3: Adding and Subtracting Algebraic Fractions
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7.3 • Adding and Subtracting Algebraic Fractions
279
Definition 7.3
If
A
C
and are rational expressions with B ϶ 0, then
B
B
A
C
AϩC
ϩ ϭ
B
B
B
and
C
AϪC
A
Ϫ ϭ
B
B
B
Thus if the denominators of two algebraic fractions are the same, we can add or subtract the
fractions by adding or subtracting the numerators and placing the result over the common
denominator. Here are some examples:
5
7
5ϩ7
12
ϩ ϭ
ϭ
x
x
x
x
8
3
8Ϫ3
5
Ϫ
ϭ
ϭ
xy
xy
xy
xy
14
15
14 ϩ 15
29
ϩ
ϭ
ϭ
2x ϩ 1
2x ϩ 1
2x ϩ 1
2x ϩ 1
3
4
3Ϫ4
Ϫ1
or
Ϫ
ϭ
ϭ
aϪ1
aϪ1
aϪ1
aϪ1
Ϫ
1
aϪ1
In the next examples, notice how we put to use our previous work with simplifying
polynomials.
1x ϩ 32 ϩ 12x Ϫ 32
xϩ3
2x Ϫ 3
3x
ϩ
ϭ
ϭ
4
4
4
4
1x
ϩ
52
Ϫ
1x
ϩ
22
xϩ5
xϩ2
xϩ5ϪxϪ2
3
Ϫ
ϭ
ϭ
ϭ
7
7
7
7
7
13x ϩ 12 ϩ 12x ϩ 32
3x ϩ 1
2x ϩ 3
5x ϩ 4
ϩ
ϭ
ϭ
xy
xy
xy
xy
213n ϩ 12
31n Ϫ 12
213n ϩ 12 Ϫ 31n Ϫ 12
6n ϩ 2 Ϫ 3n ϩ 3
3n ϩ 5
Ϫ
ϭ
ϭ
ϭ
n
n
n
n
n
It may be necessary to simplify the fraction that results from adding or subtracting two
fractions.
14x Ϫ 32 ϩ 12x ϩ 32
4x Ϫ 3
2x ϩ 3
6x
3x
ϩ
ϭ
ϭ
ϭ
8
8
8
8
4
13n
Ϫ
12
Ϫ
1n
Ϫ
52
3n Ϫ 1
nϪ5
3n Ϫ 1 Ϫ n ϩ 5
Ϫ
ϭ
ϭ
12
12
12
12
ϭ
21n ϩ 22
2n ϩ 4
nϩ2
ϭ
ϭ
12
12
6
1Ϫ2x ϩ 32 ϩ 13x Ϫ 12
Ϫ2x ϩ 3
3x Ϫ 1
xϩ2
ϩ 2
ϭ
ϭ 2
2
x Ϫ4
x Ϫ4
x2 Ϫ 4
x Ϫ4
ϭ
xϩ2
1x ϩ 221x Ϫ 22
ϭ
1
xϪ2
Recall that to add or subtract rational numbers with different denominators, we first
change them to equivalent fractions that have a common denominator. In fact, we found that
by using the least common denominator (LCD), our work was easier. Let’s carefully review
the process because it will also work with algebraic fractions in general.
280
Chapter 7 • Algebraic Fractions
Classroom Example
3
5
Add ϩ .
6
7
EXAMPLE 1
Add
3
1
ϩ .
5
4
Solution
By inspection, we see that the LCD is 20. Thus we can change both fractions to equivalent
fractions that have a denominator of 20.
3
1
3 4
1 5
12
5
17
ϩ ϭ a bϩ a bϭ
ϩ
ϭ
5
4
5 4
4 5
20
20
20
Form
of 1
Classroom Example
7
2
Subtract
Ϫ .
12
15
Form
of 1
EXAMPLE 2
Subtract
5
7
Ϫ .
18
24
Solution
If we cannot find the LCD by inspection, then we can use the prime factorization forms.
и3и3
24 ϭ 2 и 2 и 2 и 3
18 ϭ 2
v ¡ LCD ϭ 2
и 2 и 2 и 3 и 3 ϭ 72
5
7
5 4
7 3
20
21
1
Ϫ
ϭ
a bϪ
a bϭ
Ϫ
ϭϪ
18
24
18 4
24 3
72
72
72
Now let’s consider adding and subtracting algebraic fractions with different denominators.
Classroom Example
4m ϩ 3
mϩ1
Add
ϩ
.
5
6
EXAMPLE 3
Add
xϪ2
3x ϩ 1
ϩ
.
4
3
Solution
By inspection, we see that the LCD is 12.
xϪ2
3x ϩ 1
xϪ2
3
3x ϩ 1
4
ϩ
ϭ a
ba b ϩ a
ba b
4
3
4
3
3
4
ϭ
ϭ
ϭ
Classroom Example
xϪ8
xϪ4
Ϫ
.
Subtract
2
10
EXAMPLE 4
31x Ϫ 22
413x ϩ 12
ϩ
12
12
31x Ϫ 22 ϩ 413x ϩ 12
12
3x Ϫ 6 ϩ 12x ϩ 4
15x Ϫ 2
ϭ
12
12
Subtract
nϪ2
nϪ6
Ϫ
.
2
6
Solution
By inspection, we see that the LCD is 6.
nϪ2
nϪ6
nϪ2
3
nϪ6
Ϫ
ϭ a
ba b Ϫ
2
6
2
3
6
ϭ
31n Ϫ 22
6
Ϫ
1n Ϫ 62
6
7.3 • Adding and Subtracting Algebraic Fractions
ϭ
281
31n Ϫ 22 Ϫ 1n Ϫ 62
6
3n Ϫ 6 Ϫ n ϩ 6
ϭ
6
2n
6
n
ϭ
3
ϭ
Don’t forget to simplify!
It does not create any serious difficulties when the denominators contain variables; our
approach remains basically the same.
Classroom Example
2
5
Add
ϩ .
5n
6n
EXAMPLE 5
Add
3
7
ϩ .
4x
3x
Solution
By inspection, we see that the LCD is 12x.
3
7
3 3
7 4
9
28
9 ϩ 28
37
ϩ
ϭ a bϩ a bϭ
ϩ
ϭ
ϭ
4x
3x
4x 3
3x 4
12x
12x
12x
12x
Classroom Example
19
15
Subtract
Ϫ
.
24a
32a
EXAMPLE 6
Subtract
11
5
.
Ϫ
12x
14x
Solution
12x ϭ 2
14x ϭ 2
и 2 и 3 и xv
и7иx
¡ LCD ϭ 2
и 2 и 3 и 7 и x ϭ 84x
11
5
11 7
5 6
Ϫ
ϭ
a bϪ
a b
12x
14x
12x 7
14x 6
77
30
77 Ϫ 30
47
ϭ
Ϫ
ϭ
ϭ
84x
84x
84x
84x
Classroom Example
8
3
Add ϩ
.
x
xϩ5
EXAMPLE 7
Add
4
2
ϩ
.
y
yϪ2
Solution
By inspection, we see that the LCD is y1y Ϫ 22 .
y
2
4
2 yϪ2
4
ϩ
ϭ a
bϩ
a b
y
y
y
yϪ2
yϪ2
yϪ2
Form
of 1
ϭ
ϭ
ϭ
21y Ϫ 22
y1y Ϫ 22
Form
of 1
ϩ
4y
y1y Ϫ 22
21y Ϫ 22 ϩ 4y
y1y Ϫ 22
2y Ϫ 4 ϩ 4y
6y Ϫ 4
ϭ
y1y Ϫ 22
y1y Ϫ 22
282
Chapter 7 • Algebraic Fractions
Notice the final result in Example 7. The numerator, 6y Ϫ 4, can be factored into 2(3y Ϫ 2).
However, because this produces no common factors with the denominator, the fraction can
6y Ϫ 4
not be simplified. Thus the final answer can be left as
; it is also acceptable to express
y1y Ϫ 22
213y Ϫ 22
it as
.
y1y Ϫ 22
Classroom Example
6
3
Subtract
Ϫ
.
xϩ7
xϪ2
EXAMPLE 8
Subtract
4
7
Ϫ
.
xϩ2
xϩ3
Solution
By inspection, we see that the LCD is 1x ϩ 221x ϩ 32 .
4
7
4
xϩ3
7
xϩ2
Ϫ
ϭ a
ba
bϪ a
ba
b
xϩ2
xϩ3
xϩ2
xϩ3
xϩ3
xϩ2
ϭ
ϭ
41x ϩ 32
1x ϩ 221x ϩ 32
Ϫ
71x ϩ 22
1x ϩ 321x ϩ 22
41x ϩ 32 Ϫ 71x ϩ 22
1x ϩ 221x ϩ 32
ϭ
4x ϩ 12 Ϫ 7x Ϫ 14
1x ϩ 221x ϩ 32
ϭ
Ϫ3x Ϫ 2
1x ϩ 221x ϩ 32
Concept Quiz 7.3
For Problems 1–10, answer true or false.
2x
1
2x ϩ 1
1. The addition problem
is equal to
for all values of x except
ϩ
xϩ4
xϩ4
xϩ4
1
x ϭ Ϫ and x ϭ Ϫ4.
2
2. Any common denominator can be used to add rational expressions, but typically we
use the least common denominator.
10x2z
2x2
and
are equivalent fractions.
3y
15yz
4. The least common multiple of the denominators is always the lowest common
denominator.
3. The fractions
5
3
5. To simplify the expression
we could use 2x Ϫ 1 for the common
ϩ
2x Ϫ 1
1 Ϫ 2x
denominator.
1
5
3
2
, then
ϩ
ϭ
.
2
2x Ϫ 1
1 Ϫ 2x
2x Ϫ 1
Ϫ2
17
3
Ϫ
ϭ
Ϫ4
3
12
4x Ϫ 1
2x ϩ 1
x
ϩ
ϭ
5
6
5
x
3x
5x
5x
Ϫ
ϩ
ϭ
4
2
3
12
2
3
Ϫ5 Ϫ 6x
If x ϶ 0, then
.
Ϫ
Ϫ1ϭ
3x
2x
6x
6. If x ϶
7.
8.
9.
10.
7.3 • Adding and Subtracting Algebraic Fractions
Problem Set 7.3
For Problems 1–34, add or subtract as indicated. Be sure to
express your answers in simplest form. (Objective 1)
39.
y
3y
ϩ
6
4
40.
3y
7y
ϩ
4
5
1.
5
12
ϩ
x
x
2.
17
13
Ϫ
x
x
41.
8x
3x
Ϫ
3
7
42.
5y
3y
Ϫ
6
8
3.
7
5
Ϫ
3x
3x
4.
4
3
ϩ
5x
5x
43.
2x
3x
ϩ
6
5
44.
6x
7x
ϩ
9
12
5.
7
1
ϩ
2n
2n
6.
5
4
ϩ
3n
3n
45.
7n
3n
Ϫ
8
9
46.
8n
7n
Ϫ
10
15
7.
9
13
Ϫ 2
2
4x
4x
8.
12
22
Ϫ 2
2
5x
5x
47.
48.
10.
xϪ2
4
ϩ
x
x
xϪ2
xϩ1
ϩ
5
6
9.
xϩ1
3
ϩ
x
x
xϩ3
xϪ4
ϩ
5
2
49.
xϪ6
xϩ2
ϩ
9
3
50.
xϪ2
xϩ4
ϩ
4
8
11.
3
6
Ϫ
xϪ1
xϪ1
12.
8
10
Ϫ
xϩ4
xϩ4
51.
52.
14.
2x ϩ 3
3
Ϫ
x
x
2n ϩ 3
4n Ϫ 1
ϩ
4
7
13.
xϩ1
1
Ϫ
x
x
3n Ϫ 1 2n ϩ 5
ϩ
3
4
53.
54.
16.
4t Ϫ 1
8t Ϫ 5
ϩ
7
7
5n Ϫ 2
4n ϩ 7
Ϫ
12
6
15.
3t Ϫ 1
2t ϩ 3
ϩ
4
4
4n Ϫ 3
3n ϩ 5
Ϫ
6
18
55.
56.
18.
9a Ϫ 1
4a Ϫ 2
Ϫ
6
6
5x
3x
7x
Ϫ
Ϫ
2
4
6
17.
7a ϩ 2
4a Ϫ 6
Ϫ
3
3
3x
x
5x
ϩ Ϫ
4
6
8
57.
58.
20.
2n Ϫ 5
6n Ϫ 1
ϩ
10
10
4x 5 11x
ϩ Ϫ
3
9
6
19.
4n ϩ 3
6n ϩ 5
ϩ
8
8
x
3
7x
Ϫ
Ϫ
5
10
12
59.
60.
22.
2n Ϫ 6
7n Ϫ 1
Ϫ
5
5
7
5
ϩ
8x
12x
21.
3n Ϫ 7
9n Ϫ 1
Ϫ
6
6
5
1
ϩ
8x
6x
61.
62.
24.
4x ϩ 1
2x ϩ 5
Ϫ
3x
3x
11
8
Ϫ
9y
15y
23.
5x Ϫ 2
8x ϩ 3
Ϫ
7x
7x
5
7
Ϫ
6y
9y
63.
5
11
Ϫ
12x
16x2
64.
4
7
Ϫ 2
9x
6x
65.
3
2
5
Ϫ
ϩ
2x
3x
4x
66.
3
5
10
Ϫ
ϩ
4x
6x
9x
3n
67.
68.
11x Ϫ 9
8x2
3
7
ϩ
x
xϪ5
4
9
ϩ
x
xϪ8
69.
2
3
Ϫ
n
nϪ1
70.
5
7
Ϫ
n
nϩ3
71.
4
6
Ϫ
n
nϩ4
72.
8
3
Ϫ
n
nϪ9
73.
6
12
Ϫ
x
2x ϩ 1
74.
2
6
Ϫ
x
3x Ϫ 2
75.
4
6
ϩ
xϩ4 xϪ3
76.
7
8
ϩ
xϪ2
xϩ1
25.
27.
29.
31x ϩ 22
4x
61n Ϫ 12
3n
213x Ϫ 42
2
7x
ϩ
ϩ
Ϫ
61x Ϫ 12
4x
31n ϩ 22
3n
7xϪ 8
7x2
26.
28.
30.
41x Ϫ 32
5x
21n Ϫ 42
3n
ϩ
ϩ
314x Ϫ 32
8x2
21x ϩ 62
5x
41n ϩ 22
Ϫ
31.
a2
4
Ϫ
aϩ2
aϩ2
32.
n2
16
Ϫ
nϪ4
nϪ4
33.
3x
18
Ϫ
2
1x Ϫ 62
1xϪ62 2
34.
x2 ϩ 5x
4
ϩ
2
1x ϩ 12
1x ϩ 12 2
For Problems 35–80, add or subtract as indicated and express
your answers in simplest form. (Objective 3)
35.
3x
5x
ϩ
8
4
36.
5x
2x
ϩ
3
9
77.
3
9
Ϫ
xϪ2 xϩ1
78.
4
5
Ϫ
xϪ1
xϩ6
37.
7n
4n
Ϫ
12
3
38.
n
7n
Ϫ
6
12
79.
3
4
Ϫ
2x Ϫ 1
3x ϩ 1
80.
6
4
Ϫ
3x Ϫ 4
2x ϩ 3
283
284
Chapter 7 • Algebraic Fractions
Thoughts Into Words
81. Give a step-by-step description of how to do this addition problem:
83. Suppose that your friend does an addition problem as
follows:
51122 ϩ 8172
5
7
60 ϩ 56
116
29
ϩ
ϭ
ϭ
ϭ
ϭ
8
12
81122
96
96
24
3x Ϫ 1
2x ϩ 3
ϩ
6
9
3
3
and
opposites? What should
xϪ2
2Ϫx
3
3
be the result of adding
and
?
xϪ2
2Ϫx
82. Why are
Is this answer correct? What advice would you offer
your friend?
Further Investigations
9
4
ϩ
. Notice
xϪ2
2Ϫx
that the denominators x Ϫ 2 and 2 Ϫ x are opposites; that is,
Ϫ1(2 Ϫ x) = (x Ϫ 2). In such cases, add the fractions as
follows:
Consider the addition problem
9
4
9
4
Ϫ1
ϩ
ϭ
ϩ
a
b
xϪ2
2Ϫx
xϪ2
2 Ϫ x Ϫ1
9
Ϫ4
ϭ
ϩ
xϪ2
xϪ2
ϭ
9 ϩ 1Ϫ42
xϪ2
Form
of 1
5
ϭ
xϪ2
Answers to the Concept Quiz
1. False
2. True
3. True
4. True
7.4
5. True
For Problems 84–89, use this approach to help with the
additions and subtractions.
84.
7
2
Ϫ
xϪ1
1Ϫx
85.
5
1
ϩ
xϪ3
3Ϫx
86.
x
4
ϩ
xϪ4
4Ϫx
87.
Ϫ4
2
ϩ
aϪ1
1Ϫa
88.
2
1
3
ϩ
Ϫ
xϩ3
3Ϫx
x Ϫ9
89.
n
3
Ϫ
2n Ϫ 1
1 Ϫ 2n
2
6. True
7. False
8. False
9. True
10. True
Addition and Subtraction of Algebraic Fractions and
Simplifying Complex Fractions
OBJECTIVES
1
Add or subtract rational expressions for which the denominators can
be factored
2
Simplify complex fractions
In this section, we expand our work with adding and subtracting rational expressions, and we
discuss the process of simplifying complex fractions. Before we begin, however, this seems
like an appropriate time to offer a bit of advice regarding your study of algebra. Success in
algebra depends on having a good understanding of the concepts as well as being able to perform the various computations. As for the computational work, you should adopt a carefully
organized format that shows as many steps as you need in order to minimize the chances of
making careless errors. Don’t be eager to find shortcuts for certain computations before you
have a thorough understanding of the steps involved in the process. This advice is especially
appropriate at the beginning of this section.
7.4 • Addition and Subtraction of Algebraic Fractions and Simplifying Complex Fractions
285
Study Examples 1–4 very carefully. Note that the same basic procedure is followed in
solving each problem:
Step 1 Factor the denominators.
Step 2 Find the LCD.
Step 3 Change each fraction to an equivalent fraction that has the LCD as its denominator.
Step 4 Combine the numerators and place over the LCD.
Step 5 Simplify by performing the addition or subtraction.
Step 6 Look for ways to reduce the resulting fraction.
Classroom Example
4
7
ϩ .
Add 2
z
z ϩ 5z
EXAMPLE 1
Add
3
5
ϩ .
x
x ϩ 2x
2
Solution
1st denominator: x2 ϩ 2x ϭ x1x ϩ 22 v ¡ LCD is x(x ϩ 2)
2nd denominator: x
3
5
3
5 xϩ2
ϩ ϭ
ϩ a
b
x
x xϩ2
x1x ϩ 22
x ϩ 2x
2
This fraction has
the LCD as its
denominator
Classroom Example
1
8
Ϫ
.
Subtract 2
xϪ4
x Ϫ 16
Form
of 1
ϭ
51x ϩ 22
3 ϩ 51x ϩ 22
3
ϩ
ϭ
x1x ϩ 22
x1x ϩ 22
x1x ϩ 22
ϭ
3 ϩ 5x ϩ 10
5x ϩ 13
ϭ
x1x ϩ 22
x1x ϩ 22
EXAMPLE 2
Subtract
4
1
.
Ϫ
x
Ϫ
2
x Ϫ4
2
Solution
x2 Ϫ 4 ϭ 1x ϩ 221x Ϫ 22
xϪ2ϭxϪ2
v ¡ LCD is (x ϩ 2)(x Ϫ 2)
4
1
4
1
xϩ2
Ϫ
ϭ
Ϫ a
ba
b
xϪ2
1x ϩ 221x Ϫ 22
xϪ2
xϩ2
x Ϫ4
2
ϭ
ϭ
ϭ
ϭ
11x ϩ 22
4
Ϫ
1x ϩ 221x Ϫ 22
1x ϩ 221x Ϫ 22
4 Ϫ 11x ϩ 22
1x ϩ 221x Ϫ 22
Ϫx ϩ 2
1x ϩ 221x Ϫ 22
Ϫ11x Ϫ 22
1x ϩ 221x Ϫ 22
ϭϪ
1
xϩ2
ϭ
4ϪxϪ2
1x ϩ 221x Ϫ 22
Note the changing of Ϫx ϩ 2 to Ϫ1(x Ϫ 2)