Download ADDITION AND SUBTRACTION

350
(6-18)
Chapter 6
Rational Expressions
6.3
In this
section
●
Adding and Subtracting
with Identical
Denominators
●
Least Common
Denominator
●
Adding and Subtracting
with Different
Denominators
●
Shortcuts
●
Applications
ADDITION AND SUBTRACTION
We can multiply or divide any rational expressions, but we add or subtract only
rational expressions with identical denominators. So when the denominators are not
the same, we must find equivalent forms of the expressions that have identical
denominators. In this section we will review the idea of the least common denominator and will learn to use it for addition and subtraction of rational expressions.
Adding and Subtracting with Identical Denominators
It is easy to add or subtract fractions with identical denominators. For example,
1 3 4
7 7 7
and
3 2 1
.
5 5 5
In general, we have the following definition.
Addition and Subtraction of Rational Numbers
If b 0, then
a c
ac
b
b b
and
a c ac
.
b
b b
Rational expressions with identical denominators are added or subtracted in the
same manner as fractions.
E X A M P L E
helpful
1
hint
You can remind yourself of the
difference between addition
and multiplication of fractions
with a simple example: If you
and your spouse each own
17 of Microsoft, then together you own 27 of
Microsoft. If you own 17
of Microsoft, and give 17 of
your stock to your child, then
your child owns 149 of
Microsoft.
Identical denominators
Perform the indicated operations.
3
5
a) 2x 2x
x2 4x 7 x2 2x 1
c) x2 1
x2 1
5x 3 5 7x
b) x1
x1
Solution
3
5
8
a) Add the numerators.
2x 2x 2x
4
Reduce.
x
5x 3 5 7x 5x 3 5 7x
b) x1
x1
x1
2x 2
x1
2(x 1)
x1
2
Add the numerators.
Combine like terms.
Factor.
Reduce to its lowest terms.
6.3
study
tip
Studying in an environment
similar to the one in which
you will be tested can increase
your chances of recalling
information. When possible,
review for a test in the classroom in which you will take
the test.
Addition and Subtraction
(6-19)
351
c) The polynomials in the numerators are treated as if they were in parentheses:
2
2
x 2 4x 7 x 2 2x 1 x 4x 7 (x 2x 1)
2
2
2
x 1
x 1
x 1
2
x 4x 7 x 2 2x 1
x2 1
6x 6
x2 1
6(x 1)
6
(x 1)(x 1) x 1
■
Least Common Denominator
To add fractions with denominators that are not identical, we use the basic principle
of rational numbers to build up the denominators to the least common denominator (LCD). For example,
12
3
2
5
13
1 1
.
43
4 6
6 2 12 12 12
The LCD 12 is the least common multiple (LCM) of the numbers 4 and 6.
Finding the LCM for a pair of large numbers such as 24 and 126 will help you
to understand the procedure for finding the LCM for any polynomials. First factor
the numbers completely:
24 23 3
126 2 32 7
hint
126
The product of 24 and 126 is
3024 and 3024 is a common
multiple but not the least
common multiple of 24 and
126. If you divide 3024 by 6,
the greatest common factor
of 24 and 126, you get 504.
Any number that is a multiple of both 24 and 126 must have all of the factors of 24
and all of the factors of 126 in its factored form. So in the LCM we use the factors 2, 3, and 7, and for each factor we use the highest power that appears on that
factor. The highest power of 2 is 3, the highest power of 3 is 2, and the highest
power of 7 is 1. So the LCM is 23 32 7. If we write this product without
exponents, we can see clearly that it is a multiple of both 24 and 126:
2 2 2 3 3 7 504
helpful
24
504 126 4
504 24 21
The strategy for finding the LCM for a group of polynomials can be stated as
follows.
Strategy for Finding the LCM for Polynomials
1. Factor each polynomial completely. Use exponents to express repeated
factors.
2. Write the product of all of the different factors that appear in the
polynomials.
3. For each factor, use the highest power of that factor in any of the
polynomials.
352
(6-20)
Chapter 6
E X A M P L E
2
Rational Expressions
Finding the LCM
Find the least common multiple for each group of polynomials.
b) a2bc, ab3c2, a3bc
c) x2 5x 6, x2 6x 9
a) 4x2 y, 6y
Solution
a) Factor 4x2y and 6y as follows:
6y 2 3y
4x2y 22 x2y,
To get the LCM, we use 2, 3, x, and y the maximum number of times that each
appears in either of the expressions. The LCM is 22 3 x2y, or 12x2 y.
b) The expressions a2bc, ab3c2, and a3bc are already factored. To get the LCM, we
use a, b, and c the maximum number of times that each appears in any of the
expressions. The LCM is a3b3c2.
c) Factor x2 5x 6 and x2 6x 9 completely:
x2 5x 6 (x 2)(x 3),
x2 6x 9 (x 3)2
The LCM is (x 2)(x 3)2.
■
Adding and Subtracting with Different Denominators
To add or subtract rational expressions with different denominators, we must build
up each rational expression to equivalent forms with identical denominators, as we
did in Section 6.1. Of course, it is most efficient to use the LCD as in the following
examples.
E X A M P L E
3
Different denominators
Perform the indicated operations.
x 1 2x 3
5
3
a) 3
b) 6
4
a2b
ab
Solution
a) The LCD for a2b and ab3 is a2b3. To build up each denominator to a2b3, multiply the numerator and denominator of the first expression by b2, and multiply
the numerator and denominator of the second expression by a:
5
3(b2)
3
5(a)
Build up each denominator to the LCD.
2
3
2
2 a b ab
a b(b ) ab3(a)
3b2
5a
2 3 ab
a2b3
3b2 5a
Add the numerators.
a2b3
x 1 2x 3 (x 1)(2) (2x 3)(3) Build up each denominator
b) to
the LCD 12.
6
4
4(3)
6(2)
2x 2 6x 9
Distributive property
12
12
Subtract the numerators.
2x 2 (6x 9)
Note that 6x 9 is put in
12
parentheses.
2x 2 6x 9
12
4x 11
12
Remove the parentheses.
Combine like terms.
6.3
(6-21)
Addition and Subtraction
353
Before you add or subtract rational expressions, they must
CAUTION
be written with identical denominators. For multiplication and division it is not
necessary to have identical denominators.
In the next example we must first factor polynomials to find the LCD.
hint
It is not actually necessary to
identify the LCD. Once the denominators are factored, simply look at each denominator
and ask,“What factor does the
other denominator have that
is missing from this one?”
Then use the missing factor to
build up the denominator and
you will obtain the LCD.
Different denominators
Perform the indicated operations.
1
2
a) 2 2 x 1 x x
5
3
b) a2 2a
Solution
a) Because x2 1 (x 1)(x 1) and x2 x x(x 1), the LCD is
x(x 1)(x 1). The first denominator is missing the factor x, and the second
denominator is missing the factor x 1.
1
2
1
2
2 2 x 1 x x (x 1)(x 1) x (x 1)
helpful
4
E X A M P L E
Missing x
Missing x 1
1(x)
2(x 1)
(x 1)(x 1)(x) x(x 1)(x 1)
The LCD is
x(x 1)(x 1).
Build up the
denominators to
the LCD.
x
2x 2
x(x 1)(x 1) x(x 1)(x 1)
3x 2
x(x 1)(x 1)
Add the
numerators.
For this type of answer we usually leave the denominator in factored form. That
way, if we need to work with the expression further, we do not have to factor the
denominator again.
b) Because 1(2 a) a 2, we can convert the denominator 2 a to a 2.
5
3
5
3(1)
a 2 2 a a 2 (2 a)(1)
5
3
a2 a2
The LCD is a 2.
5 (3)
a2
Subtract the numerators.
8
a2
Simplify.
Note that if we had changed the denominator of the first expression to 2 a, we
would have gotten the answer
8
,
2a
but this rational expression is equivalent to the first answer.
354
(6-22)
Chapter 6
Rational Expressions
Shortcuts
Consider the following addition:
a c a(d ) c(b) ad bc
b d b(d ) d(b)
bd
The LCD is bd.
We can use this result as a rule for adding simple fractions in which the LCD is the
product of the denominators. A similar rule works for subtraction.
Adding or Subtracting Simple Fractions
If b 0 and d 0, then
a c ad bc
b d
bd
E X A M P L E
5
and
a c ad bc
.
b d
bd
Adding and subtracting simple fractions
Use the rules for adding and subtracting simple fractions to find the sums and
differences.
1
2
1
3
a) a a
c) 5 3
1 1
b) a x
2
d) x 3
Solution
a) For the numerator, compute ad bc 1 3 2 1 5. Use 2 3 or 6 for
the denominator:
1 1 5
2 3 6
1 1 1x1a xa
b) a x
ax
ax
a a 3a 5a 8a
c) 5 3
15
15
2 x 2 3x 2
d) x 3 1 3
3
CAUTION
The rules for adding or subtracting simple fractions can be applied to any rational expressions, but they work best when the LCD is the product
of the two denominators. Always make sure that the answer is in its lowest terms. If
the product of the two denominators is too large, these rules are not helpful because
then reducing can be difficult.
Applications
Rational expressions occur often in expressing rates. For example, if you can
process one application in 2 hours, then you are working at the rate of 21 of
6.3
Addition and Subtraction
(6-23)
355
an application per hour. If you can complete one task in x hours, then you
1
are working at the rate of task per hour.
x
E X A M P L E
helpful
6
hint
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.
Work rates
Susan takes an average of x hours to process a mortgage application, whereas
Betty’s average is 1 hour longer. Write a rational expression for the number of
applications that they can process in 40 hours.
Solution
The number of applications processed by Susan is the product of her rate and her
time:
1 application
40
40 hr applications
x
hr
x
The number of applications processed by Betty is the product of her rate and her
time:
1 application
40
40 hr applications
x1
x1
hr
Find the sum of the rational expressions:
40x 40 40x 80x 40
40
40
x(x 1)
x(x 1)
x1
x
So together in 40 hours they process
WARM-UPS
80x 40
x(x 1)
applications.
True or false? Explain.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
The LCM of 6 and 10 is 60. False
The LCM of 6a2b and 8ab3 is 24ab. False
The LCM of x2 1 and x 1 is x2 1. True
The LCD for the rational expressions 5 and x3 is x 1. False
x
x1
1 2 3
False
2 3 5
1
6
5 for any nonzero real number x. False
x
x
7
7 3a
3 for any a 0. True
a
a
c d 5c 3d
for any real numbers c and d. True
15
3 5
2 3 17
True
3 4 12
If Jamal uses x reams of paper in one day, then he uses 1 ream per day.
x
False
356
6. 3
(6-24)
Chapter 6
Rational Expressions
EXERCISES
Reading and Writing After reading this section, write out the
answers to these questions. Use complete sentences.
1. How do you add rational numbers?
The sum of ab and cb is (a c)b.
2. What is the least common denominator (LCD)?
The LCD is the least common multiple of the denominators.
3. What is the least common multiple?
The least common multiple (LCM) of some numbers is the
smallest number that is a multiple of all of the numbers.
4. How do we find the LCM for a group of polynomials?
To find the LCM for some polynomials, first factor completely, then use the LCM of the coefficients and every
other factor with the highest exponent that occurs on that
factor in any of the polynomials.
5. How do we add or subtract rational expressions with different denominators?
To add rational expressions with different denominators,
you must build up the expressions to equivalent expressions with the same denominator.
6. For which operations with rational expressions is it not
necessary to have identical denominators?
You do not need identical denominators for division and
multiplication.
Perform the indicated operations. Reduce answers to their lowest terms. See Example 1.
3x 5x
7. 4x
2
2
5x2 4x2
8. 3x2
3
3
x 3 3x 5 x 1
9. 2x
2x
x
9 4y 6 y 1 y
10. 3y
3y
y
3x 4 2x 6 5
11. 2x 4 2x 4 2
a3
b3
12. a 2 ab b2
ab ab
2
x2 4x 6 x 2 2 x 12
13. 2 2
x 9
x3
x 9
x2 3x 3 x2 4x 7
14. 1
x4
x4
Find the least common multiple for each group of polynomials.
See Example 2.
15.
16.
17.
18.
24, 20 120
12, 18, 22 396
10x3y, 15x 30x3y
12a3b2, 18ab5 36a3b5
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
a3b, ab4c, ab5c2 a3b5c2
x2yz, xy2z3, xy6 x2y6z3
x, x 2, x 2 x(x 2)(x 2)
y, y 5, y 2 y(y 5)(y 2)
4a 8, 6a 12 12a 24
4a 6, 2a2 3a 4a 2 6a
x2 1, x2 2x 1 (x 1)(x 1)2
y2 2y 15, y2 6y 9 (y 5)(y 3)2
x2 4x, x2 16, x2 6x 8 x(x 4)(x 4)(x 2)
z2 25, 5z 25, 5z 25 5z2 125
Perform the indicated operations. Reduce answers to its lowest
terms. See Examples 3 and 4.
1
3 17
29. 28 35 140
7
1
5
30. 36 144
48
4
1
7
31. 24
15 40
7
3 109
32. 52 40 520
3
5 3w 5z
33. 2 2 2
wz
wz
w z2
2
3 2b 3a
34. 2 2 2
a b ab
a b2
2x 3 x 2 2x 1
35. 8
6
24
a 5 3 2a a 9
36. 10
15
30
x
3x 11x
37. 2a 5a 10a
x
3x
5x
38. 6y 8y
24y
9
9 4xy
39. x 4y
4y
2
2
b
b 4ac
40. c 4a
4a
5
7 2a 14
41. a 2 a a(a 2)
2
3 x 3
42. x 1 x x(x 1)
1
2
3a b
43. a b a b (a b)(a b)
5
3
8x 4
44. x 2 x 2 (x 2)(x 2)
x
3
4x 9
45. x2 9 x 3 (x 3)(x 3)
6.3
x
6x 25
5
46. x2 25 x 5 (x 5)(x 5)
1
1
47. 0
ab ba
3
7
4
48. x5 5x x5
5
3
11
49. 2x 4 2 x 2x 4
4
7
25
50. 3x 9 3 x 3x 9
x 3
5
6
51. 2
2
x x 2 x 2x 3 (x 1)(x 2)(x 3)
3x
2
5
52. x2 4 x2 3x 10 (x 2)(x 2)(x 5)
Addition and Subtraction
2
3
76. a3 a1
1
1
77. a3 1 a3 1
1
1
78. x3 8 x3 8
(a2b3)4 (ab)3
79. (ab4)3 (a4b)2
(6-25)
a 7
(a 3)(a 1)
2
(a3 1)(a3 1)
2x3
3
(x 8)(x3 8)
b
(a b)3 a b
(ab)2
80. 2 (ab)3
ab
(a b)
5x 2 x 4
x2 3x
4
81. 3
x 1
x 1 (x 1)(x 2 x 1)
4a
3
a3
82. 2
2
a 2a 1 a 1 (a 1)
x2 7x 18
x
6
53. 2
2
x 9 x 4x 3 (x 1)(x 3)(x 3)
3x2 4x 22
2x 1
x5
54. x2 x 12 x2 5x 6 (x 2)(x 3)(x 4)
1
2
3
8x 2
55. x x 1 x 2 x(x 1)(x 2)
2
3
5
4a2 8a 2
56. a a 1 a 1 a(a 1)(a 1)
x2 25 x2 10x 25 x 2 25
83. x2 25
x2 5x
x(x 5)
Perform the following operations. Write down only the answer.
See Example 5.
1 7
1
3
1 17
58. 57. 3
5
4
20
4 12
19
a 5 3a 10
1
3
59. 60. 8
5
40
2 3
6
x x 5x
y y
y
61. 62. 3 2 6
4 3
12
a3
2
a3
3a 2 6a 5
86. 3
2
a 8 a 2 a 2a 4
a3 8
3a 2b
3b
2 3a 2
65. a 3
3
3
a3
67. 1 a
a
3 1
64. x 9
m
66. y
3
1
68. 1
x
3x
69. 1
x
a2
70. 3
a
a 2
63. b 3
2
1
71. 3 4x
3
x
8x 3
12x
x 27
9x
m 3y
3
x1
x
1
1
72. 5 5x
Perform the indicated operations.
w2 3w 6 9 w2
73. 3
w5
w5
2z 2 3z 6 z 2 5z 9 z 3
74. z2 1
z1
z2 1
1
2
x 1
75. x 2 x 3 (x 2)(x 3)
4a 2
a
x1
5x
357
a2 a 2 a2 a 1
1
84. a3 1
a2 4
a2
w2 3
2
w4
85. 3w3 81 6w 18 w2 3w 9
w2 2w 8
(w 3)(w2 3w 9)
a2 6a 9 a2 a 6
a3
87. a3 8
a2 4
a 2 2a 4
1
z3 8
88. 2
z 4 z4 16
z2
z2 2z 4
w2 3
2w
89. 2
3
w 8 w 4
w3 2w2 5w 6
(w 2)(w 2)(w2 2w 4)
x5
x1
90. x3 27 x2 9
x3 5x2 10x 6
(x2 9)(x2 3x 9)
1
1
1
91. x3 1 x 2 1 x 1
x4
x2
92. 3
x 1 x2 1
x3 x2 2x 1
(x 1)(x 1)(x2 x 1)
x 3 4x 6
(x 3 1)(x 1)
In Exercises 93–98, solve each problem. See Example 6.
93. Processing. Joe takes x hours on the average to process a
claim, whereas Ellen averages x 1 hours to process a
claim. Write a rational expression for the number of claims
that they will process while working together for an 8-hour
shift.
16x 8
x2 x
94. Roofing. Bill attaches one bundle of shingles in an average
of x minutes using a hammer, whereas Julio can attach one
bundle in an average of x 6 minutes using a pneumatic
stapler. Write a rational expression for the number of
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.