358
(6-26)
Chapter 6
Rational Expressions
bundles that they can attach while working together for
10 hours.
1200x 3600
x 2 6x
FIGURE FOR EXERCISE 94
95. Selling. George sells one magazine subscription every
20 minutes, whereas Theresa sells one every x minutes.
Write a rational expression for the number of magazine
subscriptions that they will sell when working together for
one hour.
3x 60
x
6.4
In this
section
96. Painting. Harry can paint his house by himself in 6 days.
His wife Judy can paint the house by herself in x days.
Write a rational expression for the portion of the house
that they paint when working together for 2 days.
x6
3x
97. Driving. Joan drove for 100 miles at one speed and then
increased her speed by 5 miles per hour and drove 200 additional miles. Write a rational expression for her total
travel time.
300x 500
hours
x2 5x
98. Running. Willard jogged for 3 miles at one speed and
then doubled his speed for an additional mile. Write a rational expression for his total running time.
7
hours
2x
GET TING MORE INVOLVED
99. Discussion. Explain why fractions must have common
denominators for addition but not for multiplication.
100. Discussion. Find each “infinite sum” and explain your
answer.
3
3
3
3
a) 2 3 4 . . .
10
10
10
10
9
9
9
9
b) 2 3 4 . . .
10
10
10
10
COMPLEX FRACTIONS
In this section we will use the techniques of Section 6.3 to simplify complex
fractions. As their name suggests, complex fractions are rather messy-looking
expressions.
Simplifying Complex
Fractions
Simplifying Complex Fractions
●
Simplifying Expressions with
Negative Exponents
A complex fraction is a fraction that has rational expressions in the numerator, the
denominator, or both. For example,
●
Applications
●
1 1
2 3
,
1 1
4 5
2
3 x
,
1
1
2 4
x
and
x2
x2 9
x
4
x2 6x 9 x 3
are complex fractions. In the next example we show two methods for simplifying a
complex fraction.
6.4
E X A M P L E
1
Complex Fractions
(6-27)
359
A complex fraction without variables
Simplify
1 1
2 3
.
1 1
4 5
Solution
Method A For this method we perform the computations of the numerator and
denominator separately and then divide:
calculator
1 1
5
5
9
5 20 5 2 10 50
2 3
6
1 1
9
6 20 6 9
2 3 9
27
4 5
20
close-up
You can use a calculator to
find the value of a complex
fraction.
Method B For this method we find the LCD for all of the fractions in the complex
fraction. Then we multiply the numerator and denominator of the complex fraction by
the LCD. The LCD for the denominators 2, 3, 4, and 5 is 60. So we multiply the
numerator and denominator of the complex fraction by 60:
1 1
1 1
60
30 20 50
2 3
2 3
1 1
15 12 27
1 1
60
4 5
4 5
1 60 30, 1 60 20
2
3
1
1
60 15, 60 12
4
5
■
In most cases Method B of Example 1 is the faster method for simplifying
complex fractions, and we will continue to use it.
E X A M P L E
2
A complex fraction with variables
Simplify
2
3 x
.
1
1
2 x
4
Solution
The LCD of x, x 2, and 4 is 4x 2. Multiply the numerator and denominator by 4x2 :
helpful
hint
When students see addition
or subtraction in a complex
fraction, they often convert
all of the fractions to the
same denominator. This is not
wrong, but it is not necessary.
Simply multiplying ever fraction by the LCD eliminates the
denominators of the original
fractions.
2
2
3 (4x 2)
3 x
x
1
1
1
1
2
2 2 (4x )
x
4
4
x
2
3(4x2) (4x 2)
x
1
1
2 (4x2) (4x 2)
4
x
12x2 8x
4 x2
Distributive property
■
360
(6-28)
Chapter 6
E X A M P L E
3
Rational Expressions
More complicated denominators
Simplify
study
tip
Your mood for studying
should match the mood in
which you are tested. Being
too relaxed during studying
will not match the increased
level of activation you attain
during a test. Likewise, if you
get too tensed-up during a
test, you will not do well because your test-taking mood
will not match your studying
mood.
x2
x2 9
x
4
x2 6x 9 x 3
.
Solution
Because x 2 9 (x 3)(x 3) and x 2 6x 9 (x 3)2, the LCD is
(x 3)2(x 3). Multiply the numerator and denominator by the LCD:
x2
x2
(x 3)2(x 3)
2 (x 3)(x 3)
x 9
x
4
x
4
2 (x 3)2(x 3) (x 3)2(x 3)
2
x 6x 9 x 3 (x 3)
x3
(x 2)(x 3)
Simplify.
x(x 3) 4(x 3)(x 3)
(x 2)(x 3)
Factor out x 3.
(x 3)[x 4(x 3)]
(x 2)(x 3)
(x 3)(5x 12)
■
Simplifying Expressions with Negative Exponents
Consider the expression
3a1 21
.
1 b 1
Using the definition of negative exponents, we can rewrite this expression as a
complex fraction:
3 1
3a1 21 a 2
1
1 b1
1 b
The LCD for the complex fraction is 2ab. Note that 2ab could also be obtained
from the bases of the expressions with the negative exponents. To simplify the
complex fraction, we could use Method B as we have been doing. However, it is
not necessary to rewrite the original expression as a complex fraction. The next
example shows how to use Method B with the original expression.
E X A M P L E
4
A complex fraction with negative exponents
1
1
2
Simplify the complex fraction 3a
1 .
1b
Solution
Multiply the numerator and denominator by 2ab, the LCD of the fractions.
Remember that a1 a a0 1.
3a1 21 (3a1 21)2ab
1 b1
(1 b1)2ab
3a1(2ab) 21(2ab)
Distributive property
1(2ab) b1(2ab)
6b ab
2ab 2a
■
6.4
E X A M P L E
5
Complex Fractions
(6-29)
361
A complex fraction with negative exponents
1
2
a b
Simplify the complex fraction 2 3.
ab
helpful
hint
In Examples 4, 5, and 6 we
are simplifying the expressions without first removing
the negative exponents to
gain experience in working
with negative exponents. Of
course, each expression with a
negative exponent could be
rewritten with a positive exponent and then the complex
fraction could be simplified as
in Examples 2 and 3.
ba
Solution
If we rewrote a1, b2, b2, and a3, then the denominators would be a, b2, b2,
and a3. So the LCD is a3b2. If we multiply the numerator and denominator by a3b2,
the negative exponents will be eliminated:
a1 b2
(a1 b2)a3b2
ab2 ba3 (ab2 ba3)a3b2
a1 a3b2 b2 a3b2
ab2 a3b2 ba3 a3b2
Distributive property
a2b2 a3
a4 b 3
b 2b2 b0 1
a 3a3 a0 1
Note that the positive exponents of a3b2 are just large enough to eliminate all of the
■
negative exponents when we multiply.
The next example is not exactly a complex fraction, but we can use the same
technique as in the previous example.
E X A M P L E
6
More negative exponents
Eliminate negative exponents and simplify p p1q2.
Solution
If we multiply the numerator and denominator by pq2, we will eliminate the
negative exponents:
( p p1q2) pq2
p p1q2 2
1
pq
2 2
p q 1 p pq2 p2q2
p1q2 pq2 1
pq2
■
Applications
The next example illustrates how complex fractions can occur in a problem.
E X A M P L E
7
An application of complex fractions
Eastside Elementary has the same number of students as Westside Elementary. Onehalf of the students at Eastside ride buses to school, and two-thirds of the students
at Westside ride buses to school. One-sixth of the students at Eastside are female,
and one-third of the students at Westside are female. If all of the female students
ride the buses, then what percentage of the students who ride the buses are female?
Solution
To find the required percentage, we must divide the number of females who ride the
buses by the total number of students who ride the buses. Let
x the number of students at Eastside.
362
(6-30)
Chapter 6
Rational Expressions
Because the number of students at Westside is also x, we have
1
2
x x the total number of students who ride the buses
2
3
and
1
1
x x the total number of female students.
6
3
Because all of the female students ride the buses, we can express the percentage of
riders who are female by the following rational expression:
1
1
x x
6
3
1
2
x x
2
3
Multiply the numerator and denominator by 6, the LCD for 2, 3, and 6:
16 x 13 x6 x 2x 3x 3
0.43 43%
1
2
3x 4x 7x 7
2x 3 x6
■
So 43% of the students who ride the buses are female.
WARM-UPS
True or false? Explain.
1. The LCM for 2, x, 6, and x2 is 6x3. False
2. The LCM for a b, 2b 2a, and 6 is 6a 6b. True
3. The LCD is the LCM of the denominators. True
1 1
2 3 5 3
4. True
1 6 2
1 2
1
5. 2 31 (2 3)1 False
6. (21 31)1 2 3 False
7. 2 31 51 False
8. x 21 x for any real number x. False
9.
10.
2
1
b1
To simplify a
, multiply the numerator and denominator by ab.
ab
ab2 a5b2
To simplify , multiply the numerator and denominator by
a3b a5b1
True
a5b2.
True
6. 4
EXERCISES
Reading and Writing After reading this section, write out the
answers to these questions. Use complete sentences.
1. What is a complex fraction?
A complex fraction is a fraction that contains fractions in
the numerator, denominator, or both.
2. What are the two methods for simplifying complex fractions?
One method is to perform the operations in the numerator
and then in the denominator, and then divide the results. The
other method is to multiply the numerator and the denominator by the LCD for all of the fractions.
6.4
Simplify each complex fraction. Use either method. See
Example 1.
1 1
1 1
2 3 10
3 4 35
4. 3. 1 1 3
1 1 22
4 5
5 6
2 5 1
3 6 2
5.
1 1
1
8 3 12
8
2 x 1
5 9 3
6.
1 x
2
3 5 15
3 5x
9x 21
Simplify the complex fractions. Use Method B. See Example 2.
ab
b
7.
ab
ab
a2 ab
ab
2
m n
10. 1
3
m n
m2n 2m
n 3m
m2
3 6
13. 2
4
9 m
60m 3m2
8m 36
4x 2 1
2
xy
16. 4x 2
xy 2
2xy y
2x
mn
m2
8.
m3
m n3
m n 3 n4
m2 3m
x 3y
xy
11. 1 1
x y
x 3y
xy
3
a b
9. b 1
a b
a2b 3a
a b2
3
2
t
w
12. wt
4wt
8t 12w
wt
2z
6 z
14. 1
1
3z 6
42z 12
2z
a2 b2
2
a b3
15. ab
3
ab
a2 ab
b2
1
1
22 3
xy
xy
17. 1
1
3 x y xy
xy x 2
y 2 x 2y 2
1
1
2a3b ab4
18. 1
1
22 6a b
3a4b
3ab3 6a3
a2b2 2b3
Simplify each complex fraction. See Examples 1–3.
4
x6
x x x4
x2
19. 20. 4x 4
4x 15
x x x4
x2
x2
x2
x2
x5
1
3
1 2 y1
a2
21. 22. 1
1
4 3 y1
a2
y2 y 2
2a2 3a 14
(y 1)(3y 4)
4a2 a 14
Complex Fractions
(6-31)
2
4
3x
23. 1
1
x3
4x 10
x4
w2 w3
w1
w
25. w4 w2
w
w1
6w 3
2w 2 w 4
x
2
x5
24. 2x
1
5x
x 10
3x 5
x1 x2
x2 x3
26. x3 x1
x3 x2
2x 1
2x 2 3x 3
1
3
ab ab
27. 2
4
ba ba
2b a
a 3b
3
4
2x 2x
28. 1
3
x2 x2
4
3 a1
29. 3
5 1a
3a 7
5a 2
2
4
m3 m
31. 3
1
m2 m
3m2 12m 12
(m 3)(2m 1)
3
x2
x2 1 x3 1
33. 3
x3
x2 x 1 x3 1
2x 2 4x 5
4x 2 2x 6
363
7x 2
2x 8
x x1
3 9x
30. x 2x
6 x9
2x 2 12x 6
x 2 3x 12
1
4
y 2 3y
32. 3
2
y y3
y2 11y 24
3y2 33y 54
2
3
a3 8 a2 2a 4
34. 4
a3
a2 4 a3 8
3a2 2a 8
5a2 13a 22
Simplify. See Examples 4–6.
w1 y1
35. z1 y1
yz wz
wy wz
1 x1
37. 1 x2
x
x1
a1 b1
36. a1 b1
ba
ba
4 a 2
38. 2 a 1
2a 1
a
a2 b2
39. a 1b
2
a b2
ab3
m3 n3
40. mn2
3
n m3
4
mn
364
(6-32)
Chapter 6
41. 1 a1
a1
a
x1 x2
43. x x2
1
2
x x1
2m1 3m2
45. 2
m
2m 3
a1 b1
47. ab
1
ab
x3 y3
49. 3
x y3
Rational Expressions
42. m1 a1
am
am
x x2
44. 1 x2
x2 x 1
x1
4x3 6x5
46. 2x5
2
2x 3
a2 b2
48. a2 b2
a2b2
1 8x3
51. x1 2x2 4x3
x2
(a b)2
50. a2 b2
a2b3 a3b2
ab
a 27a2
52. 1 3a1 9a2
a3
53. (x1 y1)1
xy
xy
54. (a1 b1)2
a2b2
b2 2ab a2
x3y3
Use a calculator to evaluate each complex fraction.
Round answers to four decimal places. If your calculator does fractions, then also find the fractional answer.
5 4
3 5
55. 1 5
3 6
1 3
1
12 2 4
56. 3 5
5 6
26
1.7333, 15
1
1
4 9
57. 21 31
1
0.1667, 6
5
0.1163, 43
21 31 61
58. 31 51 41
40
1.7391, 23
Solve each problem. See Example 7.
59. Racial balance. Clarksville has three elementary schools.
Northside has one-half as many students as Central, and
Southside has two-thirds as many students as Central.
One-third of the students at Northside are AfricanAmerican, three-fourths of the students at Central are
African-American, and one-sixth of the students at Southside are African-American. What percent of the city’s
elementary students are African-American?
47.4%
60. Explosive situation. All of the employees at Acme Explosives are in either development, manufacturing, or sales.
One-fifth of the employees in development are women,
one-third of the employees in manufacturing are women,
and one-half of the employees in sales are women. Use the
accompanying figure to determine the percentage of workers at Acme who are women. What percent of the women at
Acme are in sales?
38.3%, 65.2%
Distribution of Employees
at Acme Explosives
Development
1
4
Manufacturing
1
4
Sales
1
2
FIGURE FOR EXERCISE 60
61. Average speed. Mary drove from Clarksville to Leesville
at 45 miles per hour (mph). At Leesville she discovered that
she had forgotten her purse. She immediately returned to
Clarksville at 55 mph. What was her average speed for the
entire trip? (The answer is not 50 mph.)
49.5 mph
62. Average price. On her way to New York, Jenny spent
the same amount for gasoline each of the three times that
she filled up. She paid 99.9 cents per gallon the first time,
109.9 cents per gallon the second time, and 119.9 cents per
gallon the third time. What was the average price per gallon
to the nearest tenth of a cent for the gasoline that she
bought?
109.3 cents per gallon
FIGURE FOR EXERCISE 62
6.5
Solving Equations Involving Rational Expressions
GET TING MORE INVOLVED
In this
section
●
Multiplying by the LCD
●
Extraneous Roots
●
Proportions
365
b) Explain why in each case the exact value must be less
than 1.
The denominator is larger than the numerator in the first
fraction.
65. Cooperative Learning. Work with a group to simplify
the complex fraction. For what values of x is the complex
fraction undefined?
63. Cooperative learning. Write a step-by-step strategy for
simplifying complex fractions with negative exponents.
Have a classmate use your strategy to simplify some complex fractions from Exercises 35–54.
64. Discussion. a) Find the exact value of each expression.
1
1
i)
ii)
1
1
1
1
1
1
1
1
1
1
1
1 1
2
1 3
5 11
, 8 18
6.5
(6-33)
1
1
1
1
1
1
1 x
2x 1
1 2
, x 0, 1, , 3x 2
2 3
SOLVING EQUATIONS INVOLVING
RATIONAL EXPRESSIONS
Many problems in algebra are modeled by equations involving rational expressions.
In this section you will learn how to solve equations that have rational expressions,
and in Section 6.6 we will solve problems using these equations.
Multiplying by the LCD
To solve equations having rational expressions, we multiply each side of the equation by the LCD of the rational expressions.
E X A M P L E
1
An equation with rational expressions
Solve 1 1 1.
x
4
6
helpful
hint
Note that it is not necessary
to convert each fraction into
an equivalent 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 and that is
good.
Solution
The LCD for the denominators 4, 6, and x is 12x:
1 1
1
12x 12x x 4
6
3
2
1
1
1
12x 12x 12x x
4
6
12 3x 2x
Multiply each side by 12x.
Distributive property
Divide out the common factors.
12 x 0
x 12
Check 12 in the original equation. The solution set is 12.
■