bundles that they can attach while working together for 10 hours. 120 x 0 2 x 6 3 x 600
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. x 3 x 6
97. Driving. Joan drove for 100 miles at one speed and then increased her speed by 5 miles per hour and drove 200 ad-ditional miles. Write a rational expression for her total travel time. 30 x 0 2 x 5 5 x 00 hours
98. Running. Willard jogged for 3 miles at one speed and then doubled his speed for an additional mile. Write a ra-tional expression for his total running time.
2 7 x hours G E T T I N G M O R E I N VO LV E D
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. a) 1 3 0 1 3 0 2 1 3 0 3 1 3 0 . . .4 b) 1 9 0 1 9 0 2 1 9 0 3 1 9 0 . . .4 F I G U R E F O R E X E R C I S E 9 4
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 x 60
C O M P L E X F R A C T I O N S
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
A complex fraction is a fraction that has rational expressions in the numerator, the denominator, or both. For example,
, , and
are complex fractions. In the next example we show two methods for simplifying a complex fraction. x x 2 2 9 x2 6 x x 9 x 4 3 3 2 x x 1 2 1 4 1 2 1 3 1 4 1 5
6.4
I n t h i s
s e c t i o n
● Simplifying Complex Fractions● Simplifying Expressions with Negative Exponents ● Applications
E X A M P L E 1
A complex fraction without variablesSimplify .
Solution
Method A For this method we perform the computations of the numerator and denominator separately and then divide:
5 6 2 9 0 5 6 2 9 0 5 2 2 3 1 9 0 5 2 0 7
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:
3 1 0 5 2 1 0 2 5 2 0 7
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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 variablesSimplify .
Solution
The LCD of x, x2, and 4 is 4x2. Multiply the numerator and denominator by 4x2:
Distributive property 12 4 x2 x2 8x
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3(4x2) 2 x (4x 2 ) x 1 2 (4x2 ) 1 4 (4x 2 ) 3 2 x(4x 2) x 1 2 1 4
(4x 2) 3 2 x x 1 2 1 4 3 2 x x 1 2 1 4 1 2 60 30, 1 3 60 20 1 4 60 15, 1 5 60 12 1 2 1 3
60 1 4 1 5
60 1 2 1 3 1 4 1 5 5 6 2 9 0 1 2 1 3 1 4 1 5 1 2 1 3 1 4 1 5
You can use a calculator to find the value of a complex fraction.
c a l c u l a t o r
c l o s e - u p
h e l p f u l
h i n t
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 frac-tion by the LCD eliminates the denominators of the original fractions.E X A M P L E 3
More complicated denominatorsSimplify .
Solution
Because x2 9 (x 3)(x 3) and x2 6x 9 (x 3)2, the LCD is
(x 3)2(x 3). Multiply the numerator and denominator by the LCD:
Simplify. Factor out x 3. (x (x 3 2 )( ) 5 (x x 3 1 ) 2)
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Simplifying Expressions with Negative Exponents
Consider the expression
3a 1 1 b 2 1 1 .
Using the definition of negative exponents, we can rewrite this expression as a complex fraction: 3a 1 1 b 2 1 1
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 exponentsSimplify the complex fraction3a11b21
1
.
Solution
Multiply the numerator and denominator by 2ab, the LCD of the fractions. Remember that a1 a a0 1. 3a 1 1 b 2 1 1 (3 ( a 1 1 b 2 1 )2 1) a 2 b ab Distributive property 2 6 a b b a 2 b a
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3a1(2ab) 21(2ab) 1(2ab) b1(2ab) 3 a 1 2 1 1 b (x 2)(x 3) (x 3)[x 4(x 3)] (x 2)(x 3) x(x 3) 4(x 3)(x 3) (x x 3 )(x 2 3) (x 3)2(x 3) (x x 3)2 (x 3)2(x 3) x 4 3 (x 3)2(x 3) x x 2 2 9 x2 6 x x 9 x 4 3 x x 2 2 9 x2 6 x x 9 x43s t u d y
t i p
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 be-cause your test-taking mood will not match your studying mood.h e l p f u l
h i n t
In Examples 4, 5, and 6 we are simplifying the expres-sions 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 expo-nent and then the complex fraction could be simplified as in Examples 2 and 3.E X A M P L E 5
A complex fraction with negative exponentsSimplify the complex fractionaab
1 2 bb a 2 3 . 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:
a a b 1 2 b b a 2 3 (a (a b 1 2 b b a 2) 3 a ) 3 a b 3 2 b2 Distributive property a a 2b 4 2 b a 3 3
Note that the positive exponents of a3b2are just large enough to eliminate all of the
negative exponents when we multiply.
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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 exponentsEliminate negative exponents and simplify p p1q2.
Solution
If we multiply the numerator and denominator by pq2, we will eliminate the negative exponents: p p1q2 ( p p 1 1q2) p p q q 2 2 p2q p 2 q 2 1
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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 fractionsEastside Elementary has the same number of students as Westside Elementary. One-half 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.
p pq2 p2q2 p1q2 pq2 1 b2b2 b0 1 a3a3 a0 1 a1 a3b2 b2 a3b2 ab2 a3b2 ba3 a3b2
Because the number of students at Westside is also x, we have
1 2x
2
3x the total number of students who ride the buses and
1 6x
1
3x the total number of female students.
Because all of the female students ride the buses, we can express the percentage of riders who are female by the following rational expression:
Multiply the numerator and denominator by 6, the LCD for 2, 3, and 6:
3 x x 2 4 x x 3 7 x x 3 7 0.43 43%
So 43% of the students who ride the buses are female.
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True or false? Explain.
1. The LCM for 2, x, 6, and x2is 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
4. 5 6 3 2 True 5. 21 31 (2 3)1 False 6. (21 31)1 2 3 False 7. 2 31 51 False
8. x 212xfor any real number x. False
9. To simplifyaa1bb1, multiply the numerator and denominator by ab. True
10. To simplifyaab 3b 2 a a 5b 5 b 2 1
, multiply the numerator and denominator by a5b2.
True 1 2 1 3 1 1 2
1 6x 1 3x6 1 2x 2 3x
6 1 6x 1 3x 1 2x 2 3x
W A R M - U P S
E X E R C I S E S
6 . 4
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 denomi-nator by the LCD for all of the fractions.
Simplify each complex fraction. Use either method. See Example 1. 3. 1 3 0 4. 3 2 5 2 5. 8 6. 9 3 x 5 2 x 1
Simplify the complex fractions. Use Method B. See Example 2.
7. 8. 9. a a 2 a b b m m n 2 3 3 n m 4 a2 a b b 3 2 a 10. 11. 12. m n 2n 3m 2m x x 3 y y 8t w 12 t w 13. 14. 15. 60 8 m m 3 3 m 6 2 42 2 z z 12 a2 b2 ab 16. 17. 18. 2xy 2x y y x 2 y x x 2y 2 2 3 a a 2b b 2 3 2 6 b a 3 3
Simplify each complex fraction. See Examples 1–3.
19. 20. x x 2 2 x x 2 5 21. 22. (y y 2 1)( y 3y 2 4) 2 4 a a 2 2 3 a a 1 1 4 4 2 a 3 2 4 a 1 2 1 y 1 1 3 y 1 1 x x x 6 2 x 4x x 1 2 5 x x 4 4 x 4 x x 4 4 2a 1 3b a 1 b4 6a 1 2 b2 3a 1 4b x2 1 y2 x1y3 x 1 3 y x 1 y 4x2 x2 y 1 4x x y2 2 a2 a 2b3 b2 a a 3b b 6 2 z z 3 1 z 1 6 3 m 6 2 4 9 m2 w2 3t w 4w t t x xy 3y 1 x 1 y m 2 n m1 3n a 3 b b a 1 b m m 2 n m m n3 3 a b b a a b b 2 5 9x 13 1 3 5x 1 2 5 2 3 5 6 1 2 1 8 1 3 1 1 2 1 3 1 4 1 5 1 6 1 2 1 3 1 4 1 5 23. 24. 4x x 4 10 x 3x 10 5 25. 26. 2w2 6w w 3 4 2x2 2 x 3x 1 3 27. 28. 2 a b 3 a b 7 2 x x 2 8 29. 30. 3 5 a a 7 2 2 x x 2 2 3 1 x 2 x 12 6 31. 32. ( 3 m m2 3 1 )( 2 2 m m 1 1 2 ) 3y y 2 2 3 1 3 1 y y 5 2 4 4 33. 34. 2 4 x x 2 2 2 4 x x 6 5 5 a 3 2 a 2 13 2 a a 2 8 2
Simplify. See Examples 4–6.
35. w z 1 1 y y 1 1 36. a a 1 1 b b 1 1 w yz y w w z z b b a a 37. 1 1 x x 1 2 38. 4 2 a a 2 1 x x 1 2a a 1 39. a 2 a 1b b2 40. m m 3 n 2 n3 a2 a b3 b2 n3 m 4n m3 a3 2 8 a2 2 3 a 4 a2 4 4 a a 3 3 8 x2 3 1 x x 3 2 1 x2 3 x 1 x x 3 3 1 y 1 2 3 4 y 3 y y 2 3 m 2 3 m4 m 3 2 m1 3 x x 9 1 x 6 x 2 x 9 x 3 a 4 1 5 1 3 a 2 3 x 2 4 x x 1 2 x 3 2 a 1 b a 3 b b 2 a b 4 a x x 1 2 x x 2 3 x x 3 3 x x 1 2 w w 2 1 w w 3 w w 4 w w 2 1 x x 5 2 5 2 x x 1 3 2 x 4 x 1 3 1
41. 1 a1 42. m1 a1 a a 1 a a m m 43. x x 1 x x 2 2 44. 1 x x x 2 2 x2 1 x 1 x2 x x 1 1 45. 2m 1 m 2 3m2 46. 4x 3 2 x5 6x5 2m 3 2x2 3 47. a a 1 b b 1 48. a a2 2 b b 2 2 a 1 b a2b2 49. x x3 3 y y 3 3 50. a ( a 2 b b ) 2 2 x3y3 a 2 b a 3 a b 3 b2 51. x1 1 2 x 8 2 x 3 4x3 52. 1 a 3 a 2 1 7 a 9 2 a2 x 2 a 3 53. (x1 y1)1 54. (a1 b1)2 x x y y b2 a 2 2 a b b 2 a2
Use a calculator to evaluate each complex fraction. Round answers to four decimal places. If your calcula-tor does fractions, then also find the fractional answer.
55. 56. 1.7333, 2 1 6 5 0.1163, 4 5 3 57. 4 2 1 1 9 3 1 1 58. 2 3 1 1 3 5 1 1 6 4 1 1 0.1667, 1 6 1.7391, 4 2 0 3
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 African-American, three-fourths of the students at Central are African-American, and one-sixth of the students at South-side 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 Explo-sives are in either development, manufacturing, or sales.
1 1 2 1 2 3 4 3 5 5 6 5 3 4 5 1 3 5 6
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 work-ers at Acme who are women. What percent of the women at Acme are in sales?
38.3%, 65.2%
F I G U R E F O R E X E R C I S E 6 0 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
F I G U R E F O R E X E R C I S E 6 2 Distribution of Employees at Acme Explosives Sales Manufacturing Development 1 4 1 4 1 2
G E T T I N G M O R E I N VO LV E D
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 com-plex fractions from Exercises 35–54.
64. Discussion. a) Find the exact value of each expression.
i) 1 ii) 1 1 1 1 1 1 1 1 1 1 1 2 1 1 1 1 3 5 8, 1 1 1 8
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? 1 1 1 1 1 1 1 x 2 3 x x 2 1 , x 0, 1, 1 2, 2 3
S O L V I N G E Q U A T I O N S I N V O L V I N G
R A T I O N A L E X P R E S S I O N S
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 equa-tion by the LCD of the raequa-tional expressions.
E X A M P L E 1
An equation with rational expressionsSolve1x1416.
Solution
The LCD for the denominators 4, 6, and x is 12x:
12x
1x
12x
1
6
Multiply each side by 12x.
3 2 12x 1 x 12x 1 4 12x 1 6 Distributive property 12 3x 2x Divide out the common factors.
12 x 0
x 12
Check12 in the original equation. The solution set is 12.
