7.3
Addition and Subtraction
of Rational Expressions
and Functions
7.3
OBJECTIVES
1.
2.
3.
4.
Add two rational expressions
Subtract two rational expressions
Add two rational functions
Subtract two rational functions
Recall that adding or subtracting two arithmetic fractions with the same denominator is
straightforward. The same is true in algebra. To add or subtract two rational expressions
with the same denominator, we add or subtract their numerators and then write that sum or
difference over the common denominator.
Rules and Properties: Adding or Subtracting
Rational Expressions
Q
PQ
P
R
R
R
and
Q
PQ
P
R
R
R
when R 0.
Example 1
Adding and Subtracting Rational Expressions
Perform the indicated operations.
NOTE Because we have
common denominators, we
simply perform the indicated
operations on the numerators.
3
1
5
315
2a 2
2a 2
2a 2
2a 2
7
2a2
© 2001 McGraw-Hill Companies
CHECK YOURSELF 1
Perform the indicated operations.
5
4
7
2 2
3y 2
3y
3y
The sum or difference of rational expressions should always be expressed in simplest
form. Consider Example 2.
509
510
CHAPTER 7
RATIONAL EXPRESSIONS AND FUNCTIONS
Example 2
Adding and Subtracting Rational Expressions
Add or subtract as indicated.
5x
15
2
x 9
x 9
(a)
Add the numerators.
2
(b)
5x 15
x2 9
5(x 3)
5
(x 3)(x 3)
x3
Factor and divide by the
common factor.
3x y
x 3y
(3x y) (x 3y)
2x
2x
2x
Be sure to enclose the second
numerator in parentheses.
3x y x 3y
2x
Remove the parentheses by
changing each sign.
2x 4y
2(x 2y)
2x
2x
Factor and divide by the
common factor of 2.
x 2y
x
CHECK YOURSELF 2
Perform the indicated operations.
(a)
you look at the denominators
and find that the LCD is obvious
(as in Example 2).
(b)
2x 4y
5x y
3y
3y
Now, what if our rational expressions do not have common denominators? In that case,
we must use the least common denominator (LCD). The least common denominator is
the simplest polynomial that is divisible by each of the individual denominators. Each
expression in the desired sum or difference is then “built up” to an equivalent expression
having that LCD as a denominator. We can then add or subtract as before.
Although in many cases we can find the LCD by inspection, we can state an algorithm
for finding the LCD that is similar to the one used in arithmetic.
Step by Step: Finding the Least Common Denominator
NOTE Again, we see the key
role that factoring plays in the
process of working with
rational expressions.
Step 1 Write each of the denominators in completely factored form.
Step 2 Write the LCD as the product of each prime factor to the highest
power to which it appears in the factored form of any individual
denominator.
Example 3 illustrates the procedure.
© 2001 McGraw-Hill Companies
NOTE By inspection, we mean
6a
12
2
a 2a 8
a 2a 8
2
ADDITION AND SUBTRACTION OF RATIONAL EXPRESSIONS AND FUNCTIONS
SECTION 7.3
511
Example 3
Finding the LCD for Two Rational Expressions
Find the LCD for each of the following pairs of rational expressions.
5
3
2 and
4x
6xy
(a)
Factor the denominators.
NOTE You may very well be
able to find this LCD by
inspecting the numerical
coefficients and the variable
factors.
4x2 22 x2
6xy 2 3 x y
The LCD must have the factors
22 3 x2 y
and so 12x2y is the desired LCD.
7
2
and
x3
x5
(b)
Here, neither denominator can be factored. The LCD must have the factors x 3 and
x 5. So the LCD is
NOTE It is generally best to
leave the LCD in this factored
form.
(x 3)(x 5)
CHECK YOURSELF 3
Find the LCD for the following pairs of rational expressions.
3
5
and
8a3
6a2
(a)
(b)
4
3
and
x7
x5
Let’s see how factoring techniques are applied in Example 4.
Example 4
Finding the LCD for Two Rational Expressions
Find the LCD for the following pairs of rational expressions.
2
1
and 2
x x6
x 9
(a)
2
© 2001 McGraw-Hill Companies
Factoring, we have
x2 x 6 (x 2)(x 3)
and
NOTE The LCD must contain
x2 9 (x 3)(x 3)
each of the factors appearing in
the original denominators.
The LCD of the given expressions is then
NOTE Because (x 3) appears
only once in each denominator,
it appears only once in the LCD.
(x 2)(x 3)(x 3)
(b)
3
5
and 2
x 2 4x 4
x 2x 8
512
CHAPTER 7
RATIONAL EXPRESSIONS AND FUNCTIONS
Again, we factor:
x2 4x 4 (x 2)2
x2 2x 8 (x 2)(x 4)
NOTE The LCD must contain
The LCD is then
(x 2)2 as a factor because
x 2 appears twice as a factor
in the first denominator.
(x 2)2 (x 4)
CHECK YOURSELF 4
Find the LCD for the following pairs of rational expressions.
(a)
3
5
and 2
x 2x 15
x 25
(b)
5
3
and 2
y 6y 9
y y 12
2
2
Let’s look at Example 5, in which the concept of the LCD is applied in adding and subtracting rational expressions.
Example 5
Adding and Subtracting Rational Expressions
Add or subtract as indicated.
(a)
5
3
2
4xy
2x
The LCD for 2x2 and 4xy is 4x2y. We rewrite each of the rational expressions with the
LCD as a denominator.
x
we are multiplying by 1: in
x
2y
the first fraction and
in the
2y
second fraction, which is why
the resulting fractions are
equivalent to the original ones.
5
3
5x
3 2y
2
2
4xy
2x
4xy x
2x 2y
(b)
5x
6y
5x 6y
2 4x2y
4x y
4x2y
Multiply the first rational expression
2y
x
by and the second by
to form
x
2y
2
the LCD of 4x y.
3
2
a3
a
The LCD for a and a 3 is a(a 3). We rewrite each of the rational expressions with
that LCD as a denominator.
3
2
a3
a
3a
2(a 3)
a(a 3)
a(a 3)
Subtract the numerators.
3a 2(a 3)
a(a 3)
Remove the parentheses, and combine like terms.
3a 2a 6
a6
a(a 3)
a(a 3)
© 2001 McGraw-Hill Companies
NOTE Notice that in each case
ADDITION AND SUBTRACTION OF RATIONAL EXPRESSIONS AND FUNCTIONS
SECTION 7.3
513
CHECK YOURSELF 5
Perform the indicated operations.
3
4
2
2ab
5b
(a)
(b)
3
5
y2
y
Let’s proceed to Example 6, in which factoring will be required in forming the LCD.
Example 6
Adding and Subtracting Rational Expressions
Add or subtract as indicated.
5
8
2
x2 3x 4
x 16
(a)
We first factor the two denominators.
x2 3x 4 (x 1)(x 4)
x2 16 (x 4)(x 4)
We see that the LCD must be
(x 1)(x 4)(x 4)
Again, rewriting the original expressions with factored denominators gives
5
8
(x 1)(x 4)
(x 4)(x 4)
NOTE We use the facts that
© 2001 McGraw-Hill Companies
x4
1
x4
and
x1
1
x1
5(x 4)
8(x 1)
(x 1)(x 4)(x 4)
(x 4)(x 4)(x 1)
5(x 4) 8(x 1)
(x 1)(x 4)(x 4)
Now add the numerators.
5x 20 8x 8
(x 1)(x 4)(x 4)
Combine like terms in the numerator.
3x 12
(x 1)(x 4)(x 4)
Factor.
3(x 4)
(x 1)(x 4)(x 4)
Divide by the common factor x 4.
3
(x 1)(x 4)
(b)
5
3
x 5x 6
4x 12
2
Again, factor the denominators.
x2 5x 6 (x 2)(x 3)
4x 12 4(x 3)
CHAPTER 7
RATIONAL EXPRESSIONS AND FUNCTIONS
The LCD is 4(x 2)(x 3), and proceeding as before, we have
5
3
(x 2)(x 3)
4(x 3)
54
3(x 2)
4(x 2)(x 3)
4(x 2)(x 3)
20 3(x 2)
4(x 2)(x 3)
20 3x 6
3x 26
4(x 2)(x 3)
4(x 2)(x 3)
Simplify the numerator
and combine like terms.
CHECK YOURSELF 6
Add or subtract as indicated.
4
7
2
x 4
x 3x 10
(a)
2
(b)
2
5
2
3x 9
x 9
Example 7 looks slightly different from those you have seen thus far, but the reasoning
involved in performing the subtraction is exactly the same.
Example 7
Subtracting Rational Expressions
Subtract.
3
5
2x 1
3
To perform the subtraction, remember that 3 is equivalent to the fraction , so
1
3
5
3
5
2x 1
1
2x 1
The LCD for 1 and 2x 1 is just 2x 1. We now rewrite the first expression with that
denominator.
3
5
3(2x 1)
5
2x 1
(2x 1)
2x 1
6x 8
3(2x 1) 5
2x 1
2x 1
CHECK YOURSELF 7
Subtract.
4
3
3x 1
© 2001 McGraw-Hill Companies
514
ADDITION AND SUBTRACTION OF RATIONAL EXPRESSIONS AND FUNCTIONS
SECTION 7.3
515
Example 8 uses an observation from Section 7.1. Recall that
a b (b a)
1(b a)
Let’s see how this is used in adding rational expressions.
Example 8
Adding and Subtracting Rational Expressions
Add.
x2
3x 10
x5
5x
Your first thought might be to use a denominator of (x 5)(5 x). However, we can
simplify our work considerably if we multiply the numerator and denominator of the second fraction by 1 to find a common denominator.
NOTE Use
1
1
1
NOTE Notice that
(1)(5 x) x 5
The fractions now have a
common denominator, and we
can add as before.
x2
3x 10
x5
5x
x2
(1)(3x 10)
x5
(1)(5 x)
x2
3x 10
x5
x5
x 2 3x 10
x5
(x 2)(x 5)
x5
x2
CHECK YOURSELF 8
Add.
© 2001 McGraw-Hill Companies
x2
10x 21
x7
7x
The sum of two rational functions is always a rational function. Given two rational functions, f(x) and g(x), we can rename the sum, so h(x) f(x) g(x). This will always be true
for values of x for which both f and g are defined. So, for example, h(2) f(2) g(2), so long as both f(2) and g(2) exist.
Example 9 illustrates this approach.
Example 9
Adding Two Rational Functions
Given
f(x) 3x
x5
and
g(x) x
x4
CHAPTER 7
RATIONAL EXPRESSIONS AND FUNCTIONS
complete the following.
(a) Find f(1) g(1).
Because f (1) 1
1
and g(1) , then
2
3
f (1) g(1) 1
1
2
3
3
2
6
6
1
6
(b) Find h(x) f(x) g(x).
h(x) f(x) g(x)
3x
x
x5
x4
3x(x 4) x(x 5)
(x 5)(x 4)
3x2 12x x2 5x
(x 5)(x 4)
4x2 7x
(x 5)(x 4)
x 5, 4
(c) Find the ordered pair (1, h(1)).
h(1) 3
1
18
6
6.
The ordered pair is 1,
1
CHECK YOURSELF 9
Given
f(x) x
2x 5
and
g(x) 2x
3x 1
complete the following.
(a) Find f(1) g(1).
(b) Find h(x) f(x) g(x).
(c) Find the ordered pair (1, h(1)).
When subtracting rational functions, you must take particular care with the signs in the
numerator of the expression being subtracted.
© 2001 McGraw-Hill Companies
516
ADDITION AND SUBTRACTION OF RATIONAL EXPRESSIONS AND FUNCTIONS
Example 10
Subtracting Rational Functions
Given
f(x) 3x
x5
g(x) and
x2
x4
complete the following.
(a) Find f(1) g(1).
Because f (1) 1
1
and g(1) , then
2
3
f (1) g(1) 1
1
2
3
3
2
6
6
1
6
(b) Find h(x) f(x) g(x).
h(x) 3x
x2
x5
x4
3x(x 4) (x 2)(x 5)
(x 5)(x 4)
(3x2 12x) (x2 3x 10)
(x 5)(x 4)
2x2 15x 10
(x 5)(x 4)
x 5, 4
(c) Find the ordered pair (1, h(1)).
h(1) 3
1
18
6
6.
The ordered pair is 1,
1
© 2001 McGraw-Hill Companies
CHECK YOURSELF 10
Given
f(x) x
2x 5
and
g(x) complete the following.
(a) Find f(1) g(1).
(b) Find h(x) f(x) g(x).
(c) Find the ordered pair (1, h(1)).
2x 1
3x 1
SECTION 7.3
517
CHAPTER 7
RATIONAL EXPRESSIONS AND FUNCTIONS
CHECK YOURSELF ANSWERS
2
6
xy
2. (a)
; (b)
3. (a) 24a3; (b) (x 7)(x 5)
2
3y
a4
y
4. (a) (x 5)(x 5)(x 3); (b) (y 3)2(y 4)
8a 15b
2y 6
3
5x 9
5. (a)
; (b)
6. (a)
; (b)
10ab2
y(y 2)
(x 2)(x 5)
3(x 3)(x 3)
9x 1
2
7x2 11x
1 5
x , ;
7.
8. x 3
9. (a) ; (b) h(x) 3x 1
3
(2x 5)(3x 1)
3 2
2
2
5
x 11x 5
1 5
5
x , ; (c) 1, (c) 1,
10. (a) ; (b) h(x) 3
6
(2x 5)(3x 1)
3 2
6
1.
Probability and Pari-Mutual Betting. In most gambling games, payoffs are determined
by the odds. At horse and dog tracks, the odds (D) are a ratio that is calculated by taking into
account the total amount wagered (A), the amount wagered on a particular animal (a), and
the government share, called the take-out ( f ). The ratio is then rounded down to a comparison of integers like 99 to 1, 3 to 1, or 5 to 2. Below is the formula that tracks use to find odds.
D
A(1 f )
1
a
Work with a partner to complete the following.
1. Assume that the government takes 10%, and simplify the expression for D. Use this
formula to compute the odds on each horse if a total of $10,000 was bet on all the
horses and the amounts were distributed as shown in the table.
Horse
Total Amount
Wagered on This
Horse to Win
1
2
3
4
5
$5000
$1000
$2000
$1500
$ 500
Odds: Amount
Paid on Each
Dollar Bet If
Horse Wins
2. Odds can be used as a guide in determining the chance that a given horse will win.
The probability of a horse winning is related to many variables, such as track
condition, how the horse is feeling, and weather. However, the odds do reflect the
consensus opinion of racing fans and can be used to give some idea of the probability.
The relationship between odds and probability is given by the equations
P(win) 1
D1
and P(loss) 1 P(win)
or P(loss) 1 1
D1
Solve this equation for D, the odds against the horse winning. Do the probabilities for
each horse winning all add up to 1? Should they add to 1?
© 2001 McGraw-Hill Companies
518
Name
7.3
Exercises
Section
Date
In exercises 1 to 36, perform the indicated operations. Express your results in simplest
form.
ANSWERS
7
5
1.
2 2x
2x 2
11
2
2.
3 3b
3b3
1.
2.
3.
5
2
3a 7
3a 7
4.
6
3
5x 3
5x 3
3.
4.
5.
2x
6
x3
x3
6.
7w
21
w3
w3
5.
6.
7.
y2
3y 4
2y 8
2y 8
8.
x2
9
4x 12
4x 12
7.
8.
4m 7
2m 3
9.
m5
m5
3b 8
b 16
10.
b6
b6
9.
10.
11.
11.
x7
2x 2
2
x2 x 6
x x6
12.
5x 12
3x 2
2
x 2 8x 15
x 8x 15
12.
13.
13.
5
3
3x
2x
14.
4
3
5w
4w
14.
15.
© 2001 McGraw-Hill Companies
15.
6
3
2
a
a
16.
3
7
2
p
p
16.
17.
17.
2
2
m
n
18.
3
3
x
y
18.
19.
19.
3
5
3
4b2
3b
20.
4
3
2
5x3
2x
20.
519
ANSWERS
21.
21.
2
1
a
a2
22.
4
3
c
c1
23.
2
3
x1
x2
24.
4
2
y1
y3
22.
23.
24.
25. 4 25.
3
3x 2
26. 6 1
2x 3
26.
27.
27.
2w
w
w7
w2
28.
n
3n
n5
n4
29.
3x
2x
3x 2
2x 1
30.
4
3
x1
1x
31.
6
2
m7
7m
32.
5
3
a5
5a
33.
2x
3
2x 3
3 2x
34.
5
2
y2 5y 6
y2
35.
4m
1
m 3m 2
m2
36.
x
2
x 1
x1
28.
29.
30.
31.
32.
33.
34.
35.
2
2
As we saw in Section 7.2 exercises, the graphing calculator can be used to check our
work. In exercises 37 to 42, enter the first rational expression in Y1 and the second in Y2.
In Y3, you will enter either Y1 Y2 or Y1 Y2. Enter your algebraically simplified
rational expression in Y4. The graphs of Y3 and Y4 will be identical if you have correctly
simplified the expression.
36.
37.
39.
37.
6y
9
y2 8y 15
y3
38.
8a
4
a2 8a 12
a2
39.
6x
18
x 10x 24
x6
40.
21p
15
p 3p 10
p5
41.
2
3
2
z2 4
z 2z 8
42.
5
2
2
x 2 3x 10
x 25
40.
41.
42.
520
2
2
© 2001 McGraw-Hill Companies
38.
ANSWERS
In exercises 43 to 46, find (a) f(1) g(1), (b) h(x) f(x) g(x), and (c) the ordered pair
(1, h(1)).
43. f(x) 43.
(a)
(b)
3x
2x
and g(x) x1
x3
(c)
44.
4x
x4
44. f(x) and g(x) x4
x1
(a)
(b)
(c)
45. f(x) x
1
and g(x) 2
x1
x 2x 1
45.
(a)
(b)
46. f(x) x3
x2
and g(x) x4
x4
(c)
46.
In exercises 47 to 50, find (a) f(1) g(1), (b) h(x) f(x) g(x), and (c) the ordered pair
(1, h(1)).
x5
x5
47. f(x) and g(x) x5
x5
48. f(x) 2x
3x
and g(x) x4
x7
49. f(x) x9
x9
and g(x) 2
4x 36
x 18x 81
2
4x 1
50. f(x) and g(x) x5
x
(a)
(b)
(c)
47.
(a)
(b)
(c)
48.
(a)
(b)
(c)
49.
(a)
(b)
In exercises 51 to 60, evaluate each expression at the given variable value(s).
(c)
© 2001 McGraw-Hill Companies
5x 5
x3
51. 2
2
, x 4
x 3x 2
x 5x 6
50.
(b)
y3
2y 6
52. 2
2
,y3
y 6y 8
y 4
53.
54.
2m 2n
m 2n
, m 3, n 2
2
2 2
m n
m 2mn n2
(a)
(c)
51.
52.
53.
54.
w 3z
w 2z
, w 2, z 1
2 w 2wz z
w 2 z2
2
521
ANSWERS
55.
55.
1
1
2a
2
,a 4
a3
a3
a 9
56.
1
1
4
2
, m 2
m1
m3
m 2m 3
57.
3w2 16w 8
w
w1
,w 3
2
w 2w 8
w4
w2
58.
4x 2 7x 45
x2
x
, x 3
2
x 6x 5
x1
x5
59.
1
1
a2 9
, a 3
2a 5a 3
a2
a3
60.
m 2 2mn n2
2
1
, m 4, n 3
m 2 2mn 3n2
mn
mn
56.
57.
58.
59.
60.
2
Answers
6
7
y1
3
3.
5. 2
7.
9. 2
11.
2
x
3a 7
2
x2
19
3(2a 1)
2(n m)
13.
15.
17.
6x
a2
mn
9b 20
a4
5x 7
19.
21.
23.
3
12b
a(a 2)
(x 1)(x 2)
12x 5
w(3w 11)
7x
4
25.
27.
29.
31.
3x 2
(w 7)(w 2)
(3x 2)(2x 1)
m7
3m 1
15
35.
37.
33. 1
(m 1)(m 2)
y5
12
5z 14
39.
41.
x4
(z 2)(z 2)(z 4)
1
5x2 7x
1
3
x2 x 1
3
; (c) 1,
45. (a) ; (b)
43. (a) ; (b)
2 ; (c) 1,
2
(x 1)(x 3)
2
4
(x 1)
4
1.
57. 8
522
59. Undefined
51. 1
53.
49
25
55. 2
© 2001 McGraw-Hill Companies
5
20x
5
47. (a) ; (b)
; (c) 1, 6
(x 5)(x 5)
6
3
x5
3
; (c) 1, 49. (a) ; (b)
16
4(x 9)
16