rational expressions, and convert each expression into an equivalent
rational expression with the LCD as the denominator.

a) �

9
4
xy

�, �

15
2
xz
� b) �

6
5
x2

�, �

8x
1
3y
�, �

4
3
y2

�

Solution
a) Factor each denominator completely:

9xy � 32xy 15xz � 3 � 5xz

Strategy for Finding the LCD for Polynomials

1. Factor each denominator completely. Use exponent notation for repeated
factors.

2. Write the product of all of the different factors that appear in the denominators.

3. On each factor, use the highest power that appears on that factor in any of
the denominators.

E X A M P L E 3 Finding the LCD
If the given expressions were used as denominators of rational expressions, then what
would be the LCD for each group of denominators?

a) 20, 50 b) x3yz2, x5y2z, xyz5 c) a2 � 5a � 6, a2 � 4a � 4

Solution
a) First factor each number completely:

20 � 22 � 5 50 � 2 � 52

The highest power of 2 is 2, and the highest power of 5 is 2. So the LCD of 20 and
50 is 22 � 52, or 100.

b) The expressions x3yz2, x5y2z, and xyz5 are already factored. For the LCD, use the
highest power of each variable. So the LCD is x5y2z5.

c) First factor each polynomial.

a2 � 5a � 6 � (a � 2)(a � 3) a2 � 4a � 4 � (a � 2)2

The highest power of (a � 3) is 1, and the highest power of (a � 2) is 2. So the
LCD is (a � 3)(a � 2)2.

Now do Exercises 33–46

U3V Converting to the LCD
When adding or subtracting rational expressions, we must convert the expressions
into expressions with identical denominators. To keep the computations as simple as
possible, we use the least common denominator.

dug84356_ch06a.qxd 9/14/10 12:39 PM Page 403

404 Chapter 6 Rational Expressions 6-24

The LCD is 32 � 5xyz. Now convert each expression into an expression with this
denominator. We must multiply the numerator and denominator of the first rational
expression by 5z and the second by 3y:

�
9
4
xy
� � �

9
4
xy
�
�

5
5

z
z

� � �
4
2
5
0
x
z
yz

�

Same denominator

�
15

2
x

z
� � �

15
2
x
�

z
3
�

y
3y

� � �
45

6
x
y
yz

�

b) Factor each denominator completely:

6×2 � 2 � 3×2 8x3y � 23x3y 4y2 � 22y2

The LCD is 23 � 3 � x3y2 or 24x3y2. Now convert each expression into an expression
with this denominator:

�
6
5
x2
� ��

6
5
x2
�
�

4
4
x
x
y
y

2�� 2
2
4
0
x
x
3

y
y
2
2�

�
8x

1
3y
� � �

8x
1
3y
� 3

�
y
3y

� � �
24

3
x
y
3y2
�

�
4
3
y2
� � �

4
3
y2
�
�

6
6
x
x

3� � �2
1
4
8
x
x
3y

3
2�
Now do Exercises 47–58

⎫
⎪
⎬
⎪
⎭

E X A M P L E 5 Converting to the LCD
Find the LCD for the rational expressions

�
x2
5
�
x
4

� and �
x2 �

3
x � 6
�

and convert each into an equivalent rational expression with that denominator.

Solution
First factor the denominators:

x2 � 4

� (x � 2)(x � 2)

x2 � x � 6 � (x � 2)(x � 3)

The LCD is (x � 2)(x � 2)(x � 3). Now we multiply the numerator and denominator of
the first rational expression by (x � 3) and those of the second rational expression by
(x � 2). Because each denominator already has one factor of (x � 2), there is no reason to
multiply by (x � 2). We multiply each denominator by the factors in the LCD that are
missing from that denominator:

�
x2
5
�
x
4
� � �
�
x2 �

3
x � 6
� � �

Now do Exercises 59–70

3x � 6
��
(x � 2)(x � 2)(x � 3)

3(x �

2)
��

(x � 2)(x � 3)(x � 2)

Same
denominator

⎫
⎪
⎬
⎪
⎭

5×2 � 15x
��
(x � 2)(x � 2)(x � 3)

5x(x � 3)
��
(x � 2)(x � 2)(x � 3)

U Helpful Hint V

What is the difference between LCD,
GCF, CBS, and NBC? The LCD for the
denominators 4 and 6 is 12. The least
common denominator is greater than
or equal to both numbers.The GCF for
4 and 6 is 2.The greatest common fac-
tor is less than or equal to both num-
bers. CBS and NBC are TV networks.

dug84356_ch06a.qxd 9/14/10 12:39 PM Page 404

6-25 6.3 Finding the Least Common Denominator 405

Warm-Ups ▼

Fill in the blank.
1. To the denominator of a fraction, we multiply

the numerator and denominator by the same nonzero real
number.

2. The is the smallest number
that is a multiple of all denominators.

3. The LCD is the product of every factor that appears in
the factorizations, raised to the power that
appears on the factor.

True or false?

4. �
2
3

� � �
2
3
�

� 5
5

�
U1V Building Up the Denominator

Build each rational expression into an equivalent rational
expression with the indicated denominator. See Example 1.

1. � 2. �

3. �
3
4

� � �
1
?

6
� 4. �

3
7

� � �
2
?
8
�

5. 1 � �7
?

� 6. 1 � �3
?
x�

7. 2 � �
6
?

� 8. 5 � �
1
?
2
�

9. �
5
x

� � �

a
?

x
� 10. �

3
x

� � �
3
?
x
�

?
�
35

2
�
5

?
�
27

1
�
3

11. 7 � �
2

x
� 12. 6 � �

4
?
y
�

13. �
5

b
� � �
3
?

bt
� 14. �

2
7

ay
� � �

2a
?

yz
�

15. �
�

2a
9
w
z

� � �
8a

?
wz
� 16. �

�

3
7
x
yt

� � �
18

?
xyt
�

17. �
3
2
a
� � 18. �

1
7
2
b
c5
� � �

36
?
c8
�

19. �
5x

4
y2
� � �

10x
?

2y5
� 20. �

8
5
x
y
3

2
z
� � �

24x
?

5z3
�

?
�
15a3

Exercises

U Study Tips V
• Try changing subjects or tasks every hour when you study. The brain does not easily assimilate the same material hour after hour.
• You will learn more from working on a subject one hour per day than seven hours on Saturday.

6
.3

5. �
2
3
� � �
3
2
�
� 5
5
�

6. The LCD for the denominators 25 � 3 and 24 � 32 is
25 � 32.

7. The LCD for �
1
6

� and �
1
1
0
� is 60.

8. The LCD for �
x �

1
2

� and �
x �

1
2

� is x2 � 4.

9. The LCD for �
a2

1
� 1
� and �

a �
1

1
� is a2 � 1.

dug84356_ch06a.qxd 9/17/10 8:08 PM Page 405

406 Chapter 6 Rational Expressions 6-26

Build each rational expression into an equivalent rational
expression with the indicated denominator. See Example 2.

21. �
x �
5
3

� � �
2x

?
� 6
�

22. �
a �

4
5

� � �
3a �

?
15

�

23. �
2x

� 2
� � �

�8x

� 8
�

24.

�
m �

n
� � �

2n �

2m
�

25. �
5b2

8
�
a

5b
� � �

20b2 �

b3
�

26. �
�6x

�

9
� � �

18×2 �

�

27. �
x �

3
2
� � �

x2
?

� 4
�

28. �
a �

a
3
� � �
a2 �
?
9
�

29. �
x

3
�
x
1
� �

30. �
2x

�
�
7x

3
� ��

4×2 �

12x � 9
�

31. � �
y2 � y

� 20
�

32. � �
z2 � 2

z � 15
�

U2V Finding the Least Common Denominator

If the given expressions were used as denominators of rational
expressions, then what would be the LCD for each group of
denominators? See Example 3. See the Strategy for Finding the
LCD for Polynomials box on page 403.

33. 12, 16 34. 28, 42

35. 12, 18, 20 36. 24, 40, 48

37. 6a2, 15a 38. 18×2, 20xy

39. 2a4b, 3ab6, 4a3b2

40. 4m3nw, 6mn5w8, 9m6nw

z � 6
�
z � 3

y � 6
�
y � 4

?
��
x2 � 2x � 1

41. x2 � 16, x2 � 8x � 16

42. x2 � 9, x2 � 6x � 9

43. x, x � 2, x � 2

44. y, y � 5, y � 2

45. x2 � 4x, x2 � 16, 2x

46. y, y2 � 3y, 3y

U3V Converting to the LCD

Find the LCD for the given rational expressions, and convert
each rational expression into an equivalent rational expression
with the LCD as the denominator. See Example 4.

47. �
1
6

�, �
3
8

� 48.

�
1
5

2
�, �

2
3
0

�

49. , �
6
5
x
� 50. , �

1
1
0x
�

51. , �
2

1
b

� 52. , �

6
x
y
�

53. , �
6

3b
�

54. �
7

4
5
b

a
� , �

105

ab
�

55. ,

56. �
8a

3
3b9
� , �

5
2c
�

57. �
9y

x
5z
�, �

12
y

3
�, �

1
2y
�

58.

�
12

a6b
�, , �

2a
1
b3
�

Find the LCD for the given rational expressions, and convert
each rational expression into an equivalent rational expression
with the LCD as the denominator. See Example 5.

59. �
x

2
�
x
3
�, �
x
5
�
x
2
�

60. �
a

2
�
a

5
�, �

a
3
�
a
2
�

3b
�
14a3

3
�
2×5

1
�
3×2

3
�
84a

y
�
4x

2
�
3a

3
�
5x

1
�
2x

dug84356_ch06a.qxd 9/14/10 12:39 PM Page 406

6-27 6.4 Addition and Subtraction 407

61. �
a �

4
6
�, �
6 �
5
a
�

62. �
x �

4
y

�, �
2y

5
�

x
2x

�

63. ,

64. �
x2

5
�
x

1
�, �

x2 �
�
2
4

x � 1
�

65. �
w2 �

w
2
�
w
2

� 15
�, �

w2 �
�
4
2
w
w
� 5
�

66. ,

67. , ,

68. , ,
�5

�
2b2 � 3b

2b
�
2b � 3

3
�
4b2 � 9

3
�
2x � 4

x
�
x2 � 4

�5
�
6x � 12

z � 1
��
z2 � 5z � 6

z � 1
��
z2 � 6z � 8

5x
��
x2 � 6x � 9

x
�
x2 � 9

69. , ,

70. , ,

Getting More Involved

71. Discussion

Why do we learn how to convert two rational
expressions into equivalent rational expressions with
the same denominator?

72. Discussion

Which expression is the LCD for

and ?

a) 2 � 3 � x(x � 2) b) 3

6x(x � 2)

c) 36×2(x � 2)2 d) 23 � 33×3(x � 2)2

2x � 7
��
2 � 32 � x(x � 2)2

3x � 1
��
22 � 3 � x2(x � 2)

2
��
p2 � p � 20

p
��
2p2 � 11p � 12

�3
��
2p2 � 7p � 15

4
��
q2 � q � 12

3
��
2q2 � 9q � 4

2
��
2q2 � 5q � 3

In This Section

U1V Addition and Subtraction of
Rational Numbers

U2V Addition and Subtraction of
Rational Expressions

U3V Applications

6.4 Addition and Subtraction

In Section 6.3, you learned how to find the LCD and build up the denominators of
rational expressions. In this section, we will use that knowledge to add and subtract
rational expressions with different denominators.

U1V

Addition and Subtraction of Rational Numbers

We can add or subtract rational numbers (or fractions) only with identical denominators
according to the following definition.

dug84356_ch06a.qxd 9/14/10 12:39 PM Page 407

408 Chapter 6 Rational Expressions 6-28

E X A M P L E 1
Addition and Subtraction of Rational Numbers

If b � 0, then

�
a
b

� � �
b

� � �
a �

b
c

� and �
a
b

� � �
b
c
� � �
a �
b
c
�.

Adding or subtracting fractions with the same denominator
Perform the indicated operations. Reduce answers to lowest terms.

a) �

1
1
2
� � �

1
7
2
� b) �

1
4

� � �
3
4

�
Solution
a)

�
1
1
2
� � �

1
7
2
� � �

1
8
2
� � �

4�
4�

�
�
2
3
� � �
2
3

� b) �
1
4

� � �
3
4
� � �
�

2
� � ��

1
2
�
Now do Exercises 1–8

If the rational numbers have different denominators, we must convert them to
equivalent rational numbers that have identical denominators and then add or subtract.
Of course, it is most efficient to use the least common denominator (LCD), as in
Example 2.

E X A M P L E 2 Adding or subtracting fractions with different denominators
Find each sum or difference.

�
2
3

0
� � �

1
7
2
� b) �
1
6

� � �
1
4
5
�

Solution
a) Because 20 � 22 � 5 and 12 � 22 � 3, the LCD is 22 � 3 � 5, or 60. Convert each

fraction to an equivalent fraction with a denominator of 60:

�
2
3
0
� � �
1
7
2
� � �
2
3
0
�
�
3
3

� � �
1
7
2
�

�
5
5

� Build up the denominators.

� �
6
9
0
� � �

3
6
5
0
�

Simplify numerators and denominators.

� �
4
6
4
0

� Add the fractions.

� �
4
4

�
�

1
1
5

� Factor.

� �
1
1
5
1
� Reduce.

b) Because 6 � 2 � 3 and 15 � 3 � 5, the LCD is 2 � 3 � 5 or 30:

�
1
6

� � �
1
4
5
� � �

1
� 3

� � �

4
� 5

� Factor the denominators.

� �
2
1
� 3

�

5
� 5

� � �
3
4
� 5

� 2
� 2

� Build up the denominators.
U Helpful Hint V

Note how all of the operations with
rational expressions are performed
according to the rules for fractions.
So keep thinking of how you perform
operations with fractions, and you will
improve your skills with fractions and
with rational expressions.

dug84356_ch06a.qxd 9/14/10 12:39 PM Page 408

6-29 6.4 Addition and Subtraction 409

E X A M P L E 3 Rational expressions with the same denominator
Perform the indicated operations and reduce answers to lowest terms.

a) �
3
2
y
� � �

3
4
y
� b) �

x
2
�

x
2
� � �
x �
4
2

�

c) �
(x �

x2
1
�

)

(x
2
�

x
3)

��
(x �

2x
1)

�

(x
1
� 3)

�
Solution
a) �
3
2
y
� � �

3
4
y
� � �

3
6
y
� Add the fractions.

� �
2
y

� Reduce.
b) �
x
2
�
x
2
� � �
x �
4
2

� �

�
2
x

x
�

�
2
4
� Add the fractions.

� �
2(

x
x
�
�

2
2)

� Factor the numerator.

� 2 Reduce.

c) �
(x �
x2
1
�
)(x
2
�
x
3)
��
(x �
2x
1)
�
(x
1
� 3)

�� Subtract the fractions.

��
x2

(x
�

�
2x
1)
�
(x
2
�

x �
3)

1
� Remove parentheses.

� �
(x �
x2

1)
�

(x
1
� 3)

�

Combine like terms.

� �
(
(
x
x
�

� 1
1

)
)
(
(
x
x
�
�

1
3
)
)

� Factor.

� �
x
x �

�
1
3
� Reduce.

Now do Exercises 19–30

x2 � 2x � (2x � 1)
��

(x � 1)(x � 3)

� �
3
5
0
� � �

3
8
0
� Simplify the numerators and denominators.

� �
�

30
3
� Subtract.

� �
�

10
1

�
�
3
3
� Factor.

� ��
1
1
0
� Reduce.

Now do Exercises 9–18

U2V Addition and Subtraction of Rational Expressions
Rational expressions are added or subtracted just like rational numbers. We can add or
subtract only when we have identical denominators. All answers should be reduced to
lowest terms. Remember to factor first when reducing, and then divide out any com-
mon factors.

dug84356_ch06a.qxd 9/14/10 12:39 PM Page 409

When subtracting a numerator containing more than one term, be sure
to enclose it in parentheses, as in Example 3(c). Because that numerator
is a binomial, the sign of each of its terms must be changed for the
subtraction.

In Example 4, the rational expressions have different denominators.

CAUTION

410 Chapter 6 Rational Expressions 6-30

E X A M P L E 4 Rational expressions with different denominators
Perform the indicated operations.

�
2
5

x
� � �

2
3
� b) �
x

4
3y

� � �

2
y3
�

c) �
a �

6
1
� � �
a �
8
2
�

Solution
a) The LCD for 2x and 3 is 6x:

�
2
5
x
� � �
2
3

� � �
2
5

x
�

�
3
3
� � �
2
3
�
�
2
2
x
x
�

� �
1
6
5
x
� � �

4
6
x

x
�

� �
15

6
�
x

4x
�

b) The LCD is x3y3.

�
x
4
3y
� � �
x
2
y3
� � �
x
4
3y
�
�
y
y
2

2� � �x
2
y3

�
�
x
x
2
2�

� �
x
4
3

y
y
2

3� � �x
2
3

x
y
2
3�

� �
4y2

�
3y3

2×2
�

c) Because 6 � 2 � 3 and 8 � 23, the LCD is 23 � 3, or 24:

�
a �
6
1
� � �
a �
8
2

� � �
(a

6
�
�

1
4
)4

� � �
(a
8
�
�

2
3
)3

�

� �
4a

2
�
4
4

� � �
3a

2
�
4
6
�

��
4a � 4 �

24
(3a �

6)
�

��
4a � 4

2
�

4
3a � 6
�

� �
a �

24
10

�

Now do Exercises 31–46

Combine like terms.

Remove the parentheses.

Subtract the rational expressions.

Simplify numerators and
denominators.

Build up both denominators
to the LCD 24.

Add the rational expressions.

Simplify numerators and denominators.

Build up both denominators to the LCD.

Add the rational expressions.
Simplify numerators and denominators.

Build up both denominators to 6x.

U Helpful Hint V

You can remind yourself of the
difference between addition and
multiplication of fractions with a sim-
ple example: If you and your spouse
each own 1�7 of Microsoft, then
together you own 2�7 of Microsoft.
If you own 1�7 of Microsoft, and
give 1�7 of your stock to your child,
then your child owns 1�49 of
Microsoft.

dug84356_ch06a.qxd 9/14/10 12:39 PM Page 410

6-31 6.4 Addition and Subtraction 411

E X A M P L E 5

U Helpful Hint V

Once the denominators are factored
as in Example 5(a), you can simply
look at each denominator and ask,
“What factor does the other denomi-
nator(s) have that is missing from this
one?” Then use the missing factor
to build up the denominator. Repeat
until all denominators are identical,
and you will have the LCD.

E X A M P L E 6 Rational expressions with different denominators
Perform the indicated operations.

�
x
x
2 �

�
2
1
x

� � �
6
2
x
x
�

�
1
1
2
� � �
1
6
�

Solution
The LCD for x(x � 2), 6(x � 2), and 6 is 6x(x � 2).

�
x
x
2 �
�
2
1
x
� � �
6
2
x
x
�
�
1
1
2
� � �
1
6

� � �
x(

x
x
�
�

1
2)

� � �
6
2
(
x
x

�
�
1
2)
� � �
1
6
�

� �
6
6
x
(
(
x
x
�

�
1
2
)
)

� � �

x
6

(
x
2
(
x
x

�
�
1
2
)
)

� � �
1
6
x
x
(
(
x
x

�
�
2
2
)
)

�
Build up to
the LCD.

Factor
denominators.

Needs x Needs

x � 3

Rational expressions with different denominators
Perform the indicated operations:

a) �
x2 �

1
9

� � �
x2 �

2
3x

� b)

�
5 �

4
a
� � �
a �
2
5
�
Solution
a) �
x2 �
1
9
� � �
x2 �
2
3x

� ��
(x � 3)

1
(x � 3)
��

x(x
2
� 3)
� The LCD is x(x � 3)(x � 3).

��
(x � 3

)(

�

x
x
� 3)x

��
x(x

2
�
(x
3
�

)(x
3
�

)
3)
�

��
x(x � 3

x
)(x � 3)
��

x(x �
2x

3
�

)(x
6
� 3)

�

��
x(x �

3x
3
�

)(x
6
� 3)
�

b) Because �1(5 � a) � a � 5, we can get identical denominators by multiplying
only the first expression by �1 in the numerator and denominator:

�
5 �
4
a
� � �
a �
2
5

� � �
(5 �

4(�
a)

1
(�
)

)
� � �

a �
2

5
�

� �
a
�

�
4
5
� � �
a �
2
5
�
� �
a
�
�
6
5

� �4 � 2 � �6

� ��
a �

6
5
�

Now do Exercises 47–64

In Example 6, we combine three rational expressions by addition and subtraction.

We usually leave the
denominator in
factored form.

dug84356_ch06a.qxd 9/14/10 12:39 PM Page 411

412 Chapter 6 Rational Expressions 6-32

E X A M P L E 7

U3V Applications
We have seen how rational expressions can occur in problems involving rates. In
Example 7, we see an applied situation in which we add rational expressions.

� �
6x

6
(
x
x
�

�

6
2)

� � �
6
2
x
x
(x

2 �
�

x
2)

� � �
6
x

2
(x
�
�

2x
2)

�
�
� �
x2
6x
�

(x
5
�

x �
2)

6
�

� �
(x

6
�

x(
3
x
)

(
�

x �
2)

2)
� Factor.

� �
x
6
�
x
3
� Reduce.

Now do Exercises 65–70

Combine
like terms.

Combine the
numerators.

6x � 6 � 2×2 � x � x2 � 2x
��

6x(x � 2)

Simplify

numerators.

Adding work
Harry takes twice as long as Lucy to proofread a manuscript. Write a rational expression for
the amount of work they do in 3 hours working together on a manuscript.

Solution
Let x � the number of hours it would take Lucy to complete the manuscript alone and
2x � the number of hours it would take Harry to complete the manuscript alone. Make a
table showing rate, time, and work completed:

Now find the sum of each person’s work.

�
3
x

� � �
2
3
x
� � �

2
2
�
�
3
x

� � �

2
3
x
�

� �
2
6
x
� � �

2
3
x
�

� �
2
9
x
�

So in 3 hours working together they will complete �
2
9
x
� of the manuscript.

Now do Exercises 81–86

Rate Time Work

Lucy �
1
x

� �
m

h
s
r
p

� 3 hr �
3
x

� msp

Harr

y �
2

1
x
� �

m
h
s
r
p

� 3 hr �
2
3
x
� msp

dug84356_ch06a.qxd 9/14/10 12:39 PM Page 412

6-33 6.4 Addition and Subtraction 413

Warm-Ups ▼

Fill in the blank.
1. We can rational expressions only if they have

identical denominators.

2. We can any two rational expressions so that
their

denominators are identical.

True or false?

3. �
1
2

� � �
1
3
� � �
2
5
�

4. �
1
7
2
� � �

1
1
2
� � �
1
2
�

U1V Addition and Subtraction of Rational Numbers

Perform the indicated operation. Reduce each answer to lowest
terms. See Example 1.

1. �
1

1
0
� � �
1
1

0
� 2. �

1
8
� � �
3
8
�

3. �
7

8
� � �
1

8
� 4. �

4
9
� � �
1
9
�
5. �
1
6
� � �
5

6
� 6. ��

3
8
� � �
7
8
�

7. ��
7

8
� � �
1

8
� 8. ��

2
9

0
� � ��2

0
��

Perform the indicated operation. Reduce each answer to lowest
terms. See Example 2.

9. �
1

3
� � �
2

9
� 10. �

1
4
� � �
5
6
�

11. �1
7
0� � �

5
6� 12. �

5
6� � �1

3
0�

13. �
1
7
6
� � �
1
5

8
� 14. �

7
6
� � �
1
4
5
�

15. �
1

8
� � �
1
9

0
� 16. �

1
2
5
� � �
1
5
2
�

17. ��
1
6

� � ��38�� 18. ��
1
5

� � ��17��

Exercises

U Study Tips V
• When studying for a midterm or final, review the material in the order it was originally presented. This strategy will help you to see

connections between the ideas.
• Studying the oldest material first will give top priority to material that you might have forgotten.

6
.4

5. �
3
5

� � �
4
3

� � �
2
1
9
5
�

6. �
4
5

� � �
5
7

� � �
3
3
5
�

�
2
5
0
� � �

3
4

� � 1

8. For any nonzero value of x, �
2
x

� � 1 � �
3
x

�.

9. For any nonzero value of a, 1 �

�

1
a

� � �
a �

a
1

�.

10. For any value of a, a � �
1
4

� � �
4a

4
� 1
�.

dug84356_ch06a.qxd 9/17/10 8:08 PM Page 413

414 Chapter 6 Rational Expressions 6-34

U2V Addition and Subtraction of
Rational Expressions

Perform the indicated operation. Reduce each answer to lowest
terms. See Example 3.

19. �

2
1
x
� � �

2
1
x
� 20. �

3
1
y
� � �

3
2
y
�

21. �
2

w
� � �

2
7

w
� 22. �

5
3
x
y
� � �

7
3
x
y
�

23. �
a

3
�

a
5

� � �
a
1
�
5

5
� 24. �

a
a
�
�
7
4
� � �
9
a
�
�
5
4
a
�

25. �
q

q
�
�
1
4
� � �
3
q
q
�
�
4
9
�

26.

�
3 �

3
a
� � �
a �
3
5
�

27. �
h

4
(
h
h
�
�
3
1)
� � �

h
h
�
�
6
1)
�

28. �
t

2
(t
t
�
�
3
9
)
� � �

t
t
�
�
9
3)
�

29. �
(x

x
�
2 �
1)
x
(x
�
�
5
2)
��
(x �
1
1
�
)(x
2x

� 2)
�

30. �
(x �

2x
2)
�
(x
5

� 6)
��

(x
x2
�
�
2
2
)(
x
x
�
�
1
6)
�

Perform the indicated operation. Reduce each answer to
lowest terms. See Example 4.

31.

�
1
a

� � �
2
1
a
� 32. �

3
1
w
� � �

w
2

�

3. �
3
x

� � �
2
x

� 34. �
4
y

� � �
2
y

�

35. �
m
5

� � m 36. �
4
y

� � 2y

37. �
1
x

� � �
2
y

� 38.

�
2
a

� � �
3
b

�

39. �
2

3
a

� � �

5
1
a
� 40. �

6
5
y
� � �

8
3

y
�

41.

�

w �

9
3
� � �
w
1
�
2
4

� 42. �
y

1
�
0
4

� � �
y

1
�
4
2
�

43. �
4

b
a
2

� � c 44. y � �
7

3
b
�

45. �
w
2
z2
� � �

w
3
2z
� 46. �

a
1
5b
� � �

a
5
b3
�

Perform the indicated operation. Reduce each answer to lowest
terms. See Examples 5 and 6.

47. �
1
x

� � �
x �
1
2
�

48.

�
1
y

� � �
y �

2
1
�

49. �
x �

2
1
� � �
3
x
�

50. �
a �

1
1
� � �
2
a
�

51. �
a �

2
b
� � �
a �
1
b
�

52. �
x �

3
1
� � �
x �
2
1
�

53. �
x2

� x
� � �

5x
4
� 5
�

54. �
a2 �

3a
� � �

5a �

2
15
�

55. �
a2

2
�
a
9
� � �
a �
a
3
�

56. �
x2 �

x
1
� � �
x �
3
1
�

57. �
a �

4
b
� � �

b �

4
a
�

58. �
x �

2
3
� � �
3 �
3
x
�

59. �
2a

3
� 2
� � �
1 �
2
a
�

60. �
2x

5
� 4
� � �
2 �
3
x
�

61. �
x2 �

1
4
� � �

x2 � 3
3
x � 10
�

62. �
63. �

64. � �
x2 �

x �
5x
4
� 6
�

65. �
1
a

� �

�
1
b

� � �

1
c

�

66. �
1
x

� � �
x
1
2� � �x

1
3�

x � 1
��
x2 � x � 12

4
��
x2 � 2x � 3

3
��
x2 � x � 2

3x
��
x2 � 4x � 3

2x
�
x2 � 9

dug84356_ch06a.qxd 9/14/10 12:39 PM Page 414

6-35 6.4 Addition and Subtraction 415

67. �
2
x
� � �
x �
1
1
� � �
x �
1
2
�

68. �
1

a
� � �
a �
2
1
� � �
a �
3
1
�

69. �
3a

� 9
� � �

2
3
a
� � �
a2 �
4

3a
�

70. � �

Match each expression in (a)–(f) with the equivalent expression
in (A)–(F).

71. a) �
1
y

� � 2 b) �
1
y

� � �
2
y

� c) �
1
y

� � �
1
2
�

d) �
1
y

� � �
2
1
y
� e) �

2
y

� � 1 f ) �
2
y

� � 1

A) �
3
y

� B) �
2
3
y
� C) �

y �
2
2
�

D) �
y �

y
2

� E) �
y

2
�
y
2

� F) �
2y

y
� 1
�

72. a) �
1
x

� � x b) �
1
x

� � �
x
1
2� c) �

1
x
� � 1

d) �
x
1
2� � x e) x � �

1
x

� f )

�
x
1
2� � �

1
x
�

A) �
1 �

x2
x3

� B) �
1 �

x
x

� C) �
1 �

x
x2

�

D) �
1

x
�
2
x

� E) �
x2 �

x
1

� F) �
x �

x2
1
�

Perform the indicated operation. Reduce each answer to
lowest terms.

73. �
2
3
p
� �

74. �
2
3
y
� � �

2y
3
� 4
�

75. �
76. �
77. �
78. �

79. �
2

3
t

� � �

t �
2

2
� �

80. � �
2

�
n2 � n

2
�
n � 1

4
�
3n

3
�
t2 � 2t

6
�
m2 � 1

9
��
m2 � m � 2

1
��
b2 � 5b � 6

2
��
b2 � 4b � 3

12
�
w2 � 3w

4
�
w2 � w

3
��
a2 � 5a � 6

3
��
a2 � 3a � 2

1
�
2p � 8

5
�
6c

c � 4
�
2c2 � c

3
�
4c � 2

U3V Applications

Solve each problem. See Example 7.

81. Perimeter of a rectangle. Suppose that the length of a
rectangle is �3

x
� feet and its width is �

2
5
x
� feet. Find a rational

expression for the perimeter of the rectangle.

82. Perimeter of a triangle. The lengths of the sides of a

triangle are �1
x

�, �
2
1
x
�, and �

3
2
x
� meters. Find a rational expression

for the perimeter of the triangle.

1—
x

1—
2x

2—
3x

Figure for Exercise 82

83. Traveling time. Janet drove 120 miles at x mph before
6:00 A.M. After 6:00 A.M., she increased her speed by
5 mph and drove 195 additional miles. Use the fact that
T � �DR� to complete the following table.

Rate Time Distance

Before x �
m
hr

i
� 120 mi

After x � 5 �
m
hr

i
� 195 mi

Write a rational expression for her total traveling time.
Evaluate the expression for x � 60.

84. Traveling time. After leaving Moose Jaw, Hanson drove
200 kilometers at x km/hr and then decreased his speed by
20 km/hr and drove 240 additional kilometers. Make a table
like the one in Exercise 83. Write a rational expression for
his total traveling time. Evaluate the expression for x � 100.

85. House painting. Kent can paint a certain house by himself
in x days. His helper Keith can paint the same house by
himself in x � 3 days. Suppose that they work together on
the job for 2 days. To complete the table on the next page,
use the fact that the work completed is the product of the

dug84356_ch06a.qxd 9/14/10 12:39 PM Page 415

416 Chapter 6 Rational Expressions 6-36

rate and the time. Write a rational expression for the
fraction of the house that they complete by working
together for 2 days. Evaluate the expression for x � 6.

86. Barn painting. Melanie can paint a certain barn by herself
in x days. Her helper Melissa can paint the same barn by
herself in 2x days. Write a rational expression for the frac-
tion of the barn that they complete in one day by working
together. Evaluate the expression for x � 5.

Getting More Involved

87. Writing

Write a step-by-step procedure for adding rational
expressions.

88. Writing

Explain why fractions must have the same
denominator to be added. Use real-life examples.

Rate Time Work

Kent �
1
x

� �
d
jo

a
b
y

� 2 days

Keith �
x �

1
3
� �
d
jo
a
b
y
� 2 days

Photo for Exercise 86

Hundreds of years before humans even considered traveling beyond the earth,
Isaac Newton established the laws of gravity. So when Neil Armstrong made the
first human step onto the moon in 1969, he knew what amount of gravitational
force to expect. Let’s see how he knew.

Newton’s equation for the force of gravity between two objects is
F � G�m

d
1m

2
2�, where m1 and m2 are the masses of the objects (in kilograms), d is

the distance (in meters) between the centers of the two objects, and G is the
gravitational constant 6.67 10�11. To find the force of gravity for Armstrong
on earth, use 5.98 1024 kg for the mass of the earth, 6.378 106 m for the

radius of the earth, and 80 kg for Armstrong’s mass. We get

F � 6.67 10�11 �
784 Newtons.

To find the force of gravity for Armstrong on the moon, use 7.34 1022 kg for the mass of
the moon and 1.737 106 m for the radius of the moon. We get

F � 6.67 10�11 �
130 Newtons.

So the force of gravity for Armstrong on the moon was about one-sixth of the force of gravity
for Armstrong on earth. Fortunately, the moon is smaller than the earth. Walking on a planet
much larger than the earth would present a real problem in terms of gravitational force.

7.34 1022 kg � 80 kg
��

(1.737 106 m)2

5.98 1024 kg � 80 kg
��

(6.378 106 m)2

Math at Work Gravity on the Moon

dug84356_ch06a.qxd 9/14/10 12:39 PM Page 416

In This Section

U1V Complex Fractions

U2V Using the LCD to Simplify
Complex Fractions

U3V Applications
6.5 Complex Fractions

In this section, we will use the idea of least common denominator to simplify
complex fractions. Also we will see how complex fractions can arise in applications.

U1V Complex Fractions
A complex fraction is a fraction having rational expressions in the numerator,
denominator, or both. Consider the following complex fraction:

Since the fraction bar is a grouping symbol, we can compute the value of the numer-
ator, the value of the denominator, and then divide them, as shown in Example 1.

← Numerator of complex fraction

← Denominator of complex fraction

�
1
2
� � �
2
3
�
�
�
1
4

� � �
5
8

�

6-37 6.5 Complex Fractions 417

Mid-Chapter Quiz Sections 6.1 through 6.4 Chapter 6

Reduce to lowest terms.

1. �
3
8
6
4
� 2. �

8x �
8

2
�

3. �
2
w
w

2 �
�
1
2

� 4. �
2a2 �

6 �
10

3
a
a
� 12

�
Perform the indicated operation.

5. �
6
7

� � �
2
1
1
0
� 6. �

3
5
x
z
y2
� � �

8
8
x
y

2z
4

3
�

7. �
2
a
a

2 �
�
9
4

� � �
5
2

a
a
�

�

1
6
0

� 8. �

b
3

2
� � �

b
1

6
�

9. �
5
9

� � �
2
3
5

3
� 10. �

3x
8
� 9
� � �

x2 �
1

6
2
x � 9
�

11. �
s
3

2
� � �

2
s
1

2
� 12. �

m2 �
2
8
m
m � 7
� � (m � 7)

13. �
5
6

� � �
2
5
1
� 14. �

a
4
b3
� � �

a
5
2b
�

15. �
x
3
�
x
1

� � �
x2 � 2

x
x � 1
� 16. �

y �
y

5
� � �
y �
y
2
�

17. �
1
a

� � �
1
b

� � �
1
c
�

Miscellaneous.
18. What numbers(s) can’t be used in place of x in �

3
2
x
x
�
�
6
1

�?

19. Find the value of �
3
2
x
x

�
�
6
1

� when x � �2.

20. Find R(�1) if R(x) � �
6
5
x
x

2
�
�
1
3
�.

dug84356_ch06a.qxd 9/14/10 12:39 PM Page 417

418 Chapter 6 Rational Expressions 6-38

Strategy for Simplifying a Complex Fraction

1. Find the LCD for all the denominators in the complex fraction.

2. Multiply both the numerator and the denominator of the complex fraction by
the LCD. Use the distributive property if necessary.

3. Combine like terms if possible.

4. Reduce to lowest terms when possible.

E X A M P L E 2 Using the LCD to simplify a complex fraction
Use the LCD to simplify

.
�
1
2

� � �
2
3
�
—
�
1
4
� � �
5
8
�

Simplifying complex fractions
Simplify.

a) b)

Solution
a) Combine the fractions in the numerator:

�
1
2
� � �
2
3
� � �
1
2
�
�
3
3
� � �
2
3
�
�
2
2

� � �
3
6

� � �
4
6

� �

�
7
6

�

Combine the fractions in the denominator as follows:

�
1
4
� � �
5
8
� � �
1
4
�
�
2
2
� � �
5
8

� � �
2
8

� � �
5
8

� � ��
3
8

�

Now divide the numerator by the denominator:

� � �
7
6

� � ��38�� 76� � ��83�� 5168� � ��298�

b) � � � �

1
5
8
� � �

3
1
1
0
� � �

1
5
8
� � �

1
3
0
1
� � �

3
3

6
1
�

Now do Exercises 1–12

�
1
5
8
�

—

�
3
1
1
0
�

�
2
5
0
� � �
2
5
�
—

�
1
1
0
� � �

3
1
0
0
�

4 � �
2
5

�
—

�
1
1
0
� � 3

�
7
6
�

—
��

3
8�

�
1
2
� � �
2
3
�
—
�
1
4
� � �
5
8
�
4 � �
2
5
�
—
�
1
1
0
� � 3
�
1
2
� � �
2
3
�
—
�
1
4
� � �
5
8
�
E X A M P L E 1

U2V Using the LCD to Simplify Complex Fractions

A complex fraction can be simplified by performing the operations in the numerator
and denominator, and then dividing the results, as shown in Example 1. However,
there is a better method. All of the fractions in the complex fraction can be eliminated
in one step by multiplying by the LCD of all of the single fractions. The strategy for
this method is detailed in the following box and illustrated in Example 2.

dug84356_ch06b.qxd 9/14/10 12:44 PM Page 418

6-39 6.5 Complex Fractions 419

E X A M P L E 3 A complex fraction with variables
Simplify

Solution
The LCD of the denominators x, x2, and 2 is 2×2:

� Multiply the numerator and denominator by 2×2.

� Distributive property

2 � 2x

2 � �
1
x

� � 2×2

——

�
x
1
2� � 2x

2 � �
1
2

� � 2×2

�2 � �1x��(2×2)
——

��x
1
2� � �

1
2

��(2×2)
2 � �

1
x
�
—
�
x
1
2� � �
1
2
�
2 � �
1
x
�
�
�
x
1
2� � �
1
2
�
U Helpful Hint V

When students see addition or sub-
traction in a complex fraction, they
often convert all fractions to the
same denominator. This is not wrong,
but it is not necessary. Simply multi-
plying every fraction by the LCD elim-
inates the denominators of the
original fractions.

Solution
The LCD of 2, 3, 4, and 8 is 24. Now multiply the numerator and denominator of the
complex fraction by the LCD:

� Multiply the numerator and denominator by the LCD.

� Distributive property

� �
1
6
2
�

�

1
1
5
6

� Simplify.
� �
�

28
9
�

� ��
2
9
8
�

Now do Exercises 13–20

�
1
2

� � 24 � �
2
3

� � 24

——
�
1
4

� � 24 � �
5
8

� � 24

��12� � �23��24
——

��14� � �58��

24
�

1
2

� � �
2
3
�
—
�
1
4
� � �
5
8
�

We simplify a complex fraction by multiplying the numerator and denomi-
nator of the complex fraction by the LCD. Do not multiply the numerator
and denominator of each fraction in the complex fraction by the LCD.

In Example 3 we simplify a complex fraction involving variables.

CAUTION
U Calculator Close-Up V

You can check Example 2 with a
calculator as shown here.

dug84356_ch06b.qxd 9/14/10 12:44 PM Page 419

420 Chapter 6 Rational Expressions 6-40

� �
4
2
x2

�
�

x
2
2

x
� Simplify.

The numerator of this answer can be factored, but the rational expression cannot be reduced.

Now do Exercises 21–30

E X A M P L E 4 Simplifying a complex fraction
Simplify

Solution
Because x � 2 and 2 � x are opposites, we can use (x � 2)(x � 2) as the LCD. Multiply
the numerator and denominator by (x � 2)(x � 2):

�
� �
x
2
�
�
2

x
� � �1

��
�

x
3x

�
�

2
6
�

�

2x
4x

�
�
4
8
� Distributive property
� �
�
x
x
�
�

14
6

� Combine like terms.

Now do Exercises 31–46

x � 2 � 2(x � 2)
��
3(�1)(x � 2) � 4(x � 2)

�
x �
1
2

� (x � 2)(x � 2) � �
x �

2
2
� (x � 2)(x � 2)

——

———

�
2 �

3
x
� (x � 2)(x � 2) � �
x �
4
2
� (x � 2)(x � 2)
�
x �
1
2
� � �
x �
2
2
�
——
�
2 �
3
x
� � �
x �
4
2
�
�
x �
1
2
� � �
x �
2
2
�
——
�
2 �
3
x
� � �
x �
4
2
�

U3V Applications
As their name suggests, complex fractions arise in some fairly complex situations.

E X A M P L E 5 Fast-food workers
A survey of college students found that �1

2
� of the female students had jobs and �2

3
� of the male

students had jobs. It was also found that �1
4

� of the female students worked in fast-food

restaurants and �1
6

� of the male students worked in fast-food restaurants. If equal numbers of

male and female students were surveyed, then what fraction of the working students

worked in fast-food restaurants?

Solution
Let x represent the number of males surveyed. The number of females surveyed is also x.
The total number of students working in fast-food restaurants is

�
1
4

� x � �
1
6

� x.

dug84356_ch06b.qxd 9/14/10 12:44 PM Page 420

6-41 6.5 Complex Fractions 421

The total number of working students in the survey is

�
1
2

� x � �
2
3

� x.

So the fraction of working students who work in fast-food restaurants is

The LCD of the denominators 2, 3, 4, and 6 is 12. Multiply the numerator and denominator
by 12 to eliminate the fractions as follows:

� Multiply numerator and

denominator by 12.

� �
3
6
x
x

�
�

2
8
x
x

� Distributive property

� �
1
5
4
x
x

� Combine like terms.

� �
1
5
4
� Reduce.

So �
1
5
4
� (or about 36%) of the working students work in fast-food restaurants.

Now do Exercises 61–62

��14

�x � �
1
6

�x�12
——

��12

�x � �
2
3

�x�12
�
1
4

�x � �
1
6
�x
—
�
1
2
�x � �
2
3
�x
�
1
4
�x � �
1
6
�x
�
�
1
2
�x � �
2
3
�x
Warm-Ups ▼

Fill in the blank.
1. A fraction has fractions in its numerator,

denominator, or both.

2. To simplify a complex fraction, you can multiply its
and by the LCD of all of the

fractions.

True or false?
3. The LCD for the denominator 4, x, 6, and x2 is 12×3.

4. The LCD for the denominator a � b, 2b � 2a, and 6 is
6a � 6b.

5. To simplify , we multiply the numerator and

denominator by 12.

6. � �
6
3

�
�
4
2
�
7. �

8. For any real number x, � �
2
3
x
x

�
�
1
1
�.

x � �
1
2

�
—

x � �
1
3

�
�
5
6
�
—

�
1
1
2
�

�
1
2
� � �
1
3
�
—
�
1
4
� � �
1
6
�

��12� � �
1
3

��12

——

��14� � �
1
6

��12
�
1
2
� � �
1
3
�
—
�
1
4
� � �
1
6
�

dug84356_ch06b.qxd 9/14/10 12:44 PM Page 421

U1V Complex Fractions

Simplify each complex fraction. See Example 1.

1. 2. 3.

4. 5. 6.

7. 8. 9.

10. 11.

12.
U2V Using the LCD to Simplify Complex Fractions

Simplify each complex fraction. See Examples 2 and 3. See
the Strategy for Simplifying a Complex Fraction box on
page 418.

13. 14.

15.

16.

�
1
3
0
� � �

4
5
�
—
�
1
2
� � �
3
4
�
�
2
5

� � �
1
1
0
�

—
�
1
5
� � �
1
4
�
�
1
4
� � �
1
3
�
—
�
1
4
� � �
1
6
�
�
1
2
� � �
1
3
�
—
�
1
2
� � �
1
4
�

3 � �
2
9

� � �
1
6
�
——

�
1
5
8
� � �

1
3
� � 2

1 � �
1
6

� � �
2
3
�
——

1 � �
1
1
5
� � �

1
3
0
�

1 � �
1
1
2
�

—
1 � �
1
1
2
�

3 � �
1
2

�
—

5 � �
3
4

�
�
1
3

� � 1
—

�
1
6
� � 2

1 � �
1
2

�

—
2 � �

1
4
�
�
2
5

� � �
2
9

� � �
1
3
�
——
�
1
3

� � �
1
5

� �

�
1
2
5
�

�
2
5
� � �
5
6
� � �
1
2
�
——
�
1
2
� � �
1
3

� � �
1
1
5
�

�
1
3
� � �
1
4
�
—

�
1
3� � �

1
6�
�
1
2
� � �
1
3
�
—
�
1
4
� � �
1
2
�
�
1
3
� � �
5
6
�
—

�
2
3� � �

1
6�
�
1
2
� � �
1
4
�
—

�
1
2� � �

3
4�

17. 18.

19. 20.

21. 22. 23.

24. 25. 26.

27. 28.

29. 30.

Simplify each complex fraction. See Example 4.

31. 32.

33. 34.
2 � �

a �
1
3
�
——

3 � �
a �

1
3
�

1 � �
y �

3
1
�
——

3 � �
y �

1
1
�
�
x �
2
3

� � 1
——
�
x �

4
3
� � 2
�
x �
1
1
� � 1
——
�
x �
3
1
� � 3

�
2
3
w
� � �

3
4
w
�

�

�
4
1
w
� � �

9
5
w
�

�
2
3
b
� � �

1
b
�
�
�
3
4

� � �
b
1
2�

�
2
a

� � �
5
3

�
—

�
3
a

� � �
a
3
2�

�
1
2
� � �
2
x
�
—

3 � �

x
1
2�

4 � �
3
y�

—

1 � �
2
y

�

5 � �
3
a

�
—

3 � �
1
a

�
�
1
x
� � �
3
2
�
—
�
3
4

� � �
1
x

�
�
1
a
� � �
3
b
�
—
�
1
b

� � �
3
a

�
�
1
x

� � �
1
y

�

—
�
3
x

� � �
3
y

�
�
1
a
� � �
1
b
�
—
�
2
a

� � �
2
b

�
�
2
5
� � �
3
2

� � �
1
7
0
�

——
�
1
5
� � �
1
2
� � �
1
1
0
�
�
2
3
� � �
5
6
� � �
1
2
�
——
�
1
6
� � �
1
3
� � �
1
2
�

3 � �
3
5

� � �
1
1
0
�
——

2 � �
6
5

� � �
1
3
0
�

1 � �
2
3

� � �
1
2
�
——

2 � �
1
3

� � �
3
2
�
Exercises

U Study Tips V
• Stay calm and confident. Take breaks when you study. Get 6 to 8 hours of sleep every night.
• Keep reminding yourself that working hard throughout the semester will really pay off in the end.

6
.5

dug84356_ch06b.qxd 9/17/10 8:11 PM Page 422

6-43 6.5 Complex Fractions 423

35. 36.

37. 38.

39. 40.

41. 42.

43. 44.

45. 46.

Simplify each complex fraction. Reduce each answer to
lowest terms.

47. 48.

49. 50.

51. 52.
�
1
2

� � �
2
3
x
� � �
x
1
2�

—

—
�
1
2

� � �
2
1
x2
�

�
1
3

� � �
3
5

x
� � �

x
2
2�

——
�
1
3

� � �
x
3
2�

�
1
9

� � �

3
1
x

�

—
�
9
x

� � �
1
x
�
�
1
2

� � �
4
1
x
�

—
�
3
x

� � �
1
1
2x
�

�
1
3
� � �
1
y
�

—
�
3
y

� � �
3
y
�

1 � �
a
4
2�

——

1 � �
2
a

� � �
a
8
2�

�
2 �
1
x

� � �
2 �

1
x
�
——
�
x �
1
2
� � �
x �
1
2
�
�
a �
1
b
� � �
a �
1
b
�
——

�
b �

1
a

� � �
b �

1
a
�
�
x �
1
2
� � �
x �
3
3
�
——
�
x �
2
3
� � �
x �
3
2
�
�
w �
2
1

� � �
w �

3
1
�
——
�
w �
4
1
� � �
w �
5
1
�
�
y �
1
3

� � �
4
y

�
��
�
1
y
� � �
y �
2
3
�
�
m �
1
3

� � �
m
4

�
��
�
m �
3
3

� � �
m
1

�
�
1
3

� � �
9 �

2
x
�
——
�
1
6
� � �
x �
1
9
�

1 � �
a �

5
1
�

——
3 � �

1 �
2
a
�
�
x �
2
5

� � x
——

�
5
3
�
x
x
� � 1
�
3 �
1
x

� � 5
——

�
x �
1
3
� � 2

x � �
x
x

�
�
6
1
�
——

x � �
x
x
�

�
1
1
5
�

x � �
x �

4
2
�
——
x � �
x
x
�
�
1
2
�
53. 54.
55. 56.

57. 58.

59. 60.

U3V Applications

Solve each problem. See Example 5.

61. Sophomore math. A survey of college sophomores showed
that �5

6
� of the males were taking a mathematics class and �3

4
� of

the females were taking a mathematics class. One-third of the
males were enrolled in calculus, and �1

5
� of the females were

enrolled in calculus. If just as many males as females were
surveyed, then what fraction of the surveyed students taking
mathematics were enrolled in calculus? Rework this prob-
lem assuming that the number of females in the survey

was twice the number of males.

62. Commuting students. At a well-known university, �1
4

� of the

undergraduate students commute, and �1
3

� of the graduate

students commute. One-tenth of the undergraduate students

drive more than 40 miles daily, and �1
6

� of the graduate

students drive more than 40 miles daily. If there are twice
as many undergraduate students as there are graduate
students, then what fraction of the commuters drive more

than 40 miles daily?

Photo for Exercise 62

�
a2 �
a
b2
�
——
�
a �
1
b
� � �
a �
1
b
�
�
x �
x
1
�

——
�
x2 �

1
1
� � �
x �
1
1
�

�
y2 �

y2
3
�

y �
4

18
�

——
�
y2 �

y �
5y

2
� 6

�
�
a2 �

a
2
�

a �
1
24
�
——
�
a2
(
�

a �
a �

1)2
12

�

�
ab

4a
�

b5
b2

�
—

�
a
6a

�
2b

b
4�

�
2x
x
�

y2
4y

�
—
�
3x

x
�
3y

6y
�

�
a
1
�
2
5
�
—
�
a
1
�
5
2
�
�
2x

6
� 9
�

—
�
2x

9
� 3
�

dug84356_ch06b.qxd 9/14/10 12:44 PM Page 423

424 Chapter 6 Rational Expressions 6-44

Getting More Involved

63. Exploration

Simplify

, , and .

a) Are these fractions getting larger or smaller as the
fractions become more complex?

b) Continuing the pattern, find the next two complex

fractions and simplify them.

c) Now what can you say about the values of all five
complex fractions?

1 � 1

1 � �12�

1 �
1

1 � �12�

1
—
1 � �12�

64. Discussion

A complex fraction can be simplified by writing the
numerator and denominator as single fractions and then
dividing them or by multiplying the numerator and
denominator by the LCD. Simplify the complex fraction

by using each of these methods. Compare the number of
steps used in each method, and determine which method
requires fewer steps.

�
x
4
y2
� � �

x
6
y
�

—

�
x
2
2� � �x

4
2y
�

In This Section

U1V Equations with Rational
Expressions

U2V Extraneous Solutions

6.6 Solving Equations with Rational Expressions

Many problems in algebra can be solved by using equations involving rational
expressions. In this section you will learn how to solve equations that involve
rational expressions, and in Sections 6.7 and 6.8 you will solve problems using
these equations.

U1V Equations with Rational Expressions

We solved some equations involving fractions in Section 2.3. In that section, the
equations had only integers in the denominators. Our first step in solving those
equations was to multiply by the LCD to eliminate all of the denominators.

E X A M P L E 1 Integers in the denominators
Solve �1

2
� � �
x �
3
2
� � �
1
6
�.

Solution
The LCD for 2, 3, and 6 is 6. Multiply each side of the equation by 6:

�
1
2
� � �
x �
3
2
� � �
1
6

�

Original equation

6��12� � �
x �

3
2

�� 6 � �16� Multiply each side by 6.
6 � �

1
2

� � 6

2
� � �x �

�3
2

� � �6 � �
1
�6

� Distributive property

3 � 2(x � 2) � 1 Simplify.

3 � 2x � 4 � 1 Distributive property

�2x � �6 Subtract 7 from each side.

x � 3 Divide each side by �2.

U Helpful Hint V

Note that it is not necessary to
convert each fraction into an equiva-
lent fraction with a common
denominator here. Since we can
multiply both sides of an equation by
any expression we choose,we choose
to multiply by the LCD. This tactic
eliminates the fractions in one step.

dug84356_ch06b.qxd 9/14/10 12:44 PM Page 424

E X A M P L E 2 Variables in the denominators
Solve �1

x
� � �
1
6
� � �
1
4
�.

Solution
We multiply each side of the equation by 12x, the LCD for 4, 6, and x :

�
1
x
� � �
1
6
� � �
1
4

� Original equation

12x � �1x� � �
1
6

�� 12x ��14�� Multiply each side by 12x.

12x� � �
1
x�

� � 12�
2
x � �

6
1
�

� � 12�
3
x � �

4
1
�

� Distributive property

12 � 2x � 3x Simplify.

12 � x Subtract 2x from each side.

Check that 12 satisfies the original equation:

�
1
1
2
� � �
1
6

� � �
1
1
2
� � �

1
2
2
� � �

1
3
2
� � �

1
4
�

The solution to the equation is 12.

Now do Exercises 13–24

When a numerator contains a binomial, as in Example 1, the numer-
ator must be enclosed in parentheses when the denominator is
eliminated.

To solve an equation involving rational expressions, we usually multiply each side
of the equation by the LCD for all the denominators involved, just as we do for an
equation with fractions.

CAUTION

6-45 6.6 Solving Equations with Rational Expressions 425

Check x � 3 in the original equation:

�
1
2

� � �
3 �

3
2
� � �
1
2
� � �
1
3
� � �
3
6

� � �
2
6

� � �
1
6
�

Since the right-hand side of the equation is �1
6

�, you can be sure that the solution to the
equation is 3.

Now do Exercises 1–12
U Helpful Hint V

Always check your solution in the
original equation by calculating
the value of the left-hand side and
the value of the right-hand side.
If they are the same, your solution
is correct.

E X A M P L E 3 An equation with two solutions
Solve the equation �10

x
0

� �

�
x
1
�

00
5

� � 9.

dug84356_ch06b.qxd 9/14/10 12:44 PM Page 425

U2V Extraneous Solutions
In a rational expression, we can replace the variable only by real numbers that do not
cause the denominator to be 0. When solving equations involving rational expressions,
we must check every solution to see whether it causes 0 to appear in a denominator.
If a number causes the denominator to be 0, then it cannot be a solution to the equa-
tion. A number that appears to be a solution but causes 0 in a denominator is called an
extraneous solution. Since a solution to an equation is sometimes called a root to the
equation, an extraneous solution is also called an extraneous root.

426 Chapter 6 Rational Expressions 6-46

Solution
The LCD for the denominators x and x � 5 is x(x � 5):

�
10
x
0
� � �
x
1
�
00
5

� � 9

x (x � 5) � x(x � 5)�
x
1
�

00
5

� � x(x � 5)9

(x � 5)100 � x(100) � (x2 � 5x)9
100x � 500 � 100x � 9×2 � 45x

500 � 200x � 9×2 � 45x

0 � 9×2 � 155x � 500

0 � (9x � 25)(x � 20)

9x � 25 � 0 or x � 20 � 0

x � ��
2
9
5
� or x � 20

A check will show that both ��
2
9
5
� and 20 satisfy the original equation.

Now do Exercises 25–32

100
�

x
Original equation

Multiply each side by
x(x � 5).

All denominators are
eliminated.
Simplify.

Get 0 on one side.

Factor.
Zero factor property

E X A M P L E 4 An equation with an extraneous solution
Solve the equation �

x �
1
2
� � �

2x
x
� 4
� � 1.

Solution
Because the denominator 2x � 4 factors as 2(x � 2), the LCD is 2(x � 2).

2(x � 2)�
x �

1
2

� � 2(x � 2)�
2(x

x
� 2)
� � 2(x � 2) � 1 Multiply each side of the original equation by 2(x � 2).

2 � x � 2x � 4 Simplify.

2 � 3x � 4

6 � 3x

2 � x

Check 2 in the original equation:

�
2 �
1
2

� � �
2 � 2

2
� 4
� � 1

The denominator 2 � 2 is 0. So 2 does not satisfy the equation, and it is an extraneous
solution. The equation has no solutions.

Now do Exercises 33–36

dug84356_ch06b.qxd 9/14/10 12:44 PM Page 426

Always be sure to check your answers in the original equation to determine
whether they are extraneous solutions.

CAUTION

6-47 6.6 Solving Equations with Rational Expressions 427

Warm-Ups ▼

Fill in the blank.
1. The usual first step in solving an equation involving

rational expressions is to multiply by the .

2. An solution is a number that appears to be a
solution but does not check in the original equation.

True or false?
3. To solve x2 � 8x, we divide each side by x.

4. An extraneous solution is an irrational number.

5. Both 0 and 2 satisfy �
3
x

� � �
x �
5
2
� � �
2
3
�.

If the denominators of the rational expressions in an equation are not too compli-
cated, you can tell at a glance which numbers cannot be solutions. For example, the

equation �
2
x

� � �x �
3

1
� � �
x
x
�
�
2
5

� could not have 0, 1, or �5 as a solution. Any solution to this

equation must be in the domain of all three of the rational expressions in the equation.

E X A M P L E 5 Another extraneous solution
Solve the equation �1

x
� � �
x �
1
3
� � �
x
x
�
�
2
3
�.

Solution
The LCD for the denominators x and x � 3 is x(x � 3):

�
1
x
� � �
x �
1
3
� � �
x
x
�
�
2
3
� Original equation

x (x � 3) � �
1
x

� � x(x � 3) � �
x �

1
3

� � x(x � 3) � �
x
x

�
�
2
3

� Multiply each side by x(x � 3).

x � 3 � x � x(x � 2)

2x � 3 � x2 � 2x

0 � x2 � 4x � 3

0 � (x � 3)(x � 1)

x � 3 � 0 or x � 1 � 0

x � 3 or x � 1

If x � 3, then the denominator x � 3 has a value of 0. If x � 1, the original equation is
satisfied. The only solution to the equation is 1.

Now do Exercises 37–40

6. To solve �
3
x

� � �
x �
5
2
� � �
2
3

�, multiply each side by 3×2 � 6x.

7. To solve �
x �

1
1

� � 2 � �
x �

1
1

�, multiply each side by

x2 – 1.

8. The solution set to �
x �

1
1
� � 2 � �
x �
1
1

� is {�1, 1}.

9. The solution set to �
1
x

� � �
1
2

� � �
3
x

� is {4}.

dug84356_ch06b.qxd 9/14/10 12:44 PM Page 427

U1V Equations with Rational Expressions

Solve each equation. See Example 1.

1. �
2
x

� � 1 � �
4
x

� 2. �
3
x

� � 2 � �
6
x

�
3. �
3
x

� � 5 � �
2
x

� � 7 4. �
3
x

� � �
2
x

� � �
5
x

� � 11

5. �
5
y

� � �
2
3

� � �
6
y

� � �
1
3

� 6. �
6
z

� � �
5
4

� � �
2
z

� � �
3
4
�

7. �
3
4

� � �
t �

3
4

� � �
1
t
2
� 8. �

4
5

� � �
v

1
�
0
1
� � �
v
3
�
0
5
�

9. �
3
x

� � �
x �
4
1
� � �
x �

12
15

� 10. �
8
x

� � �
x
1
�
2
4

� � �
6x

2
�
4
5
�

11. �
1
5

� � �
w �

15
10

� � �
1
1
0
� � �

w �
6

1
�

12. �
q
5

� � �
q �

2
1

� � �
1
2
3
0
� � �

q �
4

1
�

Solve each equation. See Example 2.

13. �
1
x

� � �
1
2

� � 3 14. �
2
x

� � �
3
4
� � 5

15. �
1
x

� � �
2
x

� � 7 16. �
5
x

� � �
6
x

� � 12

17. �
1
x

� � �
1
2
� � �
3
4

� 18. �
3
x

� � �
1
4
� � �
5
8
�

19. �
3
2
x
� � �

2
1
x
� � �

2
7
4
� 20. �

6
1
x
� � �

8
1
x
� � �

7
1
2
�

21. �
1
2

� � �
a �
a
2
� � �
a
2
�
a
2
�

22. �
1
b

� � �
1
5
� � �
b
5
�
b
1
� � �
1
3
0
�

23. �
1
3

� � �
k

6
�

k
3

� � �
3
1
k
� � �

k
2
�

k
1

�

24. �
3
p

� � �
p

3
�

p
3

� � �
2p

2
�

p
1

� � �
5
6
�

Solve each equation. See Example 3.

25. �
2
x

� � �
x �
5
3

� 26. �
3
x

� � �
x �
4
1
�
27. �
x �
x
1
� � �
x �
6
7

� 28. �
x �

x
3
� � �
x �
2
3
�

29. �
x �

2
1
� � �
1
x
� � �
1
6

� 30. �
w �

1
1

� � �
2
1
w
� � �

4
3
0
�

31. �
a
a
2

�
�
1
4
� � �
a �
1
2

� � �
a
a

�
�
4
2
�

32. �
b
b2

�
�

1
1
7

� � �
b �
1
1

� � �
b
b

�
�
2
1
�
U2V Extraneous Solutions

Solve each equation. Watch for extraneous solutions.
See Examples 4 and 5.

33. �
x �

1
1
� � �
2
x
� � �
x �
x
1
�

34. �
4
x

� � �
x �
3
3
� � �
x �
x
3
� � �
1
3
�

35. �
x �

5
2
� � �
x �
2
3
� � �
x
x
�
�
1
3
�

36. �
y �

6
2
� � �
y �

7
8

� � �
y
y

�
�
1
8
�

37. 1 � �
y

3
�
y
2
� � �
y �
6
2
�

38. �
y �

5
3

� � �
2
y
y
�

�
7
6
� � 1

39. �
z �

z
1

� � �
z �

1
2

� � �
z2 �

2z
3
�

z �
5

2
�

40. �
z �

z
2

� � �
z �
1
5

� � �
z2 � 3

7
z � 10
�

Exercises

U Study Tips V
• The last couple of weeks of the semester is not the time to slack off. This is the time to double your efforts.
• Make a schedule and plan every hour of your time.

6
.6

dug84356_ch06b.qxd 9/17/10 8:12 PM Page 428

6-49 6.7 Applications of Ratios and Proportions 429

Miscellaneous

Solve each equation.

41. �
a
4

� � �
5
2

� 42. �
3
y

� � �
6
5

�

43. �
w
6

� � �
3
1
w
1
� 44. �

2
3
m
� � �

3
2
m
�

45. �
5
x

� � �
5
x

� 46. �
�

x
3
� � �

�
x
3
�

47. �
x �

5
3
� � �
x �
x
3

� 48. �
a �

2
4
� � �
a �
a
4
�
49. �
x �
1
2
� � �
x �
x
2

� 50. �
w

�
�
3
2
� � �
w

w
� 2
�

51. �
2x

1
� 4
� � �

x �
1
2
� � �
3
2

� 52. �
3x

7
� 9
� � �

x �
1
3
� � �
4
3
�

53. �
a2 �

3
a � 6
� � �

a2
2
� 4
� 54. �

a2 �
8
a � 6
� � �

a2
6
� 9
�

55. �
c �

4
2
� � �
2 �
1
c

� � �
c

2
�
5
6
�

56. �
x �

3
1

� � �
1 �

1
x
� � �
x2
1
�
0
1
�

7. �
x2 �

1
9
� � �
x �
3
3
� � �
x �
4
3
�
58. �
x �
3
2
� � �
x �
5
3
� � �
x2 �

1
x � 6
�

59. �
2x

3
� 4
� � �

x �
1
2
� � �

3x
1
� 1
�

60. �
2m

5
� 6
� � �

m �
1

1
� � �
m �
1
3
�

61. �
2
3
t
t

�
�
1
3

� � �
3
6
t
t

�
�
1
6
� � �
t �

t
1

�

62. �
4
3
w
w

�
�
1
6
� � �
w �
3
1

� � �
w
w

�
�
1
2
�

Applications
Solve each problem.

63. Lens equation. The focal length f for a camera lens is
related to the object distance o and the image distance i by
the formula

�
1
f

� � �
1
o

� � �
1
i

�.

See the accompanying figure. The image is in focus at dis-
tance i from the lens. For an object that is 600 mm from a
50-mm lens, use f � 50 mm and o � 600 mm to find i.

Photo for Exercise 64

Figure for Exercise 63

In This Section

U1V Ratios

U2V Proportions

6.7 Applications of Ratios and Proportions

In this section, we will use the ideas of rational expressions in ratio and proportion
problems. We will solve proportions in the same way we solved equations in
Section 6.6.

U1V Ratios
In Chapter 1 we defined a rational number as the ratio of two integers. We will now
give a more general definition of ratio. If a and b are any real numbers (not just integers),
with b � 0, then the expression �a

b
� is called the ratio of a and b or the ratio of a to b.

64. Telephoto lens. Use the formula from Exercise 63 to find
the image distance i for an object that is 2,000,000 mm from
a 250-mm telephoto lens.

dug84356_ch06b.qxd 9/14/10 12:44 PM Page 429

E X A M P L E 1 Finding equivalent ratios
Find an equivalent ratio of integers in lowest terms for each ratio.

a) �
4
2
.
.
2
1
� b) c) �

3
5
.6
�

Solution
a) Because both the numerator and the denominator have one decimal place, we will

multiply the numerator and denominator by 10 to eliminate the decimals:

�
4
2
.
.
2
1
� � �

4
2
.
.
2
1
(
(
1
1
0
0

)
)

� � �
4
2
2
1
� � �

2
2
1
1
�
�
2
1
� � �
2
1

� Do not omit the 1 in a ratio.

So the ratio of 4.2 to 2.1 is equivalent to the ratio 2 to 1.

b) This ratio is a complex fraction. We can simplify this expression using the LCD
method as shown in Section 6.5. Multiply the numerator and denominator of this
ratio by 4:

� � �
1
2
�

c) We can get a ratio of integers if we multiply the numerator and denominator by 10.

�
3
5
.6
� � �

3
5
.6
(1
(1
0
0
)
)

� � �
3
5
6
0
�

� �
1
2
8
5
� Reduce to lowest terms.

Now do Exercises 1–16

�
1
4
� � 4
�
�
1
2
� � 4
�
1
4
�
�
�
1
2
�
�
1
4
�
—
�
1
2
�

430 Chapter 6 Rational Expressions 6-50

The ratio of a to b is also written as a :b. A ratio is a comparison of two numbers. Some
examples of ratios are

�
3
4

�, �
4
2
.
.
2
1
�, , �

3
5
.6
�, and �

10
1
0

�.

Ratios are treated just like fractions. We can reduce ratios, and we can build them up.
We generally express ratios as ratios of integers. When possible, we will convert a
ratio into an equivalent ratio of integers in lowest terms.

�
1
4
�
—
�
1
2
�

In Example 2, a ratio is used to compare quantities.

E X A M P L E 2 Nitrogen to potash
In a 50-pound bag of lawn fertilizer there are 8 pounds of nitrogen and 12 pounds of
potash. What is the ratio of nitrogen to potash?

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E X A M P L E 3 Males to females
In a class of 50 students, there were exactly 20 male students. What was the ratio of males
to females in this class?

Solution
Because there were 20 males in the class of 50, there were 30 females. The ratio of males
to females was 20 to 30, or 2 to 3.

Now do Exercises 19–20

6-51 6.7 Applications of Ratios and Proportions 431

Solution
The nitrogen and potash occur in this fertilizer in the ratio of 8 pounds to 12 pounds:

�
1
8
2
� � �

2
3
�
�

4
4
�
�

� � �
2
3
�

So the ratio of nitrogen to potash is 2 to 3.

Now do Exercises 17–18

Ratios give us a means of comparing the size of two quantities. For this reason the
numbers compared in a ratio should be expressed in the same units. For example, if
one dog is 24 inches high and another is 1 foot high, then the ratio of their heights is 2
to 1, not 24 to 1.

U2V Proportions
A proportion is any statement expressing the equality of two ratios. The statement

�
a
b

� � �
d
c

� or a :b � c :d

is a proportion. In any proportion the numbers in the positions of a and d shown here
are called the extremes. The numbers in the positions of b and c as shown are called
the means. In the proportion

�
3
2
0
4
� � �

5
4
� ,

the means are 24 and 5, and the extremes are 30 and 4.

Quantities with different units
What is the ratio of length to width for a poster with a length of 30 inches and a width of
2 feet?

Solution
Because the width is 2 feet, or 24 inches, the ratio of length to width is 30 to 24. Reduce
as follows:

�
3
2
0
4
� � �
5
4
�
�
6
6
� � �
5
4
�

So the ratio of length to width is 5 to 4.

Now do Exercises 21–24

E X A M P L E 4

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432 Chapter 6 Rational Expressions 6-52

If we multiply each side of the proportion

�
a
b
� � �
d
c
�

by the LCD, bd, we get

�
a
b

� � bd � �
d
c

� � bd

a � d � b � c.

We can express this result by saying that the product of the extremes is equal to
the product of the means. We call this fact the extremes-means property or cross-
multiplying.

Extremes-Means Property (Cross-Multiplying)

Suppose a, b, c, and d are real numbers with b � 0 and d � 0. If

�
a
b
� � �
d
c

�, then ad � bc.

We use the extremes-means property to solve proportions.

E X A M P L E 5 Using the extremes-means property
Solve the proportion �3

x
� � �

x �
5

5
� for x.

Solution
Instead of multiplying each side by the LCD, we use the extremes-means property:

�
3
x
� � �
x �
5
5

� Original proportion

3(x � 5) � 5x Extremes-means property

3x � 15 � 5x Distributive property

15 � 2x

�
1
2
5
� � x

Check:

� 3 � �
1
2
5
� � �

2
5
�

� � 5 � �
2
2
5
� � �

2
5
�

So �1
2
5
� is the solution to the equation or the solution to the proportion.

Now do Exercises 25–38

5
�

�
2
2
5
�

5
�

�
1
2
5
� � 5

3
�
�
1
2
5
�
U Helpful Hint V

The extremes-means property or
cross-multiplying is nothing new.
You can accomplish the same thing
by multiplying each side of the equa-
tion by the LCD.

dug84356_ch06b.qxd 9/14/10 12:44 PM Page 432

Note that any proportion can be solved by multiplying each side by the LCD
as we did when we solved other equations involving rational expressions. The
extremes-means property gives us a shortcut for solving proportions.

6-53 6.7 Applications of Ratios and Proportions 433

E X A M P L E 6 Solving a proportion
The ratio of men to women at Brighton City College is 2 to 3. If there are 894 men, then
how many women are there?

Solution
Because the ratio of men to women is 2 to 3, we have

�
N

N
um

um
be

b
r
e
o
r
f
o
w
f m

om
en

en
��

2
3
�.

If x represents the number of women, then we have the following proportion:

�
89

x
4
� � �
2
3
�

2x � 2682 Extremes-means property

x � 1341

The number of women is 1341.

Now do Exercises 39–42

E X A M P L E 7 Solving a proportion
In a conservative portfolio the ratio of the amount invested in bonds to the amount invested
in stocks should be 3 to 1. A conservative investor invested $2850 more in bonds than she
did in stocks. How much did she invest in each category?

Solution
Because the ratio of the amount invested in bonds to the amount invested in stocks is 3 to 1,
we have

� .

If x represents the amount invested in stocks and x � 2850 represents the amount invested
in bonds, then we can write and solve the following proportion:

�
x �

x
2850
� � �

3
1
�

3x � x � 2850 Extremes-means property

2x � 2850

x � 1425

x � 2850 � 4275

So she invested $4275 in bonds and $1425 in stocks. As a check, note that these amounts
are in the ratio of 3 to 1.

Now do Exercises 43–46

3
�
1

Amount invested in bonds
��
Amount invested in stocks

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434 Chapter 6 Rational Expressions 6-54

Example 8 shows how conversions from one unit of measurement to another can
be done by using proportions.

E X A M P L E 8 Converting measurements
There are 3 feet in 1 yard. How many feet are there in 12 yards?

Solution
Let x represent the number of feet in 12 yards. There are two proportions that we can
write to solve the problem:

� �

The ratios in the second proportion violate the rule of comparing only measurements that
are expressed in the same units. Note that each side of the second proportion is actually
the ratio 1 to 1, since 3 feet � 1 yard and x feet � 12 yards. For doing conversions we can
use ratios like this to compare measurements in different units. Applying the extremes-
means property to either proportion gives

3 � 12 � x � 1,
or

x � 36.

So there are 36 feet in 12 yards.

Now do Exercises 47–50

x feet
�
12 yards

3 feet
�
1 yard

1 yard
�
12 yards

3 feet
�
x feet

Warm-Ups ▼

Fill in the blank.
1. A is a comparison of two numbers.

2. A is an equation that expresses the equality
of two ratios.

3. In �
a
b

� � �
d
c

�, b and c are the .

4. In �
a
b

� � �
d
c

�, a and d are the .

5. The property says that if �
a
b

� � �
d
c

�, then
ad � bc.

True or false?
6. The ratio of 40 men to 30 women can be expressed as

the ratio 4 to 3.

7. The ratio of 3 feet to 2 yards can be expressed as the
ratio 3 to 2.

8. The ratio of 1.5 to 2 is equivalent to the ratio 3 to 4.

9. The product of the extremes is equal to the product of
the means.

10. If �
2
x

� � �
3
5

�, then 5x � 6.

11. If 4 of the 12 members of the supreme council are
women, then the ratio of men to women is 1 to 3.

dug84356_ch06b.qxd 9/14/10 12:45 PM Page 434

U1V Ratios

For each ratio, find an equivalent ratio of integers in lowest
terms. See Example 1.

1. �
4

� 2. �
1
2
0
0
� 3. �

2
1
0
5
0
0

�

4. �
1
2
0
0
0
0
0

� 5. �
2
3
.
.
5
5
� 6. �

4
1
.
.
8
2
�

�
0

0
.3
.6
2

� 8. �
0
0
.0
.8
5

� 9. �
3
1
5
0
�

10. �
8
3
8
3
� 11. �

4
7
.5
� 12. �

2
3
.5
�

13. 14. 15.

16.

Find a ratio for each of the following, and write it as a ratio of
integers in lowest terms. See Examples 2–4.

17. Men and women. Find the ratio of men to women in a
bowling league containing 12 men and 8 women.

18. Coffee drinkers. Among 100 coffee drinkers, 36 said
that they preferred their coffee black and the rest did
not prefer their coffee black. Find the ratio of those
who prefer black coffee to those who prefer nonblack
coffee.

4
�
�
1
4
�
5
�
�
1
3
�
�
2
3
�
�
�
3
4
�
�
1
2
�
�
�
1
5
�

19. Smokers. A life insurance company found that among
its last 200 claims, there were six dozen smokers. What
is the ratio of smokers to nonsmokers in this group of
claimants?

20. Hits and misses. A woman threw 60 darts and hit the
target a dozen times. What is her ratio of hits to misses?

21. Violence and kindness. While watching television for one
week, a consumer group counted 1240 acts of violence and
40 acts of kindness. What is the violence to kindness ratio
for television, according to this group?

22. Length to width. What is the ratio of length to width for
the rectangle shown?

Exercises

U Study Tips V
• Get an early start studying for your final exams.
• If you have several final exams, it can be difficult to find the time to prepare for all of them in the last couple of days.

6
.7

Photo for Exercise 18

2.5 ftW

8 in.

Figure for Exercise 22

24. Rise and run. If the rise is �3
2

� and the run is 5, then what is
the ratio of the rise to the run?

U2V Proportions

Solve each proportion. See Example 5.

25. �
4
x

� � �
2
3

� 26. �
9
x

� � �
3
2
�

23. Rise to run. What is the ratio of rise to run for the stair-
way shown in the figure?

1 ft

8 in.

Run

Rise

Figure for Exercise 23

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436 Chapter 6 Rational Expressions 6-56

43. Basketball blowout. As the final buzzer signaled the end
of the basketball game, the Lions were 34 points ahead of
the Tigers. If the Lions scored 5 points for every 3 scored
by the Tigers, then what was the final score?

44. The golden ratio. The ancient Greeks thought that the
most pleasing shape for a rectangle was one for which the
ratio of the length to the width was approximately 8 to 5,
the golden ratio. If the length of a rectangular painting is
2 ft longer than its width, then for what dimensions would
the length and width have the golden ratio?

45. Automobile sales. The ratio of sports cars to luxury cars
sold in Wentworth one month was 3 to 2. If 20 more sports
cars were sold than luxury cars, then how many of each
were sold that month?

46. Foxes and rabbits. The ratio of foxes to rabbits in the
Deerfield Forest Preserve is 2 to 9. If there are 35 fewer
foxes than rabbits, then how many of each are there?

47. Inches and feet. If there are 12 inches in 1 foot, then how
many inches are there in 7 feet?

48. Feet and yards. If there are 3 feet in 1 yard, then how

many yards are there in 28 feet?

49. Minutes and hours. If there are 60 minutes in 1 hour, then
how many minutes are there in 0.25 hour?

50. Meters and kilometers. If there are 1000 meters in
1 kilometer, then how many meters are there in
2.33 kilometers?

51. Miles and hours. If Alonzo travels 230 miles in 3 hours,
then how many miles does he travel in 7 hours?

52. Hiking time. If Evangelica can hike 19 miles in 2 days on
the Appalachian Trail, then how many days will it take her
to hike 63 miles?

53. Force on basketball shoes. The force exerted on shoe soles
in a jump shot is proportional to the weight of the person
jumping. If a 70-pound boy exerts a force of 980 pounds on
his shoe soles when he returns to the court after a jump,
then what force does a 6 ft 8 in. professional ball player
weighing 280 pounds exert on the soles of his shoes when
he returns to the court after a jump? Use the accompanying
graph to estimate the force for a 150-pound player.

27. �
a
2

� � �
�

5
1
� 28. �

b
3
� � �
�

4
3
�

29. ��
5
9

� � �
3
x

� 30. ��
3
4

� � �
5
x
�

31. �
x �

5
2

� �

�
7
x

� 32. �
x �

4
1
� � �
2
x
�

33. �
1
x
0
� � �

x �
34

12
� 34. �

3
x
� � �
x �
2
1
�

35. �
a �

a
1
� � �
a �
a
3

� 36. �

c
c

�
�
3
1

� � �
c
c

�
�
2
3
�

37. �
m
m

�
�
1
2

� � �
m
m

�
�
3
4

� 38. �
h �

h
3

� � �
h �

h
9

�

Use a proportion to solve each problem. See Examples 6–8.

39. New shows and reruns. The ratio of new shows to reruns
on cable TV is 2 to 27. If Frank counted only eight new
shows one evening, then how many reruns were there?

40. Fast food. If four out of five doctors prefer fast food,
then at a convention of 445 doctors, how many prefer fast
food?

41. Voting. If 220 out of 500 voters surveyed said that they
would vote for the incumbent, then how many votes could
the incumbent expect out of the 400,000 voters in the
state?

Photo for Exercise 41

42. New product. A taste test with 200 randomly selected
people found that only three of them said that they would
buy a box of new Sweet Wheats cereal. How many boxes
could the manufacturer expect to sell in a country of
280 million people?

dug84356_ch06b.qxd 9/14/10 12:45 PM Page 436

6-57 6.7 Applications of Ratios and Proportions 437

54. Force on running shoes. The ratio of the force on the
shoe soles to the weight of a runner is 3 to 1. What force
does a 130-pound jogger exert on the soles of her shoes?

55. Capture-recapture. To estimate the number of trout in
Trout Lake, rangers used the capture-recapture method.
They caught, tagged, and released 200 trout. One week
later, they caught a sample of 150 trout and found that
5 of them were tagged. Assuming that the ratio of tagged
trout to the total number of trout in the lake is the same
as the ratio of tagged trout in the sample to the number
of trout in the sample, find the number of trout in
the lake.

56. Bear population. To estimate the size of the bear
population on the Keweenaw Peninsula, conservationists
captured, tagged, and released 50 bears. One year later, a
random sample of 100 bears included only 2 tagged bears.
What is the conservationist’s estimate of the size of the
bear population?

57. Fast-food waste. The accompanying figure shows the
typical distribution of waste at a fast-food restaurant
(U.S. Environmental Protection Agency, www.epa.gov).

a) What is the ratio of customer waste to food waste?

b) If a typical McDonald’s generates 67 more pounds of
food waste than customer waste per day, then how
many pounds of customer waste does it generate?

58. Corrugated waste. Use the accompanying figure to find
the ratio of waste from corrugated shipping boxes to waste
not from corrugated shipping boxes. If a typical
McDonald’s generates 81 pounds of waste per day from
corrugated shipping boxes, then how many pounds of

WASTE GENERATION AT A FAST-FOOD RESTAURANT

8%
3%
4%
7%
4%
6%

Liquids, office paper, misc.
Plastic wraps, syrup containers
Uncoated paper (napkins)
Coated paper (sandwich wrap)
Polystyrene (hot cups, lids, etc.)
Customer’s waste (Diapers, etc.)

34% Corrugated shipping boxes

34% Food waste

Figure for Exercises 57 and 58

waste per day does it generate that is not from corrugated
shipping boxes?

59. Mascara needs. In determining warehouse needs for a
particular mascara for a chain of 2000 stores, Mike Pittman
first determines a need B based on sales figures for the past
52 weeks. He then determines the actual need A from the
equation �A

B
� � k, where

k � 1 � V � C � X � D.

He uses V � 0.22 if there is a national TV ad and V � 0
if not, C � 0.26 if there is a national coupon and C � 0 if
not, X � 0.36 if there is a chain-specific ad and X � 0 if
not, and D � 0.29 if there is a special display in the chain
and D � 0 if not. (D is subtracted because less product is
needed in the warehouse when more is on display in the
store.) If B � 4200 units and there is a special display and
a national coupon but no national TV ad and no chain-
specific ad, then what is the value of A?

Getting More Involved

60. Discussion

Which of the following equations is not a proportion?
Explain.

a) �
1
2

� � �
1
2

� b) �
x �

x
2

� � �
4
5

�

c) �
4
x

� � �
9
x

� d) �
x �

8
2

� � 1 � �
x �

5
2
�

61. Discussion

Find all of the errors in the following solution to an
equation.

�
7
x
� � �
x �
8
3
� � 1

7(x � 3) � 8x � 1

7x � 3 � 8x

�x � �3

x � 3
3
4
5
2
1

0 50 100 150 200 250 300

Weight (pounds)

Fo
rc

e
(t

ho
us
an
ds
o
f
po
un
ds
)

Figure for Exercise 53

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438 Chapter 6 Rational Expressions 6-58

In This Section

U1V Formulas

U2V Uniform Motion Problems

U3V Work Problems

U4V More Rate Problems

6.8 Applications of Rational Expressions

In this section, we will study additional applications of rational expressions.

U1V Formulas
Many formulas involve rational expressions. When solving a formula of this type
for a certain variable, we usually multiply each side by the LCD to eliminate the
denominators.

E X A M P L E 1 An equation of a line
The equation for the line through (�2, 4) with slope �3

2
� can be written as

�
y
x

�
�
4
2
� � �
3
2
�.

We studied equations of this type in Chapter 3. Solve this equation for y.

Solution
To isolate y on the left-hand side of the equation, we multiply each side by x � 2:

�
y
x
�
�
4
2
� � �
3
2
� Original equation

(x � 2) � �
y
x

�
�
4
2

� � (x � 2) � �
3
2

� Multiply by x � 2.

y � 4 � �
3
2

� x � 3 Simplify.

y � �
3
2

� x � 7 Add 4 to each side.

Because the original equation is a proportion, we could have used the extremes-means
property to solve it for y.

Now do Exercises 1–10

E X A M P L E 2 Distance, rate, and time
Solve the formula �D

T
� � R for T.

Solution
Because the only denominator is T, we multiply each side by T :

�
D
T

� � R Original formula

T � �
D
T

� � T � R Multiply each side by T.

D � T

�

D
R

� � �
T
R
R
� Divide each side by R.

�
D
R

� � T Simplify.

The formula solved for T is T � �
D
R

�.

Now do Exercises 11–16

dug84356_ch06b.qxd 9/14/10 12:45 PM Page 438

E X A M P L E 4 Finding the value of a variable
In the formula of Example 1, find x if y � �3.

Solution
Substitute y � �3 into the formula, then solve for x:

�
y
x
�
�
4
2
� � �
3
2

� Original formula

�
�
x
3
�
�
2
4
� � �
3
2

� Replace y by �3.

�
x
�

�
7
2
� � �
3
2
� Simplify.

3x � 6 � �14 Extremes-means property
3x � �20

x � ��
2
3
0
�

Now do Exercises 25–34

E X A M P L E 3 Total resistance
The formula

�

R
1

� �

�
R
1

1
� � �
R
1
2
�

(from physics) expresses the relationship between different amounts of resistance in a
parallel circuit. Solve it for R2.

Solution
The LCD for R, R1, and R2 is RR1R2:

�
R
1

� � �
R
1

1
� � �
R
1

2
� Original formula

RR1R2 � �R
1

� � RR1R2 � �R
1

1
� � RR1R2 � �R

2
� Multiply each side by the LCD, RR1R2.

R1R2 � RR2 � RR1 All denominators are eliminated.

R1R2 � RR2 � RR1 Get all terms involving R2 onto the left side.

R2(R1 � R) � RR1 Factor out R2.

R2 � �R1

R
�

R1
R

� Divide each side by R1 � R.

Now do Exercises 17–24

6-59 6.8 Applications of Rational Expressions 439

In Example 3, different subscripts are used on a variable to indicate that they are
different variables. Think of R1 as the first resistance, R2 as the second resistance, and
R as a combined resistance.

U2V Uniform Motion Problems
In uniform motion problems we use the formula D � RT. In some problems in which
the time is unknown, we can use the formula T � �

D
R

� to get an equation involving
rational expressions.

dug84356_ch06b.qxd 9/14/10 12:45 PM Page 439

U3V Work Problems
If you can complete a job in 3 hours, then you are working at the rate of �1

3
� of the job

per hour. If you work for 2 hours at the rate of �1
3

� of the job per hour, then you will

complete �2
3

� of the job. The product of the rate and time is the amount of work completed.

For problems involving work, we will always assume that the work is done at a constant
rate. So if a job takes x hours to complete, then the rate is �1

x
� of the job per hour.

440 Chapter 6 Rational Expressions 6-60

E X A M P L E 5 Driving to Florida
Susan drove 1500 miles to Daytona Beach for spring break. On the way back she averaged
10 miles per hour less, and the drive back took her 5 hours longer. Find Susan’s average
speed on the way to Daytona Beach.

Solution
If x represents her average speed going there, then x � 10 is her average speed for
the return trip. See Fig. 6.1. We use the formula T � �D

R
� to make the following table.

Because the difference between the two times is 5 hours, we have

longer time � shorter time � 5.

Using the time expressions from the table, we get the following equation:

�
x
1
�

50
1
0
0

� � �
15

x
00
� � 5

x(x � 10)�
x
1
�

50
1
0
0

� � x(x � 10)�
15

x
00
� � x(x � 10)5 Multiply by x(x � 10).

1500x � 1500(x � 10) � 5×2 � 50x

15,000 � 5×2 � 50x Simplify.

3000 � x2 � 10x Divide each side by 5.

0 � x2 � 10x � 3000

(x � 50)(x � 60) � 0 Factor.

x � 50 � 0 or x � 60 � 0

x � �50 or x � 60

The answer x � �50 is a solution to the equation, but it cannot indicate the average speed
of the car. Her average speed going to Daytona Beach was 60 mph.

Now do Exercises 35–40

← Shorter time

← Longer time

D R T

Going 1500 x �
15

x
00
�

Returning 1500 x � 10 �
x
1
�

50
1
0
0
�

1500 miles

Speed � x miles per hour

Speed � x � 10 miles per hour

Figure 6.1

U Helpful Hint V

Notice that a work rate is the same
as a slope from Chapter 3. The only
difference is that the work rates here
can contain a variable.

E X A M P L E 6 Shoveling snow
After a heavy snowfall, Brian can shovel all of the driveway in 30 minutes. If his younger
brother Allen helps, the job takes only 20 minutes. How long would it take Allen to do the
job by himself?

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6-61 6.8 Applications of Rational Expressions 441

Solution
Let x represent the number of minutes it would take Allen to do the job by himself. Brian’s
rate for shoveling is �

3
1
0
� of the driveway per minute, and Allen’s rate for shoveling is �1

x
� of

the driveway per minute. We organize all of the information in a table like the table in
Example 5.

If Brian works for 20 min at the rate �
3
1
0
� of the job per minute, then he does �2

3
0
0
� or �2

3
� of the

job, as shown in Fig. 6.2. The amount of work that each boy does is a fraction of the
whole job. So the expressions for work in the last column of the table have a sum of 1:

�
2
3

� � �
2
x
0
� � 1

3x � �
2
3

� � 3x � �
2
x
0
� � 3x � 1 Multiply each side by 3x.

2x � 60 � 3x

60 � x

If it takes Allen 60 min to do the job by himself, then he works at the rate of �
6
1
0
� of the job

per minute. In 20 minutes he does �1
3

� of the job while Brian does �2
3

�. So it would take Allen

60 minutes to shovel the driveway by himself.

Now do Exercises 41–46

20
x 2

Figure 6.2

Rate Time Work

Brian �
3
1
0
� �

m
jo

i
b
n

� 20 min �2
3

� job

Allen �1
x

� �
m
jo

i
b
n

� 20 min �2
x
0
� job

Notice the similarities between the uniform motion problem in Example 5 and
the work problem in Example 6. In both cases, it is beneficial to make a table. We use
D � R � T in uniform motion problems and W � R � T in work problems. The main
points to remember when solving work problems are summarized in the following
strategy.

U Helpful Hint V

The secret to work problems is
remembering that the individual
rates or the amounts of work can be
added when people work together.
If your painting rate is 1�10 of the
house per day and your helper’s rate
is 1�5 of the house per day, then your
rate together will be 3�10 of the
house per day. In 2 days you will paint
2�10 of the house and your helper
will paint 2�5 of the house for a total
of 3�5 of the house completed.

Strategy for Solving Work Problems

1. If a job is completed in x hours, then the rate is �1
x

� job/hr.

2. Make a table showing rate, time, and work completed (W � R � T ) for each
person or machine.

3. The total work completed is the sum of the individual amounts of work
completed.

4. If the job is completed, then the total work done is 1 job.

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442 Chapter 6 Rational Expressions 6-62

U4V More Rate Problems
Rates are used in uniform motion and work problems. But rates also occur in other

problems. If you make $400 for x hours of work, then your pay rate is �40
x
0

� dollars per

hour. If you get $50 for selling x pounds of apples, then you are making money at the

rate of �5
x
0
� dollars per pound.

E X A M P L E 7 Hourly rates
Dr. Watts paid $80 to her gardener and $80 to the gardener’s helper for a total of 12 hours
labor. If the gardener makes $10 more per hour than the helper, then how many hours did
each of them work?

Solution
Let x be the number of hours for the gardener and 12 � x be the number of hours for the
helper. Make a table as follows.

Time Pay Hourly Rate

Gardener x hours 80 dollars �8
x
0
� dollars/hour

Helper 12 � x hours 80 dollars �
12

8
�
0

x
� dollars/hour

Since the gardener makes $10 more per hour, we can write the following equation.

�
12
8
�
0
x

� � 10 � �
8
x
0
�

To solve the equation multiply each side by the LCD x(12 � x).

x(12 � x)��128�0 x� � 10� � x(12 � x)�8×0� Muliply by the LCD.
80x � 10x(12 � x) � (12 � x)80 Distributive property

80x � 120x � 10×2 � 960 � 80x Distributive property

�10×2 � 280x � 960 � 0 Get 0 on the right.

x2 � 28x � 96 � 0 Divide each side by �10.

(x � 4)(x � 24) � 0 Factor.

x � 4 � 0 or x � 24 � 0

x � 4 or x � 24

12 � x � 8 12 � x � �12

Since x � 24 hours and 12 � x � �12 hours does not make sense, we must have 4 hours
for the gardener and 8 hours for the helper. Check: The gardener worked 4 hours at
$20 per hour and the helper worked 8 hours at $10 per hour. The gardener made $10 more
per hour than the helper. Note that the problem could be solved also by starting with x as
the hourly pay for the gardener and x � 10 as the hourly pay for the helper. Try it.

Now do Exercises 47–48

E X A M P L E 8 Oranges and grapefruit
Tamara bought 50 pounds of fruit consisting of Florida oranges and Texas grapefruit. She paid
twice as much per pound for the grapefruit as she did for the oranges. If Tamara bought $12
worth of oranges and $16 worth of grapefruit, then how many pounds of each did she buy?

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6-63 6.8 Applications of Rational Expressions 443

Solution
Let x represent the number of pounds of oranges and 50 � x represent the number of
pounds of grapefruit. See Fig. 6.3. Make a table.

Rate Quantity Total Cost

Oranges �1
x
2
� dollars/pound x pounds 12 dollars

Grapefruit �
50

1
�
6
x

� dollars/pound 50 � x pounds 16 dollars

Since the price per pound for the grapefruit is twice that for the oranges, we have:

2(price per pound for oranges) � price per pound for grapefruit

2��1x
2
�� 50

1
�
6
x
�

�
2
x
4
� � �

50
1
�

6
x
�

16x � 1200 � 24x Extremes-means property

40x � 1200

x � 30

50 � x � 20

If Tamara purchased 20 pounds of grapefruit for $16, then she paid $0.80 per pound. If she
purchased 30 pounds of oranges for $12, then she paid $0.40 per pound. Because $0.80 is
twice $0.40, we can be sure that she purchased 20 pounds of grapefruit and 30 pounds of
oranges.

Now do Exercises 49–50

Grapefruit

50 �

x lb

Oranges

Figure 6.3

Warm-Ups ▼
True or false?

1. The formula t � �
1

m
� t
�, solved for m is m � �

1 �
t

t
�.

2. To solve �
m
1

� � �
1
n

� � �
1
2

� for m, we multiply each side by

2mn.

3. If Fiona drives 300 miles in x hours, then her average

speed is �
30

x
0
� mph.

4. If Mike drives 20 hard bargains in x hours, then he is

driving �
2
x
0
� hard bargains per hour.

5. If Fred can paint a house in y days, then he paints �
1
y

� of

the house per day.

6. If �
1
x

� is 1 less than �
x �

2
3

�, then �
1
x

� � 1 � �
x �
2
3
�.

7. If a and b are nonzero and a � �

m
b

�, then b � am.

8. If D � RT, then T � �
D
R

�.

9. Solving P � Prt � I for P yields P � I – Prt.

10. To solve 3R � yR � m for R, we must first factor the left
side.

dug84356_ch06b.qxd 9/14/10 12:45 PM Page 443

U1V Formulas

Solve each equation for y. See Example 1.

1. �
y
x

�
�
2
1

� � 3 2. �
y
x

�
�
5
2
� � 6

3. �
y

x
�
�
1

3
� � 2 4. �

y
x
�
�
2

4
� � �2

5. �
y

x
�
�
1

6
� � ��

2
� 6. �

y
x
�
�
5
2
� � ��
1
2
�

7. �
y

x
�
�
a

b
� � m 8. �

y
x
�
�
h

k
� � a

9. �
y

x
�
�
1

4
� � ��

1
3
� 10. �
y
x
�
�
1

3
� � ��

3
4
�

Solve each formula for the indicated variable. See Examples 2
and 3.

11. A � �
C

B
� for C 12. P � �

C �

D
� for A

13. �
1

a
� � m � �

p
� for p 14. �

f
� � t � �

3
� for m

15. F � k �
m

r
1m
2

2
� for m1 16. F � �

m
r
v2

� for r

17. �
1
a
� � �
1
b
� � �
1

f
� for a 18. �

R
1
� � �
R
1
1
� � �
R
1

2
� for R

19. S � �
1 �

r
� for r 20. I � �

� r
� for R

21. �

T
1V

1
1
� � �
P

T
2V

2
� for P2 22. �

P
T
1V
1
1
� � �
P
T
2V
2

2
� for T1

23. V � �
4

3
� �r2h for h 24. h � �

S �

2�
2
r

�r2
� for S

Find the value of the indicated variable. See Example 4.

25. In the formula of Exercise 11, if A � 12 and B � 5, find C.

26. In the formula of Exercise 12, if A � 500, P � 100, and
C � 2, find D.

27. In the formula of Exercise 13, if p � 6 and m � 4, find a.

28. In the formula of Exercise 14, if m � 4 and t � 3, find f.

29. In the formula of Exercise 15, if F � 32, r � 4, m1 � 2,

and m2 � 6, find k.

30. In the formula of Exercise 16, if F � 10, v � 8, and r � 6,

find m.

31. In the formula of Exercise 17, if f � 3 and a � 2, find b.

32. In the formula of Exercise 18, if R � 3 and R1 � 5,

find R2.

33. In the formula of Exercise 19, if S � �3
2

� and r � �1
5

�, find a.

34. In the formula of Exercise 20, if I � 15, E � 3, and
R � 2, find r.

Exercises

U Study Tips V
• Establish a regular routine of eating, sleeping, and exercise.
• The ability to concentrate depends on adequate sleep, decent nutrition, and the physical well-being that comes with exercise.

6
.8

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6-65 6.8 Applications of Rational Expressions 445

U2V Uniform Motion Problems

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

35. Fast walking. Marcie can walk 8 miles in the same time
as Frank walks 6 miles. If Marcie walks 1 mile per hour
faster than Frank, then how fast does each person walk?

36. Upstream, downstream. Junior’s boat will go 15 miles per
hour in still water. If he can go 12 miles downstream in the
same amount of time as it takes to go 9 miles upstream,
then what is the speed of the current?

37. Delivery routes. Pat travels 70 miles on her milk route, and
Bob travels 75 miles on his route. Pat travels 5 miles per
hour slower than Bob, and her route takes her one-half hour
longer than Bob’s. How fast is each one traveling?

38. Ride the peaks. Smith bicycled 45 miles going east from
Durango, and Jones bicycled 70 miles. Jones averaged
5 miles per hour more than Smith, and his trip took one-half
hour longer than Smith’s. How fast was each one traveling?

2 hours. How long would it take Red to paint the fence
by himself ?

42. Envelope stuffing. Every week, Linda must stuff 1000
envelopes. She can do the job by herself in 6 hours. If

Laura helps, they get the job done in 5�12� hours. How long

would it take Laura to do the job by herself ?

43. Garden destroying. Mr. McGregor has discovered that a
large dog can destroy his entire garden in 2 hours and that
a small boy can do the same job in 1 hour. How long
would it take the large dog and the small boy working
together to destroy Mr. McGregor’s garden?

44. Draining the vat. With only the small valve open, all of
the liquid can be drained from a large vat in 4 hours. With
only the large valve open, all of the liquid can be drained
from the same vat in 2 hours. How long would it take to
drain the vat with both valves open?

Photo for Exercise 38

45. Cleaning sidewalks. Edgar can blow the leaves off the
sidewalks around the capitol building in 2 hours using a
gasoline-powered blower. Ellen can do the same job in
8 hours using a broom. How long would it take them
working together?

46. Computer time. It takes a computer 8 days to print all of
the personalized letters for a national sweepstakes. A new
computer is purchased that can do the same job in 5 days.
How long would it take to do the job with both computers

working on it?

U4V More Rate Problems

Show a complete solution to each problem. See Examples 7 and 8.

47. Repair work. Sally received a bill for a total of 8 hours
labor on the repair of her bulldozer. She paid $50 to the
master mechanic and $90 to his apprentice. If the master
mechanic gets $10 more per hour than his apprentice, then
how many hours did each work on the bulldozer?

48. Running backs. In the playoff game the ball was carried
by either Anderson or Brown on 21 plays. Anderson gained
36 yards, and Brown gained 54 yards. If Brown averaged

39. Walking and running. Raffaele ran 8 miles and then
walked 6 miles. If he ran 5 miles per hour faster than
he walked and the total time was 2 hours, then how fast
did he walk?

40. Triathlon. Luisa participated in a triathlon in which she
swam 3 miles, ran 5 miles, and then bicycled 10 miles. Luisa
ran twice as fast as she swam, and she cycled three times as
fast as she swam. If her total time for the triathlon was 1 hour
and 46 minutes, then how fast did she swim?

U3V Work Problems

Show a complete solution to each problem. See Example 6. See
the Strategy for Solving Work Problems on page 441.

41. Fence painting. Kiyoshi can paint a certain fence in
3 hours by himself. If Red helps, the job takes only

Figure for Exercise 44

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446 Chapter 6 Rational Expressions 6-66

55. Two cyclists. Ben and Jerry start from the same point and
ride their bicycles in opposite directions. If Ben rides twice
as fast as Jerry and they are 90 miles apart after four hours,
then what is the speed of each rider?

56. Catching up. A sailboat leaves port and travels due south
at an average speed of 9 mph. Four hours later a motorboat
leaves the same port and travels due south at an average
speed of 21 mph. How long will it take the motorboat to
catch the sailboat?

57. Road trip. The Griswalds averaged 45 mph on their way
to Las Vegas and 60 mph on the way back home using the
same route. Find the distance from their home to Las Vegas
if the total driving time was 70 hours.

58. Meeting cyclists. Tanya and Lebron start at the same time
from opposite ends of a bicycle trail that is 81 miles long.
Tanya averages 12 mph and Lebron averages 15 mph. How
long does it take for them to meet?

59. Filling a fountain. Pete’s fountain can be filled using a
pipe or a hose. The fountain can be filled using the pipe in
6 hours or the hose in 12 hours. How long will it take to
fill the fountain using both the pipe and the hose?

60. Mowing a lawn. Albert can mow a lawn in 40 minutes,
while his cousin Vinnie can mow the same lawn in one
hour. How long would it take to mow the lawn if Albert
and Vinnie work together?

61. Printing a report. Debra plans to use two computers to
print all of the copies of the annual report that are needed
for the year-end meeting. The new computer can do the
whole job in 2 hours while the old computer can do the
whole job in 3 hours. How long will it take to get the job
done using both computers simultaneously?

62. Installing a dishwasher. A plumber can install a
dishwasher in 50 min. If the plumber brings his apprentice
to help, the job takes 40 minutes. How long would it take
the apprentice working alone to install the dishwasher?

63. Filling a tub. Using the hot and cold water faucets together,
a bathtub fills in 8 minutes. Using the hot water faucet alone,
the tub fills in 12 minutes. How long does it take to fill the
tub using only the cold water faucet?

64. Filling a tank. A water tank has an inlet pipe and a drain
pipe. A full tank can be emptied in 30 minutes if the drain
is opened and an empty tank can be filled in 45 minutes
with the inlet pipe opened. If both pipes are accidentally
opened when the tank is full, then how long will it take to
empty the tank?

49. Apples and bananas. Bertha bought 18 pounds of fruit
consisting of apples and bananas. She paid $9 for the
apples and $2.40 for the bananas. If the price per pound of
the apples was 3 times that of the bananas, then how many
pounds of each type of fruit did she buy?

50. Fuel efficiency. Last week, Joe’s Electric Service used
110 gallons of gasoline in its two trucks. The large truck
was driven 800 miles, and the small truck was driven
600 miles. If the small truck gets twice as many miles per
gallon as the large truck, then how many gallons of
gasoline did the large truck use?

Miscellaneous

Show a complete solution to each problem.

51. Small plane. It took a small plane 1 hour longer to fly
480 miles against the wind than it took the plane to fly
the same distance with the wind. If the wind speed was
20 mph, then what is the speed of the plane in calm air?

52. Fast boat. A motorboat at full throttle takes two hours
longer to travel 75 miles against the current than it takes to
travel the same distance with the current. If the rate of the
current is 5 mph, then what is the speed of the boat at full
throttle in still water?

53. Light plane. At full throttle a light plane flies 275 miles
against the wind in the same time as it flies 325 miles with
the wind. If the plane flies at 120 mph at full throttle in
still air, then what is the wind speed?

54. Big plane. A six-passenger plane cruises at 180 mph in
calm air. If the plane flies 7 miles with the wind in the
same amount of time as it flies 5 miles against the wind,
then what is the wind speed?

Photo for Exercise 48

twice as many yards per carry as Anderson, then on how
many plays did Anderson carry the ball?

dug84356_ch06b.qxd 9/14/10 12:45 PM Page 446

6-67 Chapter 6 Summary 447

Wrap-Up

Summary

Rational Expressions Examples

Rational expression The ratio of two polynomials with the (x � 3)
denominator not equal to 0

Rational Function If a rational expression is used to determine y �
y from x, then y is a rational function of x.

Rule for reducing If a � 0 and c � 0, then � �
rational expressions

� .

(Divide out the common factors.)
�
x
x

5� � x
2 �

x
x
2

5� � �x
1
3�

Multiplication and Division of Rational Expressions Examples

Multiplication If b � 0 and d � 0, then �
a

b
� � �
d
c
� � �
b
a
d

c
� . �

x
3
3� � �x

6
5� � �

1
x
8
8�

Division If b � 0, c � 0, and d � 0, then �
a

b
� � �
d
c
� � �
a
b
� � �
d

c
�. �

x
a
3� � �x

5
9� � �x

a
3� � �

x
5
9

� � �
a
5
x6
�

(Invert the divisor and multiply.)

Addition and Subtraction of Rational Expressions Examples

Least common The LCD of a group of denominators is the 8, 12
denominator smallest number that is a multiple of all LCD � 24

of them.

Finding the least 1. Factor each denominator completely. Use 4ab3, 6a2b
common exponent notation for repeated factors.
denominator 2. Write the product of all of the different 4ab3 � 22ab3

factors that appear in the denominators. 6a2b � 2 � 3a2b
3. On each factor, use the highest power that LCD � 22 � 3a2b3 � 12a2b3

appears on that factor in any of the
denominators.

Addition and If b � 0, then � �
x

7
�
x
3
� � �
x
9
�
x
3
�

subtraction of
�
a

b
� � �
b
c
� � �
a �
b

c
� and �

a
b
� � �
b
c
� � �
a �
b

c
�.rational expressions

If the denominators are not identical, change
�
2
x

� � �

3
1
x
� � �

3
6
x
� � �

3
1
x
� � �

3
7
x
�each fraction to an equivalent fraction so that all

denominators are identical.

2x
�
x � 3

b
�
c

ab
�
ac

4x � 1
�

2x
2(4x � 1)
��

2(2x)
8x � 2
�

x � 1
�
x � 3

6Chapt er
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448 Chapter 6 Rational Expressions 6-68

Complex fraction A rational expression that has fractions in the
numerator and/or the denominator

Simplifying Multiply the numerator and denominator by � � �2

complex fractions the LCD.

Equations with Rational Expressions Examples
Solving equations Multiply each side by the LCD.

�
1
x
� � �
1
3

� � �
2
1
x
� � �

1
6
�

6x ��1x� � �
1
3

�� 6x ��2
1
x
� � �

1
6

��
6 � 2x � 3 � x

Proportion An equation expressing the equality of two ratios �
a
b

� � �
d
c
�

Extremes-means If b � 0 and d � 0, then �
x �

2
3
� � �
5
6
�

property
(cross-multiplying) �

a
b
� � �
d
c

� is equivalent to ad � bc. 2 � 6 � (x � 3)5

Cross-multiplying is a quick way to eliminate the
12 � 5x � 15

fractions in a proportion.

6 � 4
�
4 � 9

��12� � �
1
3

�� 12

��

��13� � �
3
4

�� 12
�
1
2
� � �
1
3
�
�
�
1
3
� � �
3
4
�

7. A fraction has rational expressions in its numera-
tor or denominator or both.

8. The opposite of reducing a fraction is a
fraction.

9. The smallest number that is a common multiple of a group
of denominators is the common denominator.

10. A number that appears to be a solution to an equation but
does not satisfy the equation is an root.

11. The expression a�b is the of a to b.
12. A is a statement expressing the equality of two

rational expressions.

13. The numbers a and d in a�b � c�d are the .
14. The numbers b and c in a�b � c�d are the .
15. If a�b � c�d, then ad � bc is the –

property.

Enriching Your Mathematical Word Power

Fill in the blank.

1. A expression is a ratio of two polynomials with
the denominator not equal to zero.

2. The of a rational expression is the set of all real
numbers that can be used in place of the variable.

3. If a rational expression is used to determine the value
of y from the value of x, then y is a rational
of x.

4. A rational expression is in terms when the numera-
tor and denominator have no common factors.

5. When common factors are divided out of the numerator
and denominator of a rational expression, the rational
expression is .

6. Two fractions that represent the same number are
fractions.

dug84356_ch06b.qxd 9/14/10 12:45 PM Page 448

6-69 Chapter 6 Review Exercises 449

Review Exercises

6.1 Reducing Rational Expressions
Find the domain of each rational expression.

1. �
4
x
�
2
x
�

2. �
2
x
x
�

�
9
6
�
3. �
x2 �
x �
4x
5
� 5
�

4. �
x2 �

x �
6x

2
� 8

�

Reduce each rational expression to lowest terms.

5. �
2
2
4
8
� 6. �

4
1
2
8
�

7. �
2
8
a
a

3
5
c
c
3

� 8. �
3
1
9
5
x
x

6
�

9. �
9

6
w

�
�

1
9
2

� 10. �
3
8
t
�

�

4
6
t

�

11. �
3
x2

�
�
3
1
x

� 12. �
3×2

1
�

0 �
9x

5
�
x
6
�

6.2 Multiplication and Division
Perform the indicated operation.

13. �
6
1
k
� � 3k2 14. �

15
1
abc
� � 5a3b5c2

15. �
2
3
xy
� � y2 16. 4ab � �

2
1
a4
�

17. �
a
a

2
�
�
2
9
� � �
a
a
2
�
�
3
4

� 18. �
x2

3
�
x
1
� � �
2x
6
�
x
2
�

19. �
w

3
�
w
2

� � �
4w

6w
� 8
� 20. �

2
x
y
�

�

x
2
y
x

� � �
x2 �

y2
2
�

xy
y
� y2

�

6.3 Finding the Least Common Denominator
Find the least common denominator for each group of
denominators.

21. 36, 54 22. 10, 15, 35

23. 6ab3, 8a7b2

24. 20u4v, 18uv5, 12u2v3

25. 4x, 6x � 6

26. 8a, 6a, 2a2 � 2a

27. x2 � 4, x2 � x � 2

28. x2 � 9, x2 � 6x � 9

Convert each rational expression into an equivalent rational
expression with the indicated denominator.

29. �
1

5
2
� � �

3
?
6
� 30. �

2
1
a
5
� � �

4
?
5
�

31. �
3
2
xy
� � �

15
?
x2y
� 32. �

7
3
x
z
2y
� � �

42x
?

3y8
�

33. �
y �

5
6

� � �
12 �

?
2y

� 34. �
2
�

�
3
t

� � �
2t �

?
4

�
35. �
x �
x
1
� � �
x2 �

?
1

�

36. �
t �

t
3

� �

6.4 Addition and Subtraction
Perform the indicated operation.

37. �
3
5
6
� � �

2
9
8
� 38. �

3
7
0
� � �

4
1
2
1
�

39. 3 � �
4
x

� 40. 1 � �
3
2
a
b
�

41. �
a
2
b2
� � �

a
1
2b
� 42. �

4
3
x3
� � �

6
5
x2
�

43. �
2a

9
�
a
3
� � �
3a

5
� 2
�

44. �
x �

3
2
� � �
x �
5
3
�

45. �
a �

1
8

� � �
8 �

2
a
�

46. �
x �

5
14

� � �
14

4
� x
�

47. �
2x

3
� 4
� � �
x2 �
1
4
�

48. �
x2 � 2

x
x � 3
� � �

x2
3
�

x
9

�

6.5 Complex Fractions

Simplify each complex fraction.

49. 50.

51. 52.
�
x
3
y
� � �

3
1
y
�

�
�
6
1
x
� � �

5
3
y
�

�
1
a

� � �
3
2
b
�

�

�
2
1
b
� � �

3
a
�
�
2
3
� � �
5
8
�
�
�
1
2

� � �
3
8

�
�
1
2
� � �
3
4
�
—
�
2
3
� � �
1
2
�

?
��
t2 � 2t � 15

dug84356_ch06b.qxd 9/14/10 12:45 PM Page 449

450 Chapter 6 Rational Expressions 6-70

53. 54.
55. 56.

6.6 Solving Equations with Rational Expressions
Solve each equation.

57. �
�

5
2
� � �
3
x

� 58. �
3
x

� � �
3
5
x
� � 1

59. �
a2

1
�
4
1
� � �
a �
1
1
� � �
a �
3
1

� 60. 2 � �
y �

3
5
� � �
y
2
�
y
5
�

61. z � �
2

3
�
z
z
� � �
z �
6
2

� 62. �
1
x

� � �
1
3
� � �
1
2
�

6.7 Applications of Ratios and Proportions
Solve each proportion.

63. �
3
x

� � �
2
7

� 64. �
4
x

� � �
4
x

�

65. �
w �

2
3

� � �
w
5

� 66. �
t �

3
3
� � �
t �
5
4
�

Solve each problem by using a proportion.

67. Taxis in Times Square. The ratio of taxis to private
automobiles in Times Square at 6:00 P.M. on New Year’s
Eve was estimated to be 15 to 2. If there were 60 taxis,
then how many private automobiles were there?

�
a2 � 5

6
a � 6
� � �

a �
8

2
�
��
�
a �
2
3
� � �
a �
4
2
�
�
x
x
�
�
1
3
�
———
�
x2 �

1
x � 6
� � �

x �
4

2
�
�
a �
4
1
� � �
a2

5
� 1
�

——
�
a2

1
� 1
� � �

a �
3

1
�
�
x �
1
2
� � �
x �
3
3

�
——

�
x �
2
3
� � �
x �
1
2
�

68. Student-teacher ratio. The student-teacher ratio for
Washington High was reported to be 27.5 to 1. If there are
42 teachers, then how many students are there?

69. Water and rice. At Wong’s Chinese Restaurant the secret
recipe for white rice calls for a 2 to 1 ratio of water to rice.
In one batch the chef used 28 more cups of water than rice.
How many cups of each did he use?

Photo for Exercise 67

Photo for Exercise 69

70. Oil and gas. An outboard motor calls for a fuel mixture
that has a gasoline-to-oil ratio of 50 to 1. How many pints
of oil should be added to 6 gallons of gasoline?

6.8 Applications of Rational Expressions

Solve each formula for the indicated variable.

71. �
y �

m
b

� � x for y

72. �
A
h

� � �
a �
2
b

� for a

73. F � �
mv

m
� 1
� for m

74. m � �
1 �

r
rt

� for r

75. �
y
x

�
�
1
3

� � 4 for y

76. �
y
x

�
�
3
2
� � �
�

3
1
� for y

dug84356_ch06b.qxd 9/14/10 12:45 PM Page 450

6-71 Chapter 6 Review Exercises 451

Solve each problem.

77. Making a puzzle. Tracy, Stacy, and Fred assembled a
very large puzzle together in 40 hours. If Stacy worked
twice as fast as Fred, and Tracy worked just as fast as
Stacy, then how long would it have taken Fred to assem-
ble the puzzle alone?

78. Going skiing. Leon drove 270 miles to the lodge in the
same time as Pat drove 330 miles to the lodge. If Pat
drove 10 miles per hour faster than Leon, then how fast
did each of them drive?

woman’s voice than with a man’s voice, then how many
of the 2500 calls were made by females?

81. Distribution of waste. The accompanying figure shows the
distribution of the total municipal solid waste into various
categories in 2000 (U.S. Environmental Protection Agency,
www.epa.gov). If the paper waste was 59.8 million tons
greater than the yard waste, then what was the amount of
yard waste generated?

82. Total waste. Use the information given in Exercise 81 to
find the total waste generated in 2000 and the amount of
food waste.

Miscellaneous
In place of each question mark, put an expression that makes
each equation an identity.

83. �
5
x

� � �
2
?
x
� 84. �

a
?

� � �
3
6
a
�

85. �
a �

2
5

� � �
5 �

?
a

� 86. �
a
�

�
1
7
� � �
1
?
�

87. 3 � �
?
x

� 88. 2a � �
b
?

�

89. m � �
1
2

� � ? 90. 5x � �
1
x

� � ?

91. 2a � ? � 12a 92. 10x � ? � 20×2

93. �
a
a
2

�
�
1
1
� � �
1
?

� 94. �
x2 �

?
9

� � �
x �
1
3
�

95. �
1
a

� � �
1
5

� � ? 96. �
3
7

� � �
2
b
� � ?

97. �
a
2

� � 1 � �
2
?

� 98. �
1
a

� � 1 � �
a
?

�

99. (a � b) � (�1) � ?

100. (a � 7) � (7 � a) � ?

101. � ? 102. � ?

For each expression in Exercises 103–122, either perform the
indicated operation or solve the equation, whichever is
appropriate.

103. �
1
x

� � �
2
1
x
� 104. �

1
y

� � �
3
1
y
� � 2

105. �
3
2
xy
� � �

6
1
x
� 106. �

x �
3
1
� � �
3
x
�

107. �
a �

5
5
� � �
5 �
3
a
�

108. �
x �

2
2
� � �
3
x
� � �
�
x
1
�

3a
—

�
1
2
�

�
5
1
a
�

�
2

Photo for Exercise 78

79. Merging automobiles. When Bert and Ernie merged
their automobile dealerships, Bert had 10 more cars than
Ernie. While 36% of Ernie’s stock consisted of new cars,
only 25% of Bert’s stock consisted of new cars. If they
had 33 new cars on the lot after the merger, then how
many cars did each one have before the merger?

80. Magazine sales. A company specializing in magazine
sales over the telephone found that in 2500 phone calls,
360 resulted in sales and were made by male callers, and
480 resulted in sales and were made by female callers. If
the company gets twice as many sales per call with a

2000 Total Waste Generation (before recycling)

Paper
38.1%

Yard Waste
12.1%

Plastic
10.5%

Metals
7.8%

Wood
5.3%

Food Waste
10.9%

Glass
5.5%

Other
9.8%

Figure for Exercises 81 and 82

dug84356_ch06b.qxd 9/14/10 12:45 PM Page 451

452 Chapter 6 Rational Expressions 6-72

109. �
x �

2
1
� � �
2
x
� � 1

110. �
x �

2
2
� � �
6x
1
�

4
12

�

111. �
x
�

�
3
2

� � �
5x �

9
10

�

112. �
1
3
0
� � �

5
x
�

113. �
�

1
3
� � �

�
x
2
�

114. � �
4x

x
� 8
�

115. �
2x � 2m
�

a � 3
ax � am � 3x � 3m
��

a2 � 9

x2 � 4
�

116. �
�

x
2
� � �

x �
3
2
�

117. �
x2 �

2
25

� � �
x2 � 4

1
x � 5
�

118. �
a2

4
� 1
� � �

2a
1
� 2
�

119. �
a2

�
�
3
9
� �

120. �
a2

�
�
5
4

� � �
a2 � 3

2
a � 2
�

121. �
a2

1
� 1
� � �

1 �
2

a
� � �
a �
3
1
�

122. 3 � �
x �

1
2

� � �
2
x
x
�

�
2
3
�

2
��
a2 � 5a � 6

Chapter 6 Test

What numbers cannot be used for x in each rational
expression?

1. �
2
x
x
2 �

�
1
1

� 2. �
2 �

5
3x

� 3. �
1
x

�

Perform the indicated operation. Write each answer in lowest
terms.

4. �
1
2
5
� � �

4
9

� 5. �
1
y

� � 3

6. �
a �

3
2
� � �
2 �
1
a
�
7. �
x2 �
2
4
� � �
x2 �

3
x � 2
�

8. �
(
m

m
2
�
�

1
1
)2

� � �
2
3
m

m
�
�
2
3
�

9. �
a �

3
b

� � �
b2 �

6
a2

�

10. �
5
1
a
2

2
a
b

� � �
1
2
5
a
a

3
b
b
6�
Simplify each complex fraction.

11. 12.

Solve each equation.

13. �
3
x

� � �
7
5

� 14. �
x �

x
1
� � �
3
x
� � �
1
2
�
�
2
x
� � �
x �
1
2
�
��
�
x �
1
2
� � �
3
x
�
�
2
3
� � �
4
5
�
�
�
2
5
� � �
3
2
�
15. �
1
x
� � �
1
6
� � �
1
4
�
Solve each formula for the indicated variable.

16. �
y

x
�
�
3
2
� � �
�

5
1
� for y

17. M � �
1
3

� b (c � d ) for c

Solve each problem.

18. If R(x) � �
x
1

�
�
2
x

�, then what is R(0.9)?

19. When all of the grocery carts escape from the supermarket,
it takes Reginald 12 minutes to round them up and bring
them back. Because Norman doesn’t make as much per
hour as Reginald, it takes Norman 18 minutes to do the
same job. How long would it take them working together
to complete the roundup?

20. Brenda and her husband Randy bicycled cross-country
together. One morning, Brenda rode 30 miles. By traveling
only 5 miles per hour faster and putting in one more hour,
Randy covered twice the distance Brenda covered. What
was the speed of each cyclist?

21. For a certain time period the ratio of the dollar value of
exports to the dollar value of imports for the United States
was 2 to 3. If the value of exports during that time period
was 48 billion dollars, then what was the value of imports?

dug84356_ch06b.qxd 9/14/10 12:45 PM Page 452

6-73 Chapter 6 Making Connections 453

Solve each equation.

1. 3x � 2 � 5 2. �
3
5

� x � �2

3. 2(x � 2) � 4x 4. 2(x � 2) � 2x

5. 2(x � 3) � 6x � 6 6. 2(3x � 4) � x2 � 0

7. 4x � 4×3 � 0 8. �
3
x

� � �
�

5
2
�

9. �
3
x
� � �
1
x

2
� 10. �

2
x
� � �
x �
4
2
�

11. �
1
w

8
� � �
w �

9
1

� � �
4 �

6
w

� 12. �
x �

x
1
� � �

2x
1
� 2
� � �

7
8
�

Solve each equation for y.

13. 2x � 3y � c 14. �
y
x

�
�
3
5
� � �
1
2
�

15. 2y � ay � c 16. �
A

y
� � �
C

B
�

17. �
A

y
� � �
1
3

� � �
B

y
� 18. �

A
y
� � �
1
2
� � �
1
3
�

19. 3y � 5ay � 8 20. y2 � By � 0

21. A � �
1
2

� h(b � y) 22. 2(b � y) � b

Calculate the value of b2 � 4ac for each choice of a, b, and c.

23. a � 1, b � 2, c � �15 24. a � 1, b � 8, c � 12

25. a � 2, b � 5, c � �3 26. a � 6, b � 7, c � �3

Perform each indicated operation.

27. (3x � 5) � (5x � 3) 28. (2a � 5)(a � 3)

29. x7 � x3 30. �
x �

5
3
� � �
x �
5
4
�

31. �
1
2

� � �
1
x

� 32. �
1
2

� � �
1
x
�

33. �
1
2

� � �
1
x

� 34. �
1

2
� � �
1
x
�
35. �
x �
5
3
� � �
x �
5
4

� 36. �
3
2
a
� � 2

37. (x � 8)(x � 8) 38. 3x(x2 � 7)

39. 2a5 � 5a9 40. x2 � x8

41. (k � 6)2 42. ( j � 5)2

43. (g � 3) � (3 � g) 44. (6×3 � 8×2) � (2x)

Factor each expression completely.

45. 4×4 � 12×3 � 32×2

46. 15a3 � 24a2 � 9a

47. �12b2 � 84b � 147

48. �2y2 � 288

49. by � yw � 3w � 3b

50. 2ax � 4bx � 3an � 6bn

51. �7b3 � 7

52. 2q3 � 54

Perform the indicated operations without using a calculator.
Write each answer in scientific notation.

53. (3 103)(4 104)

54. (3 103)4

55. �
8
4

1
1
0
0
1

8
5�

56. (1 103) � (1 104)

MakingConnections A Review of Chapters 1–6

dug84356_ch06b.qxd 9/14/10 12:45 PM Page 453

454 Chapter 6 Rational Expressions 6-74

Solve each problem.

57. The sum of the squares of two consecutive positive even
integers is 100. What are the integers?

58. The difference of the squares of two consecutive positive
odd integers is 32. What are the integers?

59. Present value. An investor is interested in the amount or
present value that she would have to invest today to
receive periodic payments in the future. The present value
of $1 in one year and $1 in 2 years with interest rate r
compounded annually is given by the formula

P � �
1 �

1
r

� � �
(1 �

1
r)2

�.

a) Rewrite the formula so that the right-hand side is a
single rational expression.

b) Find P if r � 7%.
c) The present value of $1 per year for the next 10 years is

given by the formula
P � �
1 �
1
r
� � �
(1 �
1
r)2
� � �
(1 �

1
r)3

� � � � � � �
(1 �

1
r)10
� .

Use this formula to find P if r � 5%.

dug84356_ch06b.qxd 9/14/10 12:45 PM Page 454

6-75 Chapter 6 Critical Thinking 455

These exercises can be solved by a variety of techniques, which may or may not require algebra. So be creative and think critically.
Explain all answers. Answers are in the Instructor’s Edition of this text.

4. Eyes and feet. A rancher has some sheep and ostriches.
His young daughter observed that the animals have a total
of 60 eyes and 86 feet. How many animals of each type
does the rancher have?

CriticalThinking For Individual or Group Work Chapter 6

1. Equilateral triangles. Consider the sequence of three
equilateral triangles shown in the accompanying figure.

a) How many equilateral triangles are there in (a) of the
accompanying figure?

b) How many equilateral triangles congruent to the one in
(a) can be found in (b) of the accompanying figure?
How many are found in (c)?

c) Suppose the sequence of equilateral triangles shown
in (a), (b), and (c) is continued. How many equilateral
triangles [congruent to the one in (a)] could be found
in the nth such figure?

2. The amazing Amber. Amber has been amazing her friends
with a math trick. Amber has a friend select a three-digit
number and reverse the digits. The friend then finds the
difference of the two numbers and reads the first two digits
of the difference (from left to right). Amber can always
tell the last digit of the difference. Explain how Amber
does this.

3. Missing proceeds. Ruth and Betty sell apples at a farmers
market. Ruth’s apples sell at 2 for $1, while Betty’s slightly
smaller apples sell at 3 for $1. When Betty leaves to pick
up her kids, they each have 30 apples and Ruth takes
charge of both businesses. To simplify things, Ruth puts all
60 of the apples together and sells them at 5 for $2. When
Betty returns, all of the apples have been sold, but they
begin arguing over how to divide up the proceeds. What
is the problem? Explain what went wrong.

Figure for Exercise 1

(a) (b) (c)

5. Evaluation nightmare. Evaluate:

6. Perfect squares. Find a positive integer such that the
integer increased by 1 is a perfect square and one-half of
the integer increased by 1 is a perfect square. Also find
the next two larger positive integers that have this same
property.

7. Multiplying primes. Find the units digit of the product of
the first 500 prime numbers.

8. Ones and zeros. Find the sum of all seven-digit numbers
that can be written using only ones or zeros.

9,876,543,210
��
9,876,543,2112 � 9,876,543,210 � 9,876,543,212

Photo for Exercise 4

dug84356_ch06b.qxd 9/14/10 12:45 PM Page 455

dug84356_ch06b.qxd 9/14/10 12:45 PM Page 456

MAT222Week 1 Discussion Grading Rubric

Grading Criteria

MAT222 Discussions Week 1
2 points possible

Content Criteria Weight

Student thoroughly addresses all elements of the discussion prompt, and demonstrates an
advanced knowledge of the topic. Student clearly understands the mathematical concepts in the
assignment, and demonstrates the prerequisite math knowledge to do the work successfully.

0.8

Writing of Math Work

Student effectively shows how the solution(s) was achieved. The math work is shown and follows
an accurate process to get to the solution.

0.8

Participation
Student responds with thorough and constructive analysis to at least two classmates, relating
the response to relevant course concepts. Student may pose pertinent follow-up thoughts or
questions about the topic, and demonstrates respect for the diverse opinions of fellow learners.
Initial post was made by Day 3 (Thursday), and responses were made by Day 7 (Monday).

0.2

Writing

Initial post contains very few, if any, errors related to grammar, spelling, and sentence structure.
Post is easy to read and understand.

0.2

Inserting Math Symbols

Throughout these courses MAT 221 and MAT 222 students are required to submit written assignments

in Microsoft Word that require the use of Mathematical symbols. Please follow the directions below in

order to insert mathematical symbols into your documents.

Inserting Symbols:

1. Click with your mouse in the space you would like the mathematical symbol to appear.

2. Click with your mouse on the “Insert” tab on the upper navigation toolbar.

3. Click with your mouse on the “Symbol” button.

4. In the dropdown menu, click with your mouse on “More Symbols”.

5. In the “Symbol” window, click on the “Subset” dropdown menu and choose the desired category

that fits the mathematical equation.

6. Select the desired symbol and click with your mouse “Insert”.

7. The symbol should appear in your document.

If you have any questions, please do not hesitate the contact your instructor.

INSTRUCTOR GUIDANCE EXAMPLE: Week One Discussion

Domains of Rational Expressions

Students, you are perfectly welcome to format your math work just as I have done in
these examples. However, the written parts of the assignment MUST be done about
your own thoughts and in your own words. You are NOT to simply copy this wording
into your posts!

Here are my given rational expressions oh which to base my work.

25×2 – 4
67

5 – 9w
9w2 – 4

The domain of a rational expression is the set of all numbers which are allowed to
substitute for the variable in the expression. It is possible that some numbers will not be
allowed depending on what the denominator has in it.

In our Real Number System division by zero cannot be done. There is no number (or
any other object) which can be the answer to division by zero so we must simply call the
attempt “undefined.” A denominator cannot be zero because in a rational number or
expression the denominator divides the numerator.

In my first expression, the denominator is a constant term, meaning there is no variable
present. Since it is impossible for 67 to equal zero, there are no excluded values for the
domain. We can say the domain (D) is the set of all Real Numbers, written in set
notation that would look like this:

D = {x| x ∈ ℜ} or even more simply as D = ℜ.

For my second expression, I need to set the denominator equal to zero to find my
excluded values for w.

9w2 – 4 = 0 I notice this is a difference of squares which I can factor.
(3w – 2)(3w + 2) = 0 Set each factor equal to zero.
3w – 2 = 0 or 3w + 2 = 0 Add or subtract 2 from both sides.
3w = 2 or 3w = -2 Divide both sides by 3.
w = 2/3 or w = -2/3 These are the excluded values for my second expression.

The domain (D) for my second expression is the set of all Reals excluding ±2/3. In set
notation, this can be written as D = {w| w ∈ ℜ, w ≠ ±2/3}

Now, both of my expressions do not have excluded values. In one expression, I have no
excluded values because there is no variable in the denominator and a non-zero number
will never just become zero. In the other expression, there are two excluded numbers
because both, if inserted in place of the variable, would cause the denominator to become
zero and thus the whole expression would become undefined.

–

2
37
–
–
K
K

� EMBED Equation.3 ��-5

� EMBED Equation.3 ��5

Sum=b

Product=a

1x (-

)

x2+0x-25=0 substitute the value of bx to facilitate factorization

The denominator is a constant term

(x2+5x)(-5x-25)=0

x (x+5)-5(x+5)=0

(x+5)(x-5) are the factors for the domain

All values of x in the expression are included as dividing by the domain will yield a solution

Thus D=� EMBED Equation.3 ��

� EMBED Equation.3 �� The Domain is Integer 2 which divides any factor in the numerator. We factorize the Numerator to obtain the factors.

� EMBED Equation.3 ��-6

� EMBED Equation.3 ��5

Sum=b

-1

Product=a x c

1x (-30)

k2+5k-6k-30=0 substitute the value of bx to facilitate factorization

(k2+5k)(-6k-30)=0

k (k+5)-6(k+5)=0

(k+5)(k-6) are the factors for the domain

k=-5 and k=6 are the excluded values in the expression since they will equal 0 in any range of values.

Thus D={K:K� EMBED Equation.3 �� EMBED Equation.3 ��,K� EMBED Equation.3 ��(-5) ,k� EMBED Equation.3 ��6}

� EMBED Equation.3 �� Factorizing the domain in the form we find factors:

2
25
2
–
x

30
2
37
–
–
K
K

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_1071350591.unknown

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_1071348848.unknown

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