In this discussion, you are assigned two rational expressions to work on. Remember to factor all polynomials completely. Remember to factor all polynomials completely. If you want to refresh your factoring skills, review sections 5.1 — 5.6 in your e-book. Read the following instructions in order and view the example to complete this discussion. Please complete the following problems according to your assigned number. (Instructors will assign each student their number.)
Your initial post should be at least 250 words in length. Support your claims with examples from required material(s) and/or other scholarly resources, and properly cite any references.
I provided a picture of the two rational expressions, the E-Book, and an example. Note- the example is just a guide but the assignment should be written in original thought with scholarly sources for ideas.
Please message me if you need additional information, my response will be swift.
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Rational Expressions
Advanced technical developments have made sports equipment faster, lighter,
and more responsive to the human body. Behind the more flexible skis, lighter
bats, and comfortable athletic shoes lies the science of biomechanics, which is the
study of human movement and the factors that influence it.
Designing and testing an athletic shoe go hand in hand. While a shoe is being
designed, it is tested in a multitude of ways, including long-term wear, rear foot
stability, and strength of materials. Testing basketball shoes usually includes an
evaluation of the force applied to the ground by the foot during running, jumping,
and landing. Many biome
chanics laboratories have a
5
6.1
Reducing Rational
Expressions
6.2
Multiplication and
Division
6.3
Finding the Least
Common Denominator
6.4
Addition and Subtraction
6.5
Complex Fractions
6.6
Solving Equations with
Rational Expressions
6.7
Applications of Ratios
and Proportions
6.8
Applications of Rational
Expressions
sure the force exerted when a
player cuts from side to side, as
well as the force against the
bottom of the shoe. Force
exerted in landing from a lay
up shot can be as high as
14 times the weight of the
body. Side-to-side force is usu
ally about 1 to 2 body weights
Force (thousands of pounds)
special platform that can mea-
4
3
2
1
0
50
100 150 200
Weight (pounds)
250
300
in a cutting movement.
In Exercises 53 and 54 of
Section 6.7 you will see how
designers of athletic shoes use
proportions to find the amount
of force on the foot and soles of
shoes for activities such as
running and jumping.
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Chapter 6 Rational Expressions
6.1
In This Section
U1V Rational Expressions and
Functions
2
U V Reducing to Lowest Terms
U3V Reducing with the Quotient
Rule for Exponents
U4V Dividing a – b by b – a
U5V Factoring Out the Opposite
of a Common Factor
6
U V Writing Rational Expressions
E X A M P L E
1
Reducing Rational Expressions
Rational expressions in algebra are similar to the rational numbers in arithmetic. In
this section, you will learn the basic ideas of rational expressions.
U1V Rational Expressions and Functions
A rational number is the ratio of two integers with the denominator not equal to 0. For
example,
0
3 -9 -7
,
,
, and
1
2
4 -6
are rational numbers. Of course, we usually write the last three of these rational num
3
bers in their simpler forms , -7, and 0. A rational expression is the ratio of two poly
2
nomials with the denominator not equal to 0. Because an integer is a monomial, a
rational number is also a rational expression. As with rational numbers, if the denom
inator is 1, it can be omitted. Some examples of rational expressions are
x 2 – 1 3a 2 + 5a – 3 3
,
,
, and 9x.
x+8
a-9
7
A rational expression involving a variable has no value unless we assign a value
to the variable. If the value of a rational expression is used to determine the value of
a second variable, then we have a rational function. For example,
x2 – 1
3a2 + 5a – 3
and w =
y=
x+8
a-9
are rational functions. We can evaluate a rational expression with or without function
notation as we did for polynomials in Chapter 5.
Evaluating a rational expression
a) Find the value of 4x – 1 for x = -3.
x+2
U Calculator Close-Up V
To evaluate the rational expression
in Example 1(a) with a calculator, first
use Y = to define the rational expres
sion. Be sure to enclose both numera
tor and denominator in parentheses.
Then find y1(-3).
3x + 2
b) If R(x) = 2x – 1, find R(4).
Solution
a) To find the value of 4x – 1 for x = -3, replace x by -3 in the rational expression:
x+2
4(-3) – 1 -13
=
= 13
-3 + 2
-1
So the value of the rational expression is 13. The Calculator Close-Up shows how to
evaluate the expression with a graphing calculator using a variable. With a scientific or
graphing calculator you could also evaluate the expression by entering (4(-3) – 1)/
(-3 + 2). Be sure to enclose the numerator and denominator in parentheses.
b) R(4) is the value of the rational expression when x = 4. To find R(4), replace x by 4
3x + 2
in R(x) = 2x – 1:
3(4) + 2
R(4) =
2(4) – 1
14
R(4) = = 2
7
So the value of the rational expression is 2 when x = 4, or R(4) = 2 (read “R of 4
is 2”).
Now do Exercises 1–6
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Reducing Rational Expressions
383
5
An expression such as 0 is undefined because the definition of rational numbers
does not allow zero in the denominator. When a variable occurs in a denominator, any
real number can be used for the variable except numbers that make the expression
undefined.
E X A M P L E
2
Ruling out values for x
Which numbers cannot be used in place of x in each rational expression?
a)
x2 – 1
x+8
b)
x+2
2x + 1
c)
x+5
x2 – 4
Solution
a) The denominator is 0 if x + 8 = 0, or x = -8. So -8 cannot be used in place of x.
(All real numbers except -8 can be used in place of x.)
1
1
b) The denominator is zero if 2x + 1 = 0, or x = -2. So we cannot use -2 in place
1
of x. (All real numbers except -2 can be used in place of x.)
c) The denominator is zero if x 2 – 4 = 0. Solve this equation:
x-2=0
x=2
x2 – 4 = 0
(x – 2)(x + 2) = 0 Factor.
or
x + 2 = 0 Zero factor property
or
x = -2
So 2 and -2 cannot be used in place of x. (All real numbers except 2 and -2 can
be used in place of x.)
Now do Exercises 7–14
In Example 2 we determined the real numbers that could not be used in place of
the variable in a rational expression. The domain of any algebraic expression in one
variable is the set of all real numbers that can be used in place of the variable. For
rational expressions, the domain must exclude any real numbers that cause the denom
inator to be zero.
E X A M P L E
3
Domain
Find the domain of each expression.
a)
x2 – 9
x+3
b)
x
x2 – x – 6
c)
x-5
4
Solution
a) The denominator is 0 if x + 3 = 0, or x = -3. So -3 can’t be used for x. The
domain is the set of all real numbers except -3, which is written in set notation
as {x 1 x e -3}.
b) The denominator is 0 if x2 – x – 6 = 0:
x2 – x – 6 = 0
(x – 3)(x + 2) = 0
x-3=0
or x + 2 = 0
x=3
or
x = -2
So -2 and 3 can’t be used in place of x. The domain is the set of all real numbers
except -2 and 3, which is written as {x 1 x e -2 and x e 3}.
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Chapter 6 Rational Expressions
c) Since the denominator is 4, the denominator can’t be 0 no matter what number is
used for x. The domain is the set of all real numbers, R.
Now do Exercises 15–22
Note that if a rational expression is used to define a function, then the domain of
the rational expression is also
called the domain of the function. For example, the
2
domain of the function y = x – 9 is the set of all real numbers except -3 or {x I x ¥ -3}.
x +3
When dealing with rational expressions in this book, we will generally assume
that the variables represent numbers for which the denominator is not zero.
U2V Reducing to Lowest Terms
Rational expressions are a generalization of rational numbers. The operations that we
perform on rational numbers can be performed on rational expressions in exactly the
same manner.
Each rational number can be written in infinitely many equivalent forms. For
example,
3
6
9
12 15
=
=
=
=
=
.
5 10 15 20 25
U Helpful Hint V
How would you fill in the blank in
3
—
5 = 10 ? Most students learn to divide
5 into 10 to get 2, and then multiply 3
by 2 to get 6. In algebra, it is better to
multiply the numerator and denomi
3
nator of 5 by 2, as shown here.
Each equivalent form of 3 is obtained from 3 by multiplying both numerator and
5
5
denominator by the same nonzero number. This is equivalent to multiplying the frac
tion by 1, which does not change its value. For example,
3 3
3 2
6
=
1=
=
5 5
5 2 10
and
3 3 3
9
=
= .
5 5 3 15
If we start with 6 and convert it into 3, we say that we are reducing 6 to lowest terms.
10
5
10
We reduce by dividing the numerator and denominator by the common factor 2:
/2 3 3
6
=
=
10 2/ 5 5
A rational number is expressed in lowest terms when the numerator and the denomi
nator have no common factors other than 1.
CAUTION We can reduce fractions only by dividing the numerator and the denom
inator by a common factor. Although it is true that
2+4
6
=
,
10 2 + 8
we cannot eliminate the 2’s, because they are not factors. Removing
them from the sums in the numerator and denominator would not result
in 3.
5
Reducing Fractions
If a e 0 and c e 0, then
ab
b
= .
ac
c
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6.1
Reducing Rational Expressions
385
To reduce rational expressions to lowest terms, we use exactly the same procedure
as with fractions:
Reducing Rational Expressions
1. Factor the numerator and denominator completely.
2. Divide the numerator and denominator by the greatest common factor.
Dividing the numerator and denominator by the GCF is often referred to as dividing
out or canceling the GCF.
E X A M P L E
4
Reducing
Reduce to lowest terms.
a)
30
42
b)
x2 – 9
6x + 18
c)
3×2 + 9x + 6
2×2 – 8
Solution
a)
30 2/ /3 5
=
Factor.
42 2/ 3/ 7
=
5
7
Divide out the GCF: 2 3 or 6.
b) Since 9 = 9 1 = 1 , it is tempting to apply that fact here. However, 9 is not a common
18 9 2 2
2
factor of the numerator and denominator of x – 9 , as it is in 9 . You must factor
6x + 18
18
the numerator and denominator completely before reducing.
(x – 3)(x + 3)
x2 – 9
=
6(x + 3)
6x + 18
Factor.
x-3
Divide out the GCF: x + 3.
6
This reduction is valid for all real numbers except -3, because that is the domain of the
original expression. If x = -3, then x + 3 = 0 and we would be dividing out 0 from the
numerator and denominator, which is prohibited in the rule for reducing fractions.
=
c)
3×2 + 9x + 6
3(x + 2)(x + 1)
=
2(x + 2)(x – 2)
2×2 – 8
Factor completely.
3x + 3
Divide out the GCF: x + 2.
2(x – 2)
This reduction is valid for all real numbers except -2 and 2, because that is the
domain of the original expression.
=
Now do Exercises 23–46
CAUTION In reducing, you can divide out or cancel common factors only. You
+3
cannot cancel x from xx +
2 , because it is not a factor of either x + 3 or
x + 2. But x is a common factor in 32xx , and 32xx = 23 .
Note that there are four ways to write the answer to Example 3(c) depending on
whether the numerator and denominator are factored. Since
3x + 3
3(x + 1)
3(x + 1)
3x + 3
=
=
=
,
2(x – 2)
2(x – 2)
2x – 4
2x – 4
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Chapter 6 Rational Expressions
any of these four rational expressions is correct. We usually give such answers with the
denominator factored and the numerator not factored. With the denominator factored
you can easily spot the values for x that will cause an undefined expression.
U3V Reducing with the Quotient Rule for Exponents
To reduce rational expressions involving exponential expressions, we use the quotient
rule for exponents from Chapter 4. We restate it here for reference.
Quotient Rule for Exponents
If a e 0, and m and n are any integers, then
am
= am-n.
an
E X A M P L E
5
Using the quotient rule in reducing
Reduce to lowest terms.
a)
3a15
6a7
b)
6x4y2
4xy5
Solution
a)
3a15
3/a15
Factor.
7 =
6a
3/ 2 a7
a15-7
Quotient rule
=
for exponents
2
=
a8
2
b)
6x4y2 2/ 3x 4y2
=
4xy5
2/ 2xy 5
3×4-1y2-5
=
2
=
Factor.
Quotient rule
for exponents
3x3y-3 3×3
= 3
2
2y
Now do Exercises 47–58
The essential part of reducing is getting a complete factorization for the numerator
and denominator. To get a complete factorization, you must use the techniques for fac
toring from Chapter 5. If there are large integers in the numerator and denominator, you
can use the technique shown in Section 5.1 to get a prime factorization of each integer.
E X A M P L E
6
Reducing expressions involving large integers
Reduce 420 to lowest terms.
616
Solution
Use the method of Section 5.1 to get a prime factorization of 420 and 616:
�
�
2 420
2 616
�
�
2 210
2 308
�
�
3 105
2 154
�
�
5 35
7 77
7
11
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6.1
387
Reducing Rational Expressions
The complete factorization for 420 is 22 3 5 7, and the complete factorization for 616
is 23 7 11. To reduce the fraction, we divide out the common factors:
420 22 3 5 7
= 3
2 7 11
616
3 5
=
2 11
15
=
22
Now do Exercises 59–66
U4V Dividing a – b by b – a
In Section 4.5 you learned that a – b = -(b – a) = -1(b – a). So if a – b is divided
by b – a, the quotient is -1:
a – b = -1(b – a)
b-a
b-a
= -1
We will use this fact in Example 7.
E X A M P L E
7
Expressions with a – b and b – a
Reduce to lowest terms.
a)
5x – 5y
4y – 4x
b)
m2 – n2
n-m
Solution
a) Factor out 5 from the numerator and 4 from the denominator and use x – y = -1:
y-x
5x – 5y 5(x – y)
5
5
=
= (-1) = 4y – 4x 4(y – x)
4
4
Another way is to factor out -5 from the numerator and 4 from the denominator
and then use y – x =1:
y-x
5x – 5y -5(y – x)
-5
5
=
=
(1) = 4(y – x)
4
4y – 4x
4
-1
b)
m2 – n2 (m – n)(m + n)
=
Factor.
n-m
n-m
m-n
= -1(m + n)
= -m – n
n-m
= -1
Now do Exercises 67–74
CAUTION We can reduce a – b to -1, but we cannot reduce a – b. There is no factor
b-a
a+b
that is common to the numerator and denominator of
a-b
a+b
or
a+b
.
a-b
U5V Factoring Out the Opposite of a Common Factor
If we can factor out a common factor, we can also factor out the opposite of that com
mon factor. For example, from -3x – 6y we can factor out the common factor 3 or
the common factor -3:
-3x – 6y = 3(-x – 2y)
or
-3x – 6y = -3(x + 2y)
To reduce an expression, it is sometimes necessary to factor out the opposite of a
common factor.
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Chapter 6 Rational Expressions
E X A M P L E
8
Factoring out the opposite of a common factor
Reduce
-3w – 3w2
w2 – 1
to lowest terms.
Solution
We can factor 3w or -3w from the numerator. If we factor out -3w, we get a common
factor in the numerator and denominator:
-3w – 3w 2
-3w(1 + w)
=
Factor.
w2 – 1
(w – 1)(w + 1)
-3w
=
Since 1 + w = w + 1, we divide out w + 1.
w-1
3w
=
Multiply numerator and denominator by -1.
1-w
The last step is not absolutely necessary, but we usually perform it to express the answer
with one less negative sign.
Now do Exercises 75–84
The main points to remember for reducing rational expressions are summarized in
the following reducing strategy.
Strategy for Reducing Rational Expressions
1. Factor the numerator and denominator completely. Factor out a common fac
tor with a negative sign if necessary.
2. Divide out all common factors. Use the quotient rule if the common factors
are powers.
U6V Writing Rational Expressions
Rational expressions occur in applications involving rates. For uniform motion, rate is
distance divided by time, R = DT . For example, if you drive 500 miles in 10 hours, your
00
rate is 510
or 50 mph. If you drive 500 miles in x hours, your rate is 500
x mph. In work prob
W
lems, rate is work divided by time, R = . For example, if you lay 400 tiles in 4 hours,
T
400
your rate is 400
4 or 100 tiles/hour. If you lay 400 tiles in x hours, your rate is x tiles/hour.
E X A M P L E
9
Writing rational expressions
Answer each question with a rational expression.
a) If a trucker drives 500 miles in x + 1 hours, then what is his average speed?
b) If a wholesaler buys 100 pounds of shrimp for x dollars, then what is the price per pound?
c) If a painter completes an entire house in 2x hours, then at what rate is she painting?
Solution
500
a) Because R = D, he is averaging x + 1 mph.
T
b) At x dollars for 100 pounds, the wholesaler is paying
or x dollars/pound.
100
c) By completing 1 house in 2x hours, her rate is
1
2x
x
100
dollars per pound
house/hour.
Now do Exercises 107–112
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6.1
389
▼
Fill in the blank.
1. A rational number is a ratio of two
with the
denominator not 0.
2. A rational expression is a ratio of two
with
the denominator not 0.
3. A rational expression is reduced to lowest terms by
the numerator and denominator by the GCF.
4. The
rule is used in reducing a ratio of
monomials.
5. The expressions a – b and b – a are
.
6. If a rational expression is used to determine y from x,
then y is a
function of x.
True or false?
7. A complete factorization of 3003 is 2 · 3 · 7 · 11 · 13.
8. A complete factorization of 120 is 23 · 3 · 5.
x+1
9. We can’t replace x by -1 or 3 in –.
x-3
2x
10. For any real number x, — = x.
2
a2 + b2
11. Reducing — to lowest terms yields a + b.
a+b
Exercises
U Study Tips V
• If you must miss class, let your instructor know. Be sure to get notes from a reliable classmate.
• Take good notes in class for yourself and your classmates. You never know when a classmate will ask to see your notes.
U1V Rational Expressions and Functions
Evaluate each rational expression. See Example 1.
3x – 3
x+5
1. Evaluate — for x = -2.
3x + 1
– for x = 5.
2. Evaluate 4x – 4
2x + 9
, find R(3).
3. If R(x) = -x
-20x – 2
-, find R(-1).
4. If R(x) = x-8
x-5
-, find R(2), R(-4), R(-3.02), and
5. If R(x) = x+3
R(-2.96). Note how a small difference in x (-3.02 to
-2.96) can make a big difference in R(x).
x – 2x – 3
2
-, find R(3), R(5), R(2.05),
6. If R(x) = x-2
and R(1.999).
Which numbers cannot be used in place of the variable in each
rational expression? See Example 2.
x
7. -x+1
7a
9. -3a – 5
2x + 3
11. -x2 – 16
p-1
13. -2
3x
8. -x-7
84
10. -3 – 2a
2y + 1
12. 2
y -y-6
m + 31
14. -5
Find the domain of each rational expression. See Example 3.
x2 + x
15. -x-2
x+4
16. -x-5
6.1
Warm-Ups
Reducing Rational Expressions
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Chapter 6 Rational Expressions
x
x + 5x + 6
U3V Reducing with the Quotient Rule
2
for Exponents
x2 + 2
18. 2
x – x – 12
19.
20.
Reduce each expression to lowest terms. Assume that all
variables represent nonzero real numbers, and use only positive
exponents in your answers. See Example 5.
x2 – 4
2
x2 – 3x
9
21.
x-5
x
22.
x2 – 3
x+9
47.
x10
x7
48.
y8
y5
49.
z3
z8
50.
w9
w12
51.
4×7
-2×5
52.
-6y3
3y9
53.
-12m9n18
8m6n16
54.
-9u9v19
6u9v14
55.
6b10c4
-8b10c7
56.
9x20y
-6x25y3
57.
30a3bc
18a7b17
58.
15m10n3
24m12np
U2V Reducing to Lowest Terms
Reduce each rational expression to lowest terms. Assume that the
variables represent only numbers for which the denominators are
nonzero. See Example 4.
23.
6
27
24.
14
21
25.
42
90
26.
42
54
27.
36a
90
28.
56y
40
78
30w
30.
6x + 2
6
32.
2x + 4y
6y + 3x
34.
35.
3b – 9
6b – 15
36.
3m + 9w
3m – 6w
37.
w2 – 49
w+7
38.
a2 – b2
a-b
39.
a -1
a2 + 2a + 1
40.
x -y
x2 + 2xy + y2
67.
3a – 2b
2b – 3a
68.
5m – 6n
6n – 5m
41.
2×2 + 4x + 2
4×2 – 4
42.
2×2 + 10x + 12
3×2 – 27
69.
h2 – t 2
t-h
70.
r 2 – s2
s-r
43.
3×2 + 18x + 27
21x + 63
44.
x 3 – 3x 2 – 4x
x 2 – 4x
71.
2g – 6h
9h2 – g2
72.
5a – 10b
4b2 – a2
45.
2a3 + 16
4a + 8
46.
w3 – 27
w2 – 3w
73.
x2 – x – 6
9 – x2
74.
1 – a2
a +a- 2
29.
31.
33.
2
Reduce each expression to lowest terms. Assume that all
variables represent nonzero real numbers, and use only positive
exponents in your answers. See Example 6.
59.
60.
68
44y
210
264
616
660
61.
62.
2w + 2
2w
231
168
936
624
63.
64.
5x – 10a
10x – 20a
630×5
300×9
96y2
108y5
65.
924a23
448a19
66.
270b75
165b12
2
U4V Dividing a – b by b – a
Reduce each expression to lowest terms. See Example 7.
2
2
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6.1
U5V Factoring Out the Opposite
of a Common Factor
103.
Reducing Rational Expressions
y3 – 2y2 – 4y + 8
y2 – 4y + 4
104.
391
mx + 3x + my + 3y
m2 – 3m – 18
Reduce each expression to lowest terms. See Example 8.
75.
-x – 6
x+6
77.
-2y – 6y
3 + 9y
76.
-5x – 20
3x + 12
78.
y – 16
-8 – 2y
2
79.
-3x – 6
3x – 6
80.
8 – 4x
-8x – 16
81.
-12a – 6
2a2 + 7a + 3
82.
-2b2 – 6b – 4
b2 – 1
83.
a -b
2b2 – 2ab
84.
x -1
x – x2
3
3
3
Reduce each expression to lowest terms. See the Strategy for
Reducing Rational Expressions box on page 388.
85.
2×12
4×8
86.
4×2
2×9
87.
2x + 4
4x
88.
2x + 4×2
4x
89.
a-4
4-a
90.
2b – 4
2b + 4
91.
2c – 4
4 – c2
92.
-2t – 4
4 – t2
93.
x2 + 4x + 4
x2 – 4
94.
3x – 6
x – 4x + 4
-2x – 4
95. 2
x + 5x + 6
2x + 2w – ax – aw
x3 – xw2
106.
x2 + ax – 4x – 4a
x2 – 16
-2x – 8
96. 2
x + 2x – 8
2q8 + q7
2q6 + q5
99.
u – 6u – 16
u2 – 16u + 64
100.
v + 3v – 18
v2 + 12v + 36
a3 – 8
2a – 4
102.
4w2 – 12w + 36
2w3 + 54
98.
2
U6V Writing Rational Expressions
Answer each question with a rational expression. Be sure to
include the units. See Example 9.
107. If Sergio drove 300 miles at x + 10 miles per hour, then
how many hours did he drive?
108. If Carrie walked 40 miles in x hours, then how fast did
she walk?
109. If x + 4 pounds of peaches cost $4.50, then what is the
cost per pound?
110. If nine pounds of pears cost x dollars, then what is the
price per pound?
111. If Ayesha can clean the entire swimming pool in x hours,
then how much of the pool does she clean per hour?
2
97.
101.
105.
2
8s12
12s – 16s5
112. If Ramon can mow the entire lawn in x 3 hours, then
how much of the lawn does he mow per hour?
6
2
Applications
Solve each problem.
113. Annual reports. The Crest Meat Company found that the
cost per report for printing x annual reports at Peppy
Printing is given by the formula
150 + 0.60x
,
x
where C(x) is in dollars.
C(x) =
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Chapter 6 Rational Expressions
given by the formula,
C(p) =
500,000
.
100 – p
a) Use the accompanying graph to estimate the cost for
removing 90% and 95% of the toxic chemicals.
b) Use the formula to find C(99.5) and C(99.9).
c) What happens to the cost as the percentage of
pollutants removed approaches 100%?
0.80
0.70
0.60
0.50
Annual cost
(hundred thousand dollars)
Cost per report (dollars)
a) Use the accompanying graph to estimate the cost per
report for printing 1000 reports.
b) Use the formula to find C(1000), C(5000), and
C(10,000).
c) What happens to the cost per report as the number of
reports gets very large?
0.40
1
2
3
4
5
Number of reports (thousands)
Figure for Exercise 113
5
4
3
2
1
0
90 91 92 93 94 95 96 97 98 99
Percentage of chemicals removed
114. Toxic pollutants. The annual cost in dollars for removing
p% of the toxic chemicals from a town’s water supply is
6.2
In This Section
U1V Multiplication of Rational
Numbers
2
U V Multiplication of Rational
Expressions
U3V Division of Rational
Numbers
4
U V Division of Rational
Expressions
5
U V Applications
Figure for Exercise 114
Multiplication and Division
In Section 6.1, you learned to reduce rational expressions in the same way that we
reduce rational numbers. In this section, we will multiply and divide rational
expressions using the same procedures that we use for rational numbers.
U1V Multiplication of Rational Numbers
Two rational numbers are multiplied by multiplying their numerators and multiplying
their denominators.
Multiplication of Rational Numbers
If b e 0 and d e 0, then
a c ac
= .
b d bd
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E X A M P L E
6.2
1
Multiplication and Division
393
Multiplying rational numbers
6
Find the product 7
14
.
15
Solution
U Helpful Hint V
The product is found by multiplying the numerators and multiplying the denominators:
Did you know that the line separating
the numerator and denominator in a
fraction is called the vinculum?
6 14
84
=
7 15 105
=
21 4
Factor the numerator and denominator
21 5
4
Divide out the GCF 21.
5
The reducing that we did after multiplying is easier to do before multiplying. First factor
all terms, reduce, and then multiply:
=
6 14 2 3/ 2 7/
=
7 15
/7
/3 5
=
4
5
Now do Exercises 1–8
U2V Multiplication of Rational Expressions
Rational expressions are multiplied just like rational numbers: factor, reduce, and then
multiply. A rational number cannot have zero in its denominator and neither can a
rational expression. Since a rational expression can have variables in its denominator,
the results obtained in Examples 2 and 3 are valid only for values of the variable(s)
that would not cause a denominator to be 0.
E X A M P L E
2
Multiplying rational expressions
Find the indicated products.
a)
9x 10y
5y 3xy
b)
-8xy4 15z
3z3 2x5y3
Solution
a)
9x 10y 3 3//x 2 5//y
=
Factor.
5y 3xy
5//y
3//xy
=
b)
6
y
-8xy4 15z
-2 2 2/xy4 /3 5z
Factor.
3
5 3=
3z
2x y
3/z3
2/x5y3
=
-20xy4z
z3x5y3
Reduce.
=
-20y
z2x4
Quotient rule
Now do Exercises 9–18
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Chapter 6 Rational Expressions
E X A M P L E
3
Multiplying rational expressions
Find the indicated products.
a)
2x – 2y
2x
4
x2 – y2
b)
x 2 + 7x + 12
x
2x + 6
x2 – 16
c)
a+b
8a2
2
6a
a + 2ab + b2
Solution
a)
2x – 2y
2/(x – y)
2x
=
4
x2 – y2
2/ 2/
=
b)
c)
2/ x
(x – y)(x + y)
x
x+y
Factor.
Reduce.
x2 + 7x + 12
x
(x + 3) (x + 4)
=
2
2(x + 3)
2x + 6
x – 16
x
=
2(x – 4)
a+b
8a2
a + b 2 4a2
2
2=
6a
a + 2ab + b
2 3a (a + b)2
4a
=
3(a + b)
x
(x – 4)(x + 4)
Factor.
Reduce.
Factor.
Reduce.
Now do Exercises 19–26
U3V Division of Rational Numbers
By the definition of division, a quotient is found by multiplying the dividend by the re
c
d
ciprocal of the divisor. If the divisor is a rational number , its reciprocal is simply .
d
c
Division of Rational Numbers
If b e 0, c e 0, and d e 0, then
a c a d
� =
.
b d b c
E X A M P L E
4
Dividing rational numbers
Find each quotient.
a) 5 �
1
2
b)
6
3
�
7 14
b)
6
3
6 14 2 3/ 2 /7
�
=
=
=4
7 14 7 3
/7
3/
Solution
a) 5 �
1
= 5 2 = 10
2
Now do Exercises 27–34
U4V Division of Rational Expressions
We divide rational expressions in the same way we divide rational numbers: Invert the
divisor and multiply.
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6-15
E X A M P L E
6.2
5
Multiplication and Division
395
Dividing rational expressions
Find each quotient.
5
5
�
a)
3x 6x
b)
x7
� (2x 2)
2
c)
4 – x2
x-2
�
x2 + x x 2 – 1
Solution
U Helpful Hint V
A doctor told a nurse to give a patient
half of the usual 500-mg dose of a
drug. The nurse stated in court,
“dividing in half means dividing by
1/2, which means multiply by 2.” The
nurse was in court because the
patient got 1000 mg instead of
250 mg and died (true story).
Dividing a quantity in half and divid
ing by one-half are not the same.
a)
5
5
5 6x
� =
3x 6x 3x 5
=
Invert the divisor and multiply.
5/ 2 3//x
Factor.
/3/x
/5
=2
b)
x7
x7 1
� (2x 2) =
2
2 2x 2
=
c)
Divide out the common factors.
x5
4
Invert and multiply.
Quotient rule
4 – x2 x2 – 1
4 – x2
x-2
� 2
= 2
2
x +x x-2
x +x x -1
Invert and multiply.
-1
=
(2 – x) (2 + x) (x + 1)(x – 1)
Factor.
x(x + 1)
x-2
=
-1(2 + x)(x – 1)
x
2-x
= -1
x-2
=
-1(x2 + x – 2)
x
Simplify.
=
-x 2 – x + 2
x
Now do Exercises 35–48
We sometimes write division of rational expressions using the fraction bar. For
example, we can write
a+b
3
a + b 1 as —.
�
1
3
6
6
No matter how division is expressed, we invert the divisor and multiply.
E X A M P L E
6
Division expressed with a fraction bar
Find each quotient.
a+b
3
a) —
1
6
x2 – 1
2
b) —
x-1
3
a2 + 5
3
c) —
2
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6-16
Chapter 6 Rational Expressions
Solution
a+b
3
a) —
1
6
a+b
3
1
6
Rewrite as division.
a+b 6
3
1
Invert and multiply.
a + b 2 3p
Factor.
p3
1
(a + b)2
Reduce.
2a + 2b
x2 – 1
2
b) —
x-1
3
x2 – 1
2
x-1
3
x2 – 1
3
2
x-1
Rewrite as division.
Invert and multiply.
(x – 1)(x + 1)
3
Factor.
x-1
2
3x + 3
Reduce.
2
U Helpful Hint V
In Section 6.5 you will see another
technique for finding the quotients in
Example 6.
a2 + 5
3
c) —
2
a2 + 5
3
2
a2 + 5 1
3
2
Rewrite as division.
a2 + 5
6
Now do Exercises 49–56
U5V Applications
We saw in Section 6.1 that rational expressions can be used to represent rates. Note
that there are several ways to write rates. For example, miles per hour is written mph,
mi/hr, or mi. The last way is best when doing operations with rates because it helps us
hr
reconcile our answers. Notice how hours “cancels” when we multiply miles per hour
and hours in Example 7, giving an answer in miles, as it should be.
E X A M P L E
7
Using rational expressions with uniform motion
Shasta drove 200 miles on I-10 in x hours before lunch.
a) Write a rational expression for her average speed before lunch.
b) She drives for 3 hours after lunch at the same average speed. Write a rational
expression for her distance after lunch.
Solution
a) Because R
D
,
T
her rate before lunch is
200 miles
x hours
or
200
x
mph.
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6-17
6.2
Multiplication and Division
b) Because D = R T, her distance after lunch is the product of
3 hours (her time):
D=
200
x
397
mph (her rate) and
200 mi
600
3 hr =
mi
x hr
x
Now do Exercises 77–78
The amount of work completed is the product of rate and time, W = R T. So if a
machine washes cars at the rate of 12 per hour and it works for 3 hours, the amount of
W
work completed is 36 cars washed. Note that the rate is given by R = T .
E X A M P L E
8
Using rational expressions with work
It takes x minutes to fill a bathtub.
a) Write a rational expression for the rate at which the tub is filling.
b) Write a rational expression for the portion of the tub that is filled in 10 minutes.
Solution
W
a) The work completed in this situation is 1 tub being filled. Because R = T , the rate
1 tub
1
at which the tub is filling is x min or x tub/min.
b) Because W = R T, the work completed in 10 minutes or the portion of the tub that
1
is filled in 10 minutes is the product of x tub/min (the rate) and 10 minutes (the time):
W=
1 tub
10
10 min =
tub
x min
x
Now do Exercises 79–80
Warm-Ups
▼
Fill in the blank.
1.
2.
expressions are multiplied by multiplying their
numerators and multiplying their denominators.
can be done before multiplying rational
expressions.
3. To
rational expressions, invert the divisor and
multiply.
True or false?
4. One-half of one-fourth is one-sixth.
2 5
10
5.
=
3 7
21
6. The product of
x-7
6
and
is -2.
3
7-x
7. Dividing by 2 is equivalent to multiplying by
8. For any real number a,
9.
2
1
4
� =
3
2
3
a
a
�3= .
3
9
1
.
2
6.2
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Exercises
U Study Tips V
• Personal issues can have a tremendous effect on your progress in any course. If you need help, get it.
• Most schools have counseling centers that can help you to overcome personal issues that are affecting your studies.
24.
12
4x + 10
8 35
3.
15 24
25.
16a + 8 2a2 + a – 1
4a2 – 1
5a2 + 5
25 56
6.
48 35
26.
6x – 18
4x 2 + 4x + 1
2x – 5x – 3
6x + 3
U1V Multiplication of Rational Numbers
Perform the indicated operation. See Example 1.
2
1.
3
5
6
3
2.
4
3 8
4.
4 21
7. 24
2
5
12 51
5.
17 10
7
20
8.
3
35
10
Perform the indicated operation. See Example 4.
27.
Perform the indicated operation. See Example 2.
2x
3
5×2
11.
6
5
4x
3y
7
21
2y
9x
12.
10
5
x2
10.
3
x
2
U3V Division of Rational Numbers
U2V Multiplication of Rational Expressions
9.
(4×2 + 20x + 25)
1
4
1
2
30. 32
33.
40
3
28.
1
4
1
6
1
2
29. 12
5 15
7 14
22
34.
9
9
31.
12
32.
13.
5a 3ab
12b 55a
14.
3m 35p
7p 6mp
U4V Division of Rational Expressions
15.
-2×6 21a2
7a5
6x
16.
5z3w -6y5
-9y3 20z9
35.
x2
4
17.
15t3y5
24t5w3y2
20w7
18. 22x2y3z
37.
5x 2
3
39.
8m3
n4
41.
y-6
2
43.
x 2 + 4x + 4
8
(x + 2)3
16
Perform the indicated operation. See Example 5.
6×5
33y3z4
Perform the indicated operation. See Example 3.
x
2
36.
3
2a2
6
2a
10x
21
38.
4u2
3v
14u
15v6
(12mn2)
40.
2p4
3q3
(4pq5)
6-y
6
42.
4-a
5
19.
2x + 2y
7
20.
3
a +a
21.
3a + 3b
10a
15
a2 – b2
44.
a2 + 2a + 1
3
a2 – 1
a
22.
b3 + b
10
2
5
b +b
45.
t 2 + 3t – 10
t 2 – 25
(4t – 8)
46.
w2 – 7w + 12
w2 – 4w
2
15
6x + 6y
2a + 2
6
23. (x2 – 6x + 9)
3
x-3
3
4
(w2 – 9)
a2 – 16
3
2
5
15
2
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6-19
6.2
47. (2×2 – 3x – 5) �
48. (6y2 – y – 2) �
2x – 5
x-1
2y + 1
3y – 2
67.
2mn4 3m5n7
� 2 4
6mn2
mn
68.
rt2
rt2
2� 32
rt
rt
69.
3×2 + 16x + 5
x2
2
x
9x – 1
70.
x 2 + 6x + 5
x4
x
3x + 3
71.
a2 – 2a + 4
a2 – 4
72.
w2 – 1
w-1
(w – 1)2 w2 + 2w + 1
73.
2×2 + 19x – 10
4×2 – 1
� 2
2
x – 100
2x – 19x – 10
74.
x3 – 1
9x 2 + 9x + 9
�
x2 – x
x2 + 1
Perform the indicated operation. See Example 6.
x – 2y
5
49. —
1
10
3m + 6n
50. —8—
3
4
399
(a + 2)3
2a + 4
x2 – 4
12
51. —
x-2
6
6a2 + 6
5
52. —
6a + 6
5
x2 + 9
3
53. —
5
1
a-3
54. —
4
75.
9 + 6m + m2
9 – 6m + m2
76.
3x + 3w + bx + bw
x2 – w2
x 2 – y2
55. —
x-y
9
x 2 + 6x + 8
56. ——
x+2
x+1
U5V Applications
m2 – 9
m + mk + 3m + 3k
2
6 – 2b
9 – b2
Solve each problem. Answers could be rational expressions.
Be sure to give your answers with appropriate units.
See Examples 7 and 8.
77. Marathon run. Florence ran 26.2 miles in x hours in the
Boston Marathon.
a) Write a rational expression for her average speed.
Miscellaneous
Perform the indicated operation.
57.
x-1
9
3
1-x
58.
2x – 2y
1
3
y-x
59.
3a + 3b 1
a
3
60.
a-b
2b – 2a
b
61. a
1
2
Multiplication and Division
62.
2
5
2g
3h
1
h
63.
6y
� (2x)
3
64.
8x
� (18x)
9
65.
a3b4 a5b7
-2ab2 ab
66.
-2a2 20a
3a2 15a3
b) She runs at the same average speed for 12 hour in the
Cripple Creek Fun Run. Write a rational expression for
her distance at Cripple Creek.
78. Driving marathon. Felix drove 800 miles in x hours on
Monday.
a) Write a rational expression for his average speed.
b) On Tuesday he drove for 6 hours at the same average
speed. Write a rational expression for his distance on
Tuesday.
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Chapter 6 Rational Expressions
82. Area of a triangle. If the base of a triangle is 8x + 16 yards
and its height is 1 yards, then what is the area of the
79. Filling the tank. Chantal filled her empty gas tank in x
minutes.
a) Write a rational expression for the rate at which she
filled her tank.
x+2
triangle?
b) Write a rational expression for the portion of the tank
that is filled in 2 minutes.
1
—
x+2
yd
8x + 16 yd
80. Magazine sales. Henry sold 120 magazine subscriptions in
x days.
a) Write a rational expression for the rate at which he sold
the subscriptions.
Figure for Exercise 82
Getting More Involved
83. Discussion
Evaluate each expression.
b) Suppose that he continues to sell at the same rate for
5 more days. Write a rational expression for the number
of magazines sold in those 5 days.
c) One-half of
81. Area of a rectangle. If the length of a rectangular flag is
x meters and its width is 5 meters, then what is the area of
x
the rectangle?
b) One-third of 4
4x
3
d) One-half of
3x
2
84. Exploration
Let R =
5
—
x
1
a) One-half of 4
6×2 + 23x + 20
2x + 5 .
and H =
24×2 + 29x – 4
8x – 1
a) Find R when x = 2 and x = 3. Find H when x = 2 and
x = 3.
b) How are these values of R and H related and why?
m
xm
Figure for Exercise 81
6.3
In This Section
U1V Building Up the Denominator
U2V Finding the Least Common
Denominator
3
U V Converting to the LCD
Finding the Least Common Denominator
Every rational expression can be written in infinitely many equivalent forms.
Because we can add or subtract only fractions with identical denominators, we
must be able to change the denominator of a fraction. You have already learned
how to change the denominator of a fraction by reducing. In this section, you will
learn the opposite of reducing, which is called building up the denominator.
U1V Building Up the Denominator
To convert the fraction
2
3
into an equivalent fraction with a denominator of 21,
we factor 21 as 21 = 3 7. Because
2
3
already has a 3 in the denominator, multiply
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6.3
Finding the Least Common Denominator
401
the numerator and denominator of 2 by the missing factor 7 to get a denominator
3
of 21:
2 2 7 14
=
=
3 3 7 21
For rational expressions the process is the same. To convert the rational
expression
5
x+3
into an equivalent rational expression with a denominator of x2 – x – 12, first factor
x2 – x – 12:
x2 – x – 12 = (x + 3)(x – 4)
From the factorization we can see that the denominator x + 3 needs only a factor of
x – 4 to have the required denominator. So multiply the numerator and denominator
by the missing factor x – 4:
5
5(x – 4)
5x – 20
=
=
x + 3 (x + 3)(x – 4) x2 – x – 12
E X A M P L E
1
Building up the denominator
Build each rational expression into an equivalent rational expression with the indicated
denominator.
a) 3 =
?
12
b)
3
?
=
w wx
c)
2
?
3=
3y
12y8
Solution
a) Because 3 = 3, we get a denominator of 12 by multiplying the numerator and
1
denominator by 12:
3 3 12 36
3= =
=
1 1 12 12
b) Multiply the numerator and denominator by x:
3
3 x
3x
=
=
w w x wx
c) Note that 12y8 = 3y3 4y5. So to build 3y3 up to 12y8 multiply by 4y5:
2
2 4y5
8y5
3=
3
5=
3y
3y 4y
12y8
Now do Exercises 1–20
In Example 2 we must factor the original denominator before building up the
denominator.
E X A M P L E
2
Building up the denominator
Build each rational expression into an equivalent rational expression with the indicated
denominator.
a)
7
?
=
3x – 3y 6y – 6x
b)
x-2
?
=
x + 2 x2 + 8x + 12
402
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6-22
Chapter 6 Rational Expressions
U Helpful Hint V
Notice that reducing and building up
are exactly the opposite of each
other. In reducing you remove a
factor that is common to the numera
tor and denominator, and in building
up you put a common factor into the
numerator and denominator.
Solution
a) Because 3x – 3y = 3(x – y), we factor -6 out of 6y – 6x. This will give a factor
of x – y in each denominator:
3x – 3y = 3(x – y)
6y – 6x = -6(x – y) = -2 3(x – y)
To get the required denominator, we multiply the numerator and denominator
by -2 only:
7
7(-2)
=
3x – 3y (3x – 3y)(-2)
-14
=
6y – 6x
b) Because x2 + 8x + 12 = (x + 2)(x + 6), we multiply the numerator and
denominator by x + 6, the missing factor:
x – 2 (x – 2)(x + 6)
=
x + 2 (x + 2)(x + 6)
=
x2 + 4x – 12
x2 + 8x + 12
Now do Exercises 21–32
CAUTION When building up a denominator, both the numerator and the denomina
tor must be multiplied by the appropriate expression.
U2V Finding the Least Common Denominator
We can use the idea of building up the denominator to convert two fractions with
different denominators into fractions with identical denominators. For example,
5
6
1
4
and
can both be converted into fractions with a denominator of 12, since 12 = 2 6
and 12 = 3 4:
5 5 2 10
=
=
6 6 2 12
1 1 3
3
=
=
4 4 3 12
The smallest number that is a multiple of all of the denominators is called the least
common denominator (LCD). The LCD for the denominators 6 and 4 is 12.
To find the LCD in a systematic way, we look at a complete factorization of each
denominator. Consider the denominators 24 and 30:
24 = 2 2 2 3 = 23 3
30 = 2 3 5
Any multiple of 24 must have three 2’s in its factorization, and any multiple of 30
must have one 2 as a factor. So a number with three 2’s in its factorization will have
enough to be a multiple of both 24 and 30. The LCD must also have one 3 and one 5
in its factorization. We use each factor the maximum number of times it appears in
either factorization. So the LCD is 23 3 5:
24
2
3
�
dug84356_ch06a.qxd
�
3 5 = 2 2 2 3 5 = 120
30
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6.3
Finding the Least Common Denominator
403
If we omitted any one of the factors in 2 2 2 3 5, we would not have a
multiple of both 24 and 30. That is what makes 120 the least common denominator.
To find the LCD for two polynomials, we use the same strategy.
Strategy for Finding the LCD for Polynomials
1. Factor each denominator completely. Use exponent notation for repeated
factors.
2. Write the product of all of the different factors that appear in the denominators.
3. On each factor, use the highest power that appears on that factor in any of
the denominators.
E X A M P L E
3
Finding the LCD
If the given expressions were used as denominators of rational expressions, then what
would be the LCD for each group of denominators?
c) a2 + 5a + 6, a2 + 4a + 4
b) x3yz2, x5y2z, xyz5
a) 20, 50
Solution
a) First factor each number completely:
50 = 2 52
20 = 22 5
The highest power of 2 is 2, and the highest power of 5 is 2. So the LCD of 20 and
50 is 22 52, or 100.
b) The expressions x 3yz 2, x 5y 2z, and xyz 5 are already factored. For the LCD, use the
highest power of each variable. So the LCD is x5y2z 5.
c) First factor each polynomial.
a2 + 5a + 6 = (a + 2)(a + 3)
a2 + 4a + 4 = (a + 2)2
The highest power of (a + 3) is 1, and the highest power of (a + 2) is 2. So the
LCD is (a + 3)(a + 2)2.
Now do Exercises 33–46
U3V Converting to the LCD
When adding or subtracting rational expressions, we must convert the expressions
into expressions with identical denominators. To keep the computations as simple as
possible, we use the least common denominator.
E X A M P L E
4
Converting to the LCD
Find the LCD for the rational expressions, and convert each expression into an equivalent
rational expression with the LCD as the denominator.
a)
4
2
,
9xy 15xz
b)
5
1
3
,
,
6×2 8x3y 4y2
Solution
a) Factor each denominator completely:
9xy = 32xy
15xz = 3 5xz
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Chapter 6 Rational Expressions
U Helpful Hint V
What is the difference between LCD,
GCF, CBS, and NBC? The LCD for the
denominators 4 and 6 is 12. The least
common denominator is greater than
or equal to both numbers.The GCF for
4 and 6 is 2. The greatest common fac
tor is less than or equal to both num
bers. CBS and NBC are TV networks.
The LCD is 32 5xyz. Now convert each expression into an expression with this
denominator. We must multiply the numerator and denominator of the first rational
expression by 5z and the second by 3y:
⎫
⎪
⎬ Same denominator
2
2 3y
6y ⎪
=
=
15xz 15xz 3y 45xyz ⎭
4
4 5z
20z
=
=
9xy 9xy 5z 45xyz
b) Factor each denominator completely:
6x 2 = 2 3x 2
8x3y = 23x3y
4y2 = 22y 2
The LCD is 23 3 x3y2 or 24x3y2. Now convert each expression into an expression
with this denominator:
5 4xy2
20xy2
5
2=
2
2=
24x3y2
6x 6x 4xy
1
1 3y
3y
=
3 =
3
8x y 8x y 3y 24x3y2
3 6×3
18×3
3
=
=
4y2 4y2 6×3 24x3y2
Now do Exercises 47–58
E X A M P L E
5
Converting to the LCD
Find the LCD for the rational expressions
5x
x2 – 4
and
3
x2 + x – 6
and convert each into an equivalent rational expression with that denominator.
Solution
First factor the denominators:
x2 – 4 = (x – 2)(x + 2)
x2 + x – 6 = (x – 2)(x + 3)
The LCD is (x – 2)(x + 2)(x + 3). Now we multiply the numerator and denominator of
the first rational expression by (x + 3) and those of the second rational expression by
(x + 2). Because each denominator already has one factor of (x – 2), there is no reason to
multiply by (x – 2). We multiply each denominator by the factors in the LCD that are
missing from that denominator:
5x
5×2 + 15x
5x(x + 3)
=
=
x -4
(x – 2)(x + 2)(x + 3)
(x – 2)(x + 2)(x + 3)
2
3
3x + 6
3(x + 2)
=
=
x2 + x – 6 (x – 2)(x + 3)(x + 2)
(x – 2)(x + 2)(x + 3)
⎫
⎪ Same
⎬ denominator
⎪
⎭
Now do Exercises 59–70
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6-25
6.3
405
▼
Fill in the blank.
1. To
the denominator of a fraction, we multiply
the numerator and denominator by the same nonzero real
number.
2. The
is the smallest number
that is a multiple of all denominators.
3. The LCD is the product of every factor that appears in
the factorizations, raised to the
power that
appears on the factor.
2 2+5
5. — = -3 3+5
6. The LCD for the denominators 25 · 3 and 24 · 32 is
25 · 32.
1
1
7. The LCD for — and — is 60.
10
6
1
1
8. The LCD for — and — is x2 – 4.
x-2
x+2
1
1
– and — is a2 – 1.
9. The LCD for a2 – 1
a-1
True or false?
2
2·5
4. — = -3
3·5
Exercises
U Study Tips V
• Try changing subjects or tasks every hour when you study. The brain does not easily assimilate the same material hour after hour.
• You will learn more from working on a subject one hour per day than seven hours on Saturday.
U1V Building Up the Denominator
Build each rational expression into an equivalent rational
expression with the indicated denominator. See Example 1.
1
?
1. – = 3
27
2
?
2. – = 5
35
3
?
3. — = -4
16
3
?
4. — = -7
28
?
5. 1 = -7-
?
6. 1 = -3x-
?
7. 2 = -6
?
8. 5 = -12
5
?
9. — = -x
ax
x
?
10. — = -3
3x
?
11. 7 = -2x
?
12. 6 = -4y
5
?
13. — = -b
3bt
7
?
14. — = -2ay
2ayz
-9z
?
15. — = -2aw 8awz
-7yt
?
16. — = -3x
18xyt
?
2
17. — = -3
3a 15a
7b
?
18. –5 = –8
12c
36c
4
?
19. –2 = -5xy
10x 2y 5
?
5y2
20. -= -8x3z 24x5z3
6.3
Warm-Ups
Finding the Least Common Denominator
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Chapter 6 Rational Expressions
Build each rational expression into an equivalent rational
expression with the indicated denominator. See Example 2.
41. x2 – 16, x 2 + 8x + 16
5
?
21.
=
x+3
2x + 6
43. x, x + 2, x – 2
44. y, y – 5, y + 2
45. x 2 – 4x, x 2 – 16, 2x
4
?
22.
=
a-5
3a – 15
23.
42. x2 – 9, x 2 + 6x + 9
46. y, y 2 – 3y, 3y
5
?
=
2x + 2 -8x – 8
U3V Converting to the LCD
24.
3
?
=
m – n 2n – 2m
Find the LCD for the given rational expressions, and convert
each rational expression into an equivalent rational expression
with the LCD as the denominator. See Example 4.
25.
8a
?
=
20b2 – 20b3
5b2 – 5b
47.
1 3
,
6 8
48.
5 3
,
12 20
26.
5x
?
=
-6x – 9
18×2 + 27x
49.
1 5
,
2x 6x
50.
3
1
,
5x 10x
27.
3
?
= 2
x+2
x -4
51.
2 1
,
3a 2b
52.
y x
,
4x 6y
28.
a
?
= 2
a+3 a -9
53.
5
3
,
84a 63b
29.
3x
?
= 2
x + 2x + 1
x+1
54.
-7x
?
30.
= 2
2x – 3
4x – 12x + 9
31.
32.
55.
4b
6
,
75a 105ab
1
3
,
3x 2 2x 5
?
y-6
= 2
y-4
y + y – 20
56.
?
z-6
= 2
z – 2z – 15
z+3
57.
x
y
1
,
, 2
5
3
9y z 12x 6x y
58.
5
1
3b
,
3,
6
12a b 14a 2ab3
U2V Finding the Least Common Denominator
If the given expressions were used as denominators of rational
expressions, then what would be the LCD for each group of
denominators? See Example 3. See the Strategy for Finding the
LCD for Polynomials box on page 403.
33. 12, 16
34. 28, 42
35. 12, 18, 20
36. 24, 40, 48
2
2
37. 6a , 15a
4
8a b
,
5
6a2c
59.
2x
5x
,
x-3 x+2
60.
2a
3a
,
a-5 a+2
3 2
39. 2a b, 3ab , 4a b
40. 4m3nw, 6mn5w8, 9m6nw
3 9
Find the LCD for the given rational expressions, and convert
each rational expression into an equivalent rational expression
with the LCD as the denominator. See Example 5.
38. 18x , 20xy
6
3
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6.4
61.
4
5
,
a-6 6-a
62.
4
5x
,
x – y 2y – 2x
63.
x
5x
,
x 2 – 9 x 2 – 6x + 9
5x
-4
, 2
64. 2
x – 1 x – 2x + 1
65.
w +2
-2w
,
w2 – 2w – 15 w2 – 4w – 5
66.
z-1
z+1
,
z2 + 6z + 8 z2 + 5z + 6
67.
Addition and Subtraction
69.
2
3
4
,
,
2q 2 – 5q – 3 2q 2 + 9q + 4 q 2 + q – 12
70.
p
-3
2
,
,
2p2 + 7p – 15 2p2 – 11p + 12 p2 + p – 20
407
Getting More Involved
71. Discussion
Why do we learn how to convert two rational
expressions into equivalent rational expressions with
the same denominator?
x
3
-5
,
,
6x – 12 x 2 – 4 2x + 4
72. Discussion
68.
Which expression is the LCD for
3
2b
-5
,
,
4b – 9 2b + 3 2b2 – 3b
2
3x – 1
22 3 x2(x + 2)
6.4
In This Section
U1V Addition and Subtraction of
Rational Numbers
2
U V Addition and Subtraction of
Rational Expressions
U3V Applications
and
2x + 7
?
2 32 x(x + 2)2
a) 2 3 x(x + 2)
b) 36x(x + 2)
c) 36×2(x + 2)2
d) 23 33×3(x + 2)2
Addition and Subtraction
In Section 6.3, you learned how to find the LCD and build up the denominators of
rational expressions. In this section, we will use that knowledge to add and subtract
rational expressions with different denominators.
U1V Addition and Subtraction of Rational Numbers
We can add or subtract rational numbers (or fractions) only with identical denominators
according to the following definition.
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Chapter 6 Rational Expressions
Addition and Subtraction of Rational Numbers
If b e 0, then
a
c
a+c
a
c
a-c
+ =
and
– =
.
b
b
b
b
b
b
E X A M P L E
1
Adding or subtracting fractions with the same denominator
Perform the indicated operations. Reduce answers to lowest terms.
a)
1
7
+
12 12
b)
1 3
4 4
b)
1 3 -2
1
– =
=4 4
4
2
Solution
a)
1
7
8
4/ 2 2
+
=
=
=
12 12 12 4/ 3 3
Now do Exercises 1–8
If the rational numbers have different denominators, we must convert them to
equivalent rational numbers that have identical denominators and then add or subtract.
Of course, it is most efficient to use the least common denominator (LCD), as in
Example 2.
E X A M P L E
2
Adding or subtracting fractions with different denominators
Find each sum or difference.
a)
U Helpful Hint V
Note how all of the operations with
rational expressions are performed
according to the rules for fractions.
So keep thinking of how you perform
operations with fractions, and you will
improve your skills with fractions and
with rational expressions.
3
7
+
20 12
b)
1
4
6 15
Solution
a) Because 20 = 22 5 and 12 = 22 3, the LCD is 22 3 5, or 60. Convert each
fraction to an equivalent fraction with a denominator of 60:
3
7
3 3
7 5
+
=
+
20 12 20 3 12 5
9
35
=
+
60 60
44
=
60
4 11
=
4 15
11
=
15
Build up the denominators.
Simplify numerators and denominators.
Add the fractions.
Factor.
Reduce.
b) Because 6 = 2 3 and 15 = 3 5, the LCD is 2 3 5 or 30:
1
4
1
4
=
6 15 2 3 3 5
=
Factor the denominators.
1 5
4 2
Build up the denominators.
2 3 5 3 5 2
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6.4
Addition and Subtraction
=
5
8
30 30
Simplify the numerators and denominators.
=
-3
30
Subtract.
=
-1 3
10 3
Factor.
1
10
Reduce.
=-
409
Now do Exercises 9–18
U2V Addition and Subtraction of Rational Expressions
Rational expressions are added or subtracted just like rational numbers. We can add or
subtract only when we have identical denominators. All answers should be reduced to
lowest terms. Remember to factor first when reducing, and then divide out any com
mon factors.
E X A M P L E
3
Rational expressions with the same denominator
Perform the indicated operations and reduce answers to lowest terms.
a)
2
4
+
3y 3y
b)
2x
4
+
x+2 x+2
c)
2x + 1
x2 + 2x
(x – 1)(x + 3) (x – 1)(x + 3)
Solution
a)
2
4
6
+ =
Add the fractions.
3y 3y 3y
=
b)
2
y
Reduce.
2x
4
2x + 4
+
=
x+2 x+2
x+2
=
Add the fractions.
2(x + 2)
Factor the numerator.
x+2
=2
Reduce.
x 2 + 2x – (2x + 1)
x + 2x
2x + 1
=
(x – 1)(x + 3)
(x – 1)(x + 3)
(x – 1)(x + 3)
2
c)
Subtract the fractions.
=
x2 + 2x – 2x – 1
(x – 1)(x + 3)
Remove parentheses.
=
x2 – 1
(x – 1)(x + 3)
Combine like terms.
=
(x – 1)(x + 1)
(x – 1)(x + 3)
Factor.
=
x+1
x+3
Reduce.
Now do Exercises 19–30
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6-30
Chapter 6 Rational Expressions
CAUTION When subtracting a numerator containing more than one term, be sure
to enclose it in parentheses, as in Example 3(c). Because that numerator
is a binomial, the sign of each of its terms must be changed for the
subtraction.
In Example 4, the rational expressions have different denominators.
E X A M P L E
4
Rational expressions with different denominators
Perform the indicated operations.
U Helpful Hint V
You can remind yourself of the
difference between addition and
multiplication of fractions with a simple example: If you and your spouse
each own 1/7 of Microsoft, then
together you own 2/7 of Microsoft.
If you own 1/7 of Microsoft, and
give 1/7 of your stock to your child,
then your child owns 1/49 of
Microsoft.
a)
5
2
+
2x 3
c)
a+1 a-2
6
8
b)
4
2
+
x 3y xy 3
Solution
a) The LCD for 2x and 3 is 6x:
5
2
5 3
2 2x
+ =
+
2x 3 2x 3 3 2x
Build up both denominators to 6x.
=
15 4x
+
6x 6x
Simplify numerators and denominators.
=
15 + 4x
6x
Add the rational expressions.
b) The LCD is x 3y 3.
4
2
4 y2
2 x2
+ 3= 3
2 + xy3 x 2
3
x y xy
xy y
Build up both denominators to the LCD.
=
4y2
2×2
3 3+ 3 3
xy
xy
Simplify numerators and denominators.
=
4y 2 + 2x 2
x3y3
Add the rational expressions.
c) Because 6 = 2 3 and 8 = 23, the LCD is 23 3, or 24:
a + 1 a – 2 (a + 1)4 (a – 2)3
=
6
8
6 4
8 3
Build up both denominators
to the LCD 24.
=
4a + 4 3a – 6
24
24
Simplify numerators and
denominators.
=
4a + 4 – (3a – 6)
24
Subtract the rational expressions.
=
4a + 4 – 3a + 6
24
Remove the parentheses.
=
a + 10
24
Combine like terms.
Now do Exercises 31–46
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E X A M P L E
6.4
5
Addition and Subtraction
411
Rational expressions with different denominators
Perform the indicated operations:
a)
U Helpful Hint V
1
2
+ 2
x – 9 x + 3x
b)
2
4
2
5-a a-5
Solution
1
2
1
2
The LCD is x(x – 3)(x + 3).
+
=
+
x2 – 9 x2 + 3x (x – 3)(x + 3) x(x + 3)
Needs x
�
a)
�
Once the denominators are factored
as in Example 5(a), you can simply
look at each denominator and ask,
“What factor does the other denomi
nator(s) have that is missing from this
one?” Then use the missing factor
to build up the denominator. Repeat
until all denominators are identical,
and you will have the LCD.
Needs x – 3
=
1 x
2(x – 3)
+
(x – 3)(x + 3)x x(x + 3)(x – 3)
=
x
2x – 6
+
x(x – 3)(x + 3) x(x – 3)(x + 3)
=
3x – 6
x(x – 3)(x + 3)
We usually leave the
denominator in
factored form.
b) Because -1(5 – a) = a – 5, we can get identical denominators by multiplying
only the first expression by -1 in the numerator and denominator:
4
2
4(-1)
2
=
5 – a a – 5 (5 – a)(-1) a – 5
=
-4
2
a-5 a-5
=
-6
-4 – 2 = -6
a-5
6
=a-5
Now do Exercises 47–64
In Example 6, we combine three rational expressions by addition and subtraction.
E X A M P L E
6
Rational expressions with different denominators
Perform the indicated operations.
x+1
2x + 1
1
+
x2 + 2x 6x + 12 6
Solution
The LCD for x(x + 2), 6(x + 2), and 6 is 6x(x + 2).
x+1
2x + 1
1
x+1
2x + 1
1
+
– =
+
x 2 + 2x 6x + 12 6 x(x + 2) 6(x + 2) 6
=
6(x + 1)
x(2x + 1) 1x(x + 2)
+
6x(x + 2) 6x(x + 2) 6x(x + 2)
Factor
denominators.
Build up to
the LCD.
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Chapter 6 Rational Expressions
=
6x + 6
2×2 + x
x2 + 2x
+
6x(x + 2) 6x(x + 2) 6x(x + 2)
Simplify
numerators.
=
6x + 6 + 2×2 + x – x2 – 2x
6x(x + 2)
Combine the
numerators.
=
x2 + 5x + 6
6x(x + 2)
=
(x + 3)(x + 2)
6x(x + 2)
Factor.
=
x+3
6x
Reduce.
Combine
like terms.
Now do Exercises 65–70
U3V Applications
We have seen how rational expressions can occur in problems involving rates. In
Example 7, we see an applied situation in which we add rational expressions.
E X A M P L E
7
Adding work
Harry takes twice as long as Lucy to proofread a manuscript. Write a rational expression for
the amount of work they do in 3 hours working together on a manuscript.
Solution
Let x = the number of hours it would take Lucy to complete the manuscript alone and
2x = the number of hours it would take Harry to complete the manuscript alone. Make a
table showing rate, time, and work completed:
Rate
Time
Work
Lucy
1 msp
x hr
3 hr
3
msp
x
Harry
1 msp
2x hr
3 hr
3
msp
2x
Now find the sum of each person’s work.
3
3
2 3
3
+ =
+
x 2x 2 x 2x
=
6
3
+
2x 2x
9
2x
So in 3 hours working together they will complete
=
9
2x
of the manuscript.
Now do Exercises 81–86
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6-33
6.4
413
▼
Fill in the blank.
1. We can
rational expressions only if they have
identical denominators.
2. We can
any two rational expressions so that
their denominators are identical.
3
4
29
5. — + — = -5
3
15
4
5
3
6. — – — = -5
7
35
5
3
7. — + — = 1
20
4
2
3
8. For any nonzero value of x, — + 1 = –.
x
x
1
a+1
9. For any nonzero value of a, 1 + — = –.
a
a
1
4a – 1
10. For any value of a, a – — = –.
4
4
True or false?
1
1
2
3. — + — = -2
3
5
1
7
1
4. — – — = -2
12
12
Exercises
U Study Tips V
• When studying for a midterm or final, review the material in the order it was originally presented. This strategy will help you to see
connections between the ideas.
• Studying the oldest material first will give top priority to material that you might have forgotten.
U1V Addition and Subtraction of Rational Numbers
Perform the indicated operation. Reduce each answer to lowest
terms. See Example 1.
1
1
1. — + -10 10
1 3
2. — + -8 8
7 1
3. — – -8 8
4 1
4. — – -9 9
1 5
5. — – -6 6
3 7
6. — – -8 8
7 1
7. — + -8 8
9
3
8. — + –20
20
(
)
Perform the indicated operation. Reduce each answer to lowest
terms. See Example 2.
1 2
9. — + -3 9
1 5
10. — + -4 6
7
5
-11. -10 + 6
5
3
12. -6- + -1-0
7
5
13. — + -16 18
7
4
14. — + -6 15
1
9
15. — – -8 10
2
5
16. — – -15 12
( )
3
1
17. — – –6
8
( )
1
1
18. — – –5
7
6.4
Warm-Ups
Addition and Subtraction
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6-34
Chapter 6 Rational Expressions
Perform the indicated operation. Reduce each answer to lowest
terms. See Examples 5 and 6.
U2V Addition and Subtraction of
Rational Expressions
Perform the indicated operation. Reduce each answer to lowest
terms. See Example 3.
47.
1
1
+
x
x+2
19.
1
1
+
2x 2x
20.
1
2
+
3y 3y
48.
1
2
+
y
y+1
21.
3
7
+
2w 2w
22.
5x 7x
+
3y 3y
49.
2
3
x+1 x
23.
15
3a
+
a+5 a+5
24.
a + 7 9 – 5a
+
a-4
a-4
50.
1
2
a-1 a
25.
q – 1 3q – 9
q-4
q-4
51.
2
1
+
a-b a+b
26.
3-a a-5
3
3
52.
3
2
+
x+1 x-1
4h – 3
h-6
27.
h(h + 1) h(h + 1)
53.
3
4
x 2 + x 5x + 5
2t – 9
t-9
28.
t(t – 3) t(t – 3)
54.
3
2
a + 3a 5a + 15
55.
2a
a
+
a2 – 9 a – 3
56.
x
3
+
x
1
x -1
57.
4
4
+
a-b b-a
58.
2
3
+
x-3 3-x
x2 – x – 5
1 – 2x
29.
+
(x + 1)(x + 2) (x + 1)(x + 2)
2x – 5
x – 2x + 1
+
(x – 2)(x + 6) (x – 2)(x + 6)
2
30.
Perform the indicated operation. Reduce each answer to
lowest terms. See Example 4.
2
2
31.
1
1
+
a 2a
32.
1
2
+
3w w
59.
3
2
2a – 2 1 – a
33.
x x
+
3 2
34.
y y
+
4 2
60.
5
3
2x – 4 2 – x
35.
m
+m
5
36.
y
+ 2y
4
61.
1
3
x 2 – 4 x2 – 3x – 10
37.
1 2
+
x y
38.
2
3
+
a
b
62.
2x
3x
+ 2
x2 – 9
x + 4x + 3
39.
3
1
+
2a 5a
40.
3
5
6y 8y
63.
3
4
+ 2
x2 + x – 2
x + 2x – 3
41.
w-3 w-4
9
12
42.
y+4 y-2
10
14
64.
x+4
x-1
+ 2
x – x – 12
x + 5x + 6
43.
b2
-c
4a
44. y +
3
7b
65.
1
1
1
+ +
a
b
c
45.
2
3
+
wz2 w2z
46.
1
5
a5b ab3
66.
1
1
1
+ 2+ 3
x
x
x
2
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6.4
415
Addition and Subtraction
2
1
1
+
x x-1 x+2
U3V Applications
68.
1
2
3
+
a a+1 a-1
69.
5
3
4
– + 2
3a – 9 2a a – 3a
81. Perimeter of a rectangle. Suppose that the length of a
rectangle is 3 feet and its width is 5 feet. Find a rational
x
2x
expression for the perimeter of the rectangle.
70.
5
3
c-4
– 2
6c
4c + 2
2c + c
67.
Solve each problem. See Example 7.
82. Perimeter of a triangle. The lengths of the sides of a
triangle are 1, 1 , and
Match each expression in (a)–(f) with the equivalent expression
in (A)–(F).
71. a)
d)
A)
D)
72. a)
d)
A)
D)
1
+2
y
1
1
+
y 2y
3
y
y+2
y
1
-x
x
1
-x
x2
1 – x3
x2
1-x
x2
1
2
+
y
y
2
e)
+1
y
3
B)
2y
y+2
E)
2y
1
1
b)
– 2
x
x
1
e) x x
1-x
B)
x
x2 – 1
E)
x
b)
1
1
+
y
2
y
f) + 1
2
y+2
C)
2
2y + 1
F)
y
1
c)
-1
x
1
1
f) 2 x
x
1 – x2
C)
x
x-1
F)
x2
x 2x
c)
1
—
x
Figure for Exercise 82
83. Traveling time. Janet drove 120 miles at x mph before
6:00 A.M. After 6:00 A.M., she increased her speed by
5 mph and drove 195 additional miles. Use the fact that
T = D to complete the following table.
R
Rate
1
3
73.
2p + 8
2p
After
x+5
3
3
+ 2
a2 + 3a + 2
a + 5a + 6
76.
4
12
+ 2
w2 + w
w – 3w
77.
2
1
– 2
b2 + 4b + 3
b + 5b + 6
78.
9
6
– 2
m -m-2
m -1
79.
3
3
2
– 2
t + 2t
2t
t+2
80.
4
2
2
+
+ 2
3n
n+1
n +n
Time
mi
hr
x
75.
2
—
3x
1
—
2x
Before
3
3
2y
2y + 4
meters. Find a rational expression
for the perimeter of the triangle.
Perform the indicated operation. Reduce each answer to
lowest terms.
74.
2
3x
Distance
120 mi
mi
hr
195 mi
Write a rational expression for her total traveling time.
Evaluate the expression for x = 60.
84. Traveling time. After leaving Moose Jaw, Hanson drove
200 kilometers at x km/hr and then decreased his speed by
20 km/hr and drove 240 additional kilometers. Make a table
like the one in Exercise 83. Write a rational expression for
his total traveling time. Evaluate the expression for x = 100.
2
85. House painting. Kent can paint a certain house by himself
in x days. His helper Keith can paint the same house by
himself in x + 3 days. Suppose that they work together on
the job for 2 days. To complete the table on the next page,
use the fact that the work completed is the product of the
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6-36
Chapter 6 Rational Expressions
Rate
Time
Kent
1 job
x day
2 days
Keith
1 job
2 days
x + 3 day
Work
rate and the time. Write a rational expression for the
fraction of the house that they complete by working
together for 2 days. Evaluate the expression for x = 6.
Photo for Exercise 86
Getting More Involved
86. Barn painting. Melanie can paint a certain barn by herself
in x days. Her helper Melissa can paint the same barn by
herself in 2x days. Write a rational expression for the frac
tion of the barn that they complete in one day by working
together. Evaluate the expression for x = 5.
87. Writing
Write a step-by-step procedure for adding rational
expressions.
88. Writing
Explain why fractions must have the same
denominator to be added. Use real-life examples.
Math at Work
Gravity on the Moon
Hundreds of years before humans even considered traveling beyond the earth,
Isaac Newton established the laws of gravity. So when Neil Armstrong made the
first human step onto the moon in 1969, he knew what amount of gravitational
force to expect. Let’s see how he knew.
Newton’s equation for the force of gravity between two objects is
2
F = G m1m
, where m1 and m2 are the masses of the objects (in kilograms), d is
d2
the distance (in meters) between the centers of the two objects, and G is the
gravitational constant 6.67 � 10-11. To find the force of gravity for Armstrong
on earth, use 5.98 � 1024 kg for the mass of the earth, 6.378 � 106 m for the
radius of the earth, and 80 kg for Armstrong’s mass. We get
F = 6.67 � 10-11
5.98 � 1024 kg 80 kg
� 784 Newtons.
(6.378 � 106 m)2
To find the force of gravity for Armstrong on the moon, use 7.34 � 1022 kg for the mass of
the moon and 1.737 � 106 m for the radius of the moon. We get
F = 6.67 � 10-11
7.34 � 1022 kg 80 kg
� 130 Newtons.
(1.737 � 106 m)2
So the force of gravity for Armstrong on the moon was about one-sixth of the force of gravity
for Armstrong on earth. Fortunately, the moon is smaller than the earth. Walking on a planet
much larger than the earth would present a real problem in terms of gravitational force.
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6-37
Mid-Chapter Quiz
Reduce to lowest terms.
36
1.
84
3.
w2 – 1
2w + 2
2.
8x – 2
8
4.
2a2 – 10a + 12
6 – 3a
Perform the indicated operation.
6 21
3xy2
5.
6.
7 10
5z
7.
9.
a2 – 9
2a + 4
5a + 10
2a – 6
5
25
�
9
33
s2
s2
11.
+
21
3
Sections 6.1 through 6.4
8.
10.
8x2z3
8y4
m2 – 8m + 7
12.
� (m – 7)
2m
6.5
In This Section
U1V Complex Fractions
U2V Using the LCD to Simplify
Complex Fractions
417
Chapter 6
13.
5
5
6
21
14.
5
4
3 + 2
ab
ab
15.
x
3x
+
x + 1 x2 + 2x + 1
16.
y
y
y+5
y+2
17.
1
1
1
+ +
a
b
c
b2
b6
�
21
3
3x – 9
x2 – 6x + 9
�
8
12
6.5
Miscellaneous.
3x – 6
18. What numbers(s) can’t be used in place of x in
?
2x + 1
19. Find the value of
3x – 6
when x = -2.
2x + 1
20. Find R(-1) if R(x) =
6×2 + 3
.
5x – 1
Complex Fractions
In this section, we will use the idea of least common denominator to simplify
complex fractions. Also we will see how complex fractions can arise in applications.
Complex Fractions
U3V Applications
U1V Complex Fractions
A complex fraction is a fraction having rational expressions in the numerator,
denominator, or both. Consider the following complex fraction:
1 2
+
2 3
1 5
4 8
← Numerator of complex fraction
← Denominator of complex fraction
Since the fraction bar is a grouping symbol, we can compute the value of the numer
ator, the value of the denominator, and then divide them, as shown in Example 1.
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6-38
Chapter 6 Rational Expressions
E X A M P L E
1
Simplifying complex fractions
Simplify.
1 2
+
2 3
a) —
1 5
4 8
2
4
5
b) —
1
+3
10
Solution
a) Combine the fractions in the numerator:
1 2 1 3 2 2 3 4
+
+
+
2 3 2 3 3 2 6 6
Combine the fractions in the denominator as follows:
1 5 1 2 5 2 5
4 8 4 2 8 8 8
Now divide the numerator by the denominator:
2
4
5
b) —
1
+3
10
7
1 2
+
2 3
6
7
—
—
3
6
1 5
8
4 8
20 2
18
18
5
5
5
—
—
30
31
1
5
+
10
10 10
3
8
7
6
31
10
18 10
5 31
7
6
3
8
8
3
56
18
28
9
36
31
Now do Exercises 1–12
U2V Using the LCD to Simplify Complex Fractions
A complex fraction can be simplified by performing the operations in the numerator
and denominator, and then dividing the results, as shown in Example 1. However,
there is a better method. All of the fractions in the complex fraction can be eliminated
in one step by multiplying by the LCD of all of the single fractions. The strategy for
this method is detailed in the following box and illustrated in Example 2.
Strategy for Simplifying a Complex Fraction
1. Find the LCD for all the denominators in the complex fraction.
2. Multiply both the numerator and the denominator of the complex fraction by
the LCD. Use the distributive property if necessary.
3. Combine like terms if possible.
4. Reduce to lowest terms when possible.
E X A M P L E
2
Using the LCD to simplify a complex fraction
Use the LCD to simplify
1 2
+
2 3
—.
1 5
4 8
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6-39
6.5
Complex Fractions
419
U Calculator Close-Up V
Solution
You can check Example 2 with a
calculator as shown here.
The LCD of 2, 3, 4, and 8 is 24. Now multiply the numerator and denominator of the
complex fraction by the LCD:
1 2
+
2 3
—
1 5
4 8
1 2
+ 24
2 3
——
1
4
Multiply the numerator and denominator by the LCD.
5
24
8
1
2
24 +
24
2
3
——
1
5
24
24
8
4
Distributive property
12 + 16
6 15
Simplify.
28
9
28
9
Now do Exercises 13–20
CAUTION We simplify a complex fraction by multiplying the numerator and denomi
nator of the complex fraction by the LCD. Do not multiply the numerator
and denominator of each fraction in the complex fraction by the LCD.
In Example 3 we simplify a complex fraction involving variables.
E X A M P L E
3
A complex fraction with variables
Simplify
2
1
x2
1
x
.
1
2
U Helpful Hint V
Solution
When students see addition or subtraction in a complex fraction, they
often convert all fractions to the
same denominator. This is not wrong,
but it is not necessary. Simply multiplying every fraction by the LCD eliminates the denominators of the
original fractions.
The LCD of the denominators x, x 2, and 2 is 2x 2:
1
2
x
—
1 1
x2 2
2
1
(2×2)
x
——
1
x2
Multiply the numerator and denominator by 2×2.
1
(2×2)
2
1
2 2×2
2×2
x
——
1
1
2×2
2×2
x2
2
Distributive property
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6-40
Chapter 6 Rational Expressions
4×2 2x
Simplify.
2 x2
The numerator of this answer can be factored, but the rational expression cannot be reduced.
Now do Exercises 21–30
E X A M P L E
4
Simplifying a complex fraction
Simplify
1
2
x 2 x+2
—— .
3
4
+
2 x x+2
Solution
Because x 2 and 2 x are opposites, we can use (x 2)(x + 2) as the LCD. Multiply
the numerator and denominator by (x 2)(x + 2):
1
2
1
2
(x 2)(x + 2)
(x 2)(x + 2)
x 2 x+2
x 2
x+2
—— —————
3
4
3
4
(x 2)(x + 2) +
(x 2)(x + 2)
+
2 x
x+2
2 x x+2
x + 2 2(x 2)
3( 1)(x + 2) + 4(x 2)
x
x
2
2
x
1
x + 2 2x + 4
3x 6 + 4x 8
Distributive property
x+6
14
Combine like terms.
Now do Exercises 31–46
U3V Applications
As their name suggests, complex fractions arise in some fairly complex situations.
E X A M P L E
5
Fast-food workers
A survey of college students found that 1 of the female students had jobs and 2 of the male
3
2
students had jobs. It was also found that 1 of the female students worked in fast-food
4
restaurants and 1 of the male students worked in fast-food restaurants. If equal numbers of
6
male and female students were surveyed, then what fraction of the working students
worked in fast-food restaurants?
Solution
Let x represent the number of males surveyed. The number of females surveyed is also x.
The total number of students working in fast-food restaurants is
1
1
x + x.
4
6
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6-41
6.5
Complex Fractions
421
The total number of working students in the survey is
1
2
x + x.
2
3
So the fraction of working students who work in fast-food restaurants is
1
1
x+ x
4
6
2
1
x+ x
3
2
.
The LCD of the denominators 2, 3, 4, and 6 is 12. Multiply the numerator and denominator
by 12 to eliminate the fractions as follows:
1
1
x+ x
4
6
—
2
1
x+ x
3
2
1
1
x + x 12
6
4
—— Multiply numerator and denominator by 12.
2
1
x + x 12
3
2
3x + 2x
6x + 8x
5x
14x
5
14
So
5
14
Distributive property
Combine like terms.
Reduce.
(or about 36%) of the working students work in fast-food restaurants.
Now do Exercises 61–62
Warm-Ups
▼
Fill in the blank.
1. A
fraction has fractions in its numerator,
denominator, or both.
2. To simplify a complex fraction, you can multiply its
and
by the LCD of all of the
fractions.
True or false?
3. The LCD for the denominator 4, x, 6, and x2 is 12×3.
4. The LCD for the denominator a
6a 6b.
b, 2b
2a, and 6 is
1 1
+
2 3
5. To simplify — , we multiply the numerator and
1 1
4 6
denominator by 12.
1 1
+ 12
2 3
6+4
3 2
6. ——
1 1
12
4 6
1 1
+
2 3
7. —
1 1
4 6
5
6
—
1
12
1
x
2
8. For any real number x, —
1
x+
3
2x 1
.
3x + 1
6.5
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Exercises
U Study Tips V
• Stay calm and confident. Take breaks when you study. Get 6 to 8 hours of sleep every night.
• Keep reminding yourself that working hard throughout the semester will really pay off in the end.
U1V Complex Fractions
Simplify each complex fraction. See Example 1.
1 5
1 1
1 1
�� � ��
�� � ��
�� � ��
3
6
2 3
2 4
2. —
3. —
1. —
1 3
2 1
1 1
�� � ��
�� � ��
�� � ��
2 4
3 6
4 2
1 1
�� � ��
3 4
4. —
1 1
�� � ��
3 6
2 5 1
�� � �� � ��
5 6 2
5. ——
1 1
1
�� � �� � ��
2 3 15
2 2 1
�� � �� � ��
5 9 3
6. ——
1 1
2
�� � �� � ��
3 5 15
1
1 � ��
2
7. —
1
2 � ��
4
1
�� � 1
3
8. —
1
�� � 2
6
1
3 � ��
2
9. —
3
5 � ��
4
1
1 � ��
12
10. —
1
1 � ��
12
1 2
1 � �� � ��
6 3
11. ——
1
3
1 � �� � ��
15 10
2 1
3 � �� � ��
9 6
12. ——
5
1
�� � �� � 2
18 3
U2V Using the LCD to Simplify Complex Fractions
Simplify each complex fraction. See Examples 2 and 3. See
the Strategy for Simplifying a Complex Fraction box on
page 418.
1 1
�� � ��
2 3
—
13.
1 1
�� � ��
2 4
1 1
�� � ��
4 3
—
14.
1 1
�� � ��
4 6
2
1
�� � ��
5 10
15. —
1 1
�� � ��
5 4
3
4
�� � ��
10 5
16. —
1 3
�� � ��
2 4
2 1
1 � �� � ��
3 2
17. ——
1 3
2 � �� � ��
3 2
1
3
3 � �� � ��
5 10
18. ——
6
3
2 � �� � ��
5 10
2 5 1
�� � �� � ��
3 6 2
19. ——
1 1 1
�� � �� � ��
6 3 2
2 3
7
�� � �� � ��
5 2 10
20. ——
1 1
1
�� � �� � ��
5 2 10
1 1
�� � ��
a b
21. —
2 2
�� � ��
a b
1 1
�� � ��
x y
22. —
3 3
�� � ��
x y
1 3
�� � ��
a b
23. —
1 3
�� � ��
b a
1 3
�� � ��
x 2
24. —
3 1
�� � ��
4 x
3
5 � ��
a
25. —
1
3 � ��
a
3
4 � ��y
26. —
2
1 � ��
y
1 2
�� � ��
2 x
—
27.
1
3 � ��2
x
2 5
�� � ��
a 3
—
28.
3
3
�� � ��2
a a
3
1
�� � ��
2b b
29. �
3
1
�� � ��2
4 b
3
4
�� � ��
2w 3w
30. �
1
5
�� � ��
4w 9w
Simplify each complex fraction. See Example 4.
1
�� � 1
x�1
31. ——
3
�� � 3
x�1
2
�� � 1
x�3
32. ——
4
�� � 2
x�3
3
1 � ��
y�1
33. ——
1
3 � ��
y�1
1
2 � ��
a�3
34. ——
1
3 � ��
a�3
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6-43
6.5
4
x+
x 2
35. ——
x+1
x
x 2
x 6
x
x 1
36. ——
x + 15
x
x 1
2x
9
6
53. —
2x 3
9
2x
4y
2
1
2
5
3 x
37. ——
1
2
x 3
x
x 5
38. ——
3x
1
5 x
5
1
a 1
39. ——
2
3
1 a
1
2
3 9 x
40. ——
1
1
6 x 9
4
m
m 3
41.
3
1
+
m 3
m
1
y+3
1
2
42.
1
y
4
y
2
y+3
3
w 1 w+1
43. ——
4
5
+
w+1 w 1
1
3
x+2 x+3
44. ——
2
3
+
x+3 x+2
1
1
a b a+b
45. ——
1
1
+
b a b+a
1
1
2+x 2 x
46. ——
1
1
x+2 x 2
Simplify each complex fraction. Reduce each answer to
lowest terms.
1 1
4
+
1
3 y
a2
48. —
47. ——
y 3
2
8
1+
3 y
a a2
1
1
+
2
4x
49. —
x
1
3 12x
1
1
+
9
3x
50. —
x 1
9 x
1
5
2
+
3 3x x2
51. ——
1 3
3 x2
1
3
1
+
2 2x x2
52. ——
1
1
2 2×2
xy
55. —
3x 6y
x3y
a2 + 2a 24
a+1
57. ——
a2 a 12
(a + 1)2
x
x+1
59. ——
1
1
x2 1 x 1
Complex Fractions
423
a
5
12
54. —
a+2
15
ab + b2
4ab5
56. —
a+b
6a2b4
y2
3y 18
y2 4
58. ——
y2 + 5y + 6
y 2
a
2
a
b2
60. ——
1
1
+
a+b a b
U3V Applications
Solve each problem. See Example 5.
61. Sophomore math. A survey of college sophomores showed
that 5 of the males were taking a mathematics class and 3 of
6
4
the females were taking a mathematics class. One-third of the
males were enrolled in calculus, and 1 of the females were
5
enrolled in calculus. If just as many males as females were
surveyed, then what fraction of the surveyed students taking
mathematics were enrolled in calculus? Rework this prob
lem assuming that the number of females in the survey
was twice the number of males.
62. Commuting students. At a well-known university, 1 of the
4
undergraduate students commute, and 1 of the graduate
3
students commute. One-tenth of the undergraduate students
drive more than 40 miles daily, and 1 of the graduate
6
students drive more than 40 miles daily. If there are twice
as many undergraduate students as there are graduate
students, then what fraction of the commuters drive more
than 40 miles daily?
Photo for Exercise 62
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Chapter 6 Rational Expressions
Getting More Involved
64. Discussion
A complex fraction can be simplified by writing the
numerator and denominator as single fractions and then
dividing them or by multiplying the numerator and
denominator by the LCD. Simplify the complex fraction
63. Exploration
Simplify
1
1
1
—,
, and
.
1
1
1
1+2 1+
1+
1
1
1+
1+2
1
1+
2
a) Are these fractions getting larger or smaller as the
fractions become more complex?
b) Continuing the pattern, find the next two complex
4
6
xy2 xy
—
2
4
2+ 2
x
xy
by using each of these methods. Compare the number of
steps used in each method, and determine which method
requires fewer steps.
fractions and simplify them.
c) Now what can you say about the values of all five
complex fractions?
6.6
In This Section
U1V Equations with Rational
Expressions
2
U V Extraneous Solutions
Solving Equations with Rational Expressions
Many problems in algebra can be solved by using equations involving rational
expressions. In this section you will learn how to solve equations that involve
rational expressions, and in Sections 6.7 and 6.8 you will solve problems using
these equations.
U1V Equations with Rational Expressions
We solved some equations involving fractions in Section 2.3. In that section, the
equations had only integers in the denominators. Our first step in solving those
equations was to multiply by the LCD to eliminate all of the denominators.
E X A M P L E
1
Integers in the denominators
Solve 1
x
2
2
3
1
.
6
Solution
U Helpful Hint V
Note that it is not necessary to
convert each fraction into an equiva
lent 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.
The LCD for 2, 3, and 6 is 6. Multiply each side of the equation by 6:
1 x 2
2
3
1 x 2
6
2
3
6
2
1
2
3
6�
Original equation
6
1
6
Multiply each side by 6.
2
�6
1
�6
Distributive property
2)
1
2x + 4
1
x
�3
2(x
3
1
6
2x
x
Simplify.
Distributive property
6
3
Subtract 7 from each side.
Divide each side by
2.
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6.6
U Helpful Hint V
Check x
Always check your solution in the
original equation by calculating
the value of the left-hand side and
the value of the right-hand side.
If they are the same, your solution
is correct.
Solving Equations with Rational Expressions
425
3 in the original equation:
1
2
3
2
1
2
3
1
3
3
6
2
6
1
6
Since the right-hand side of the equation is 1, you can be sure that the solution to the
6
equation is 3.
Now do Exercises 1–12
CAUTION When a numerator contains a binomial, as in Example 1, the numer
ator must be enclosed in parentheses when the denominator is
eliminated.
To solve an equation involving rational expressions, we usually multiply each side
of the equation by the LCD for all the denominators involved, just as we do for an
equation with fractions.
E X A M P L E
2
Variables in the denominators
Solve 1 + 1
x
6
1
.
4
Solution
We multiply each side of the equation by 12x, the LCD for 4, 6, and x :
1 1
+
x 6
12x
1 1
+
x 6
2
1
1
+�
12x
6�
�x
12x�
12 + 2x
12
1
4
Original equation
12x
3
�
12x
1
4
Multiply each side by 12x.
1
Distributive property
4�
3x
Simplify.
x
Subtract 2x from each side.
Check that 12 satisfies the original equation:
1
1
+
12 6
1
2
+
12 12
3
12
1
4
The solution to the equation is 12.
Now do Exercises 13–24
E X A M P L E
3
An equation with two solutions
Solve the equation 100 +
x
100
x+5
9.
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Chapter 6 Rational Expressions
Solution
The LCD for the denominators x and x + 5 is x(x + 5):
100
100
+
9
x
x+5
100
100
x(x + 5)
+ x(x + 5)
x(x + 5)9
+5
x
x
9x + 25
(x + 5)100 + x(100)
(x2 + 5x)9
100x + 500 + 100x
9×2 + 45x
500 + 200x
9×2 + 45x
0
or
25
9
x
x
25
9
A check will show that both
Multiply each side by
x(x + 5).
All denominators are
eliminated.
Simplify.
0
9×2
155x
500
Get 0 on one side.
0
(9x + 25)(x
20)
Factor.
20
0
x
20
or
Original equation
Zero factor property
and 20 satisfy the original equation.
Now do Exercises 25–32
U2V Extraneous Solutions
In a rational expression, we can replace the variable only by real numbers that do not
cause the denominator to be 0. When solving equations involving rational expressions,
we must check every solution to see whether it causes 0 to appear in a denominator.
If a number causes the denominator to be 0, then it cannot be a solution to the equa
tion. A number that appears to be a solution but causes 0 in a denominator is called an
extraneous solution. Since a solution to an equation is sometimes called a root to the
equation, an extraneous solution is also called an extraneous root.
E X A M P L E
4
An equation with an extraneous solution
Solve the equation
1
x
x
2
2x
4
+ 1.
Solution
Because the denominator 2x
2(x
2)
1
x
2
2(x
4 factors as 2(x
2)
2
x + 2x
2
3x
6
3x
2
x
x
2(x
4
2)
+ 2(x
2), the LCD is 2(x
2) 1
2).
Multiply each side of the
original equation by 2(x 2).
Simplify.
4
Check 2 in the original equation:
1
2
+1
2 2 2 2 4
The denominator 2 2 is 0. So 2 does not satisfy the equation, and it is an extraneous
solution. The equation has no solutions.
Now do Exercises 33–36
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6.6
427
Solving Equations with Rational Expressions
If the denominators of the rational expressions in an equation are not too compli
cated, you can tell at a glance which numbers cannot be solutions. For example, the
2
3
x 2
equation x + x 1 x + 5 could not have 0, 1, or 5 as a solution. Any solution to this
equation must be in the domain of all three of the rational expressions in the equation.
E X A M P L E
5
Another extraneous solution
Solve the equation 1 +
x
1
x
x
x
3
2
.
3
Solution
The LCD for the denominators x and x 3 is x(x
1
x 2
1
+
x x 3 x 3
1
x
1
x(x 3)
+ x(x 3)
x(x 3)
x
x
x 3
x
3+x
2x
x
3
0
or
x
3
or
x
3
x(x
2)
2
2x
x
0
x
4x + 3
0
(x
3)(x
1
0
x
1
2
3):
Original equation
2
3
Multiply each side by x(x
3).
1)
If x 3, then the denominator x 3 has a value of 0. If x
satisfied. The only solution to the equation is 1.
1, the original equation is
Now do Exercises 37–40
CAUTION Always be sure to check your answers in the original equation to determine
whether they are extraneous solutions.
Warm-Ups
▼
Fill in the blank.
1. The usual first step in solving an equation involving
rational expressions is to multiply by the
.
2. An
solution is a number that appears to be a
solution but does not check in the original equation.
6. To solve
7. To solve
3
5
+
x x 2
1
x
1
2
, multiply each side by 3×2
3
1
, multiply each side by
x+1
+2
x2 – 1.
True or false?
2
3. To solve x
8x, we divide each side by x.
4. An extraneous solution is an irrational number.
5. Both 0 and 2 satisfy
3
5
+
x x 2
2
.
3
8. The solution set to
9. The solution set to
1
x
1
1 1
+
x 2
+2
1
is { 1, 1}.
x +1
3
is {4}.
x
6x.
6.6
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Exercises
U Study Tips V
• The last couple of weeks of the semester is not the time to slack off. This is the time to double your efforts.
• Make a schedule and plan every hour of your time.
U1V Equations with Rational Expressions
Solve each equation. See Example 3.
Solve each equation. See Example 1.
x
5
25. — = -2 x+3
x
4
26. — = -3 x+1
x
6
27. — = -x+1 x+7
x
2
28. — = -x+3 x-3
2
1 1
29. — = — + -x+1 x 6
1
1
3
30. — – — = -w + 1 2w 40
x
x
1. — + 1 = -2
4
x
x
2. — + 2 = -3
6
x
x
3. — – 5 = — – 7
3
2
x x x
4. — – — = — – 11
3 2 5
y 2 y 1
5. — – — = — + -5 3 6 3
z 5 z 3
6. — + — = — – -6 4 2 4
t
3 t-4
7. — – — = -4
3
12
4 v-1 v-5
8. — – — = -5
10
30
x x + 1 x + 15
9. — + — = -3
4
12
x x + 4 6x + 5
10. — + — = -8
12
24
1
1 w + 10
w+1
11. — – — = — – -10
5
15
6
a-1
1
a+4
31. -+ — = -2
a -4 a-2 a+2
1
b + 17
b-2
32. — — = -b2 – 1 b + 1 b – 1
U2V Extraneous Solutions
q q – 1 13 q + 1
12. — – — = — – -5
2
20
4
Solve each equation. Watch for extraneous solutions.
See Examples 4 and 5.
Solve each equation. See Example 2.
1
x
2
33. — + — = -x-1 x x-1
1 1
13. — + — = 3
x 2
2 3
14. — + — = 5
x 4
1 2
15. — + — = 7
x x
5 6
16. — + — = 12
x x
1 1 3
17. — + — = -x 2 4
3 1 5
18. — + — = -x 4 8
2
1
7
19. — + — = -3x 2x 24
1
1
1
20. — – — = -6x 8x 72
1 a-2 a+2
21. — + — = -2
a
2a
1 1 b-1
3
22. — + — = — + -b 5
5b
10
4
3
x
1
34. — + — = — – -x x-3 x-3 3
5
2
x-1
35. — + — = -x+2 x-3 x-3
6
7
y-1
36. — + — = -y-2 y-8 y-8
3y
6
37. 1 + — = -y-2 y-2
5
y+7
38. — = — + 1
y – 3 2y – 6
1 k+3
1
k-1
23. — – — = — – -3
6k
3k
2k
z
1
2z + 5
39. — – — = -z + 1 z + 2 z2 + 3z + 2
3 p + 3 2p – 1 5
24. — – — = — – -6
p
3p
2p
1
z
7
40. — – — = z – 2 z + 5 z2 + 3z – 10
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6.7
Applications of Ratios and Proportions
Miscellaneous
Applications
Solve each equation.
Solve each problem.
a
4
w
43.
6
5
45.
x
x
47.
41.
49.
51.
53.
55.
56.
57.
58.
59.
60.
61.
62.
5
2
3w
11
x
5
3 x
3
5
x
1
x
x+2 x+2
1
1
3
+
2x 4 x 2 2
3
2
a2 a 6 a2 4
4
1
25
c 2 2 c c+6
3
1
10
x + 1 1 x x2 1
1
3
4
+
x2 9 x + 3 x 3
3
5
1
x 2 x + 3 x2 + x 6
3
1
1
2x + 4 x + 2 3x + 1
5
1
1
2m + 6 m + 1 m + 3
2t 1 3t 1
t
+
3t + 3 6t + 6 t + 1
4w 1 w 1 w 1
3w + 6
3
w+2
42.
44.
46.
48.
50.
52.
54.
y 6
3 5
2m 3m
3
2
3
x
x
3
a+4 a+4
2
a
3
w
w+2 w+2
7
1
4
3x 9 x 3 3
8
6
a2 + a 6 a2 9
U1V Ratios
U2V Proportions
63. Lens equation. The focal length f for a camera lens is
related to the object distance o and the image distance i by
the formula
1 1 1
+ .
f o i
See the accompanying figure. The image is in focus at distance i from the lens. For an object that is 600 mm from a
50-mm lens, use f 50 mm and o 600 mm to find i.
o
i
Figure for Exercise 63
64. Telephoto lens. Use the formula from Exercise 63 to find
the image distance i for an object that is 2,000,000 mm from
a 250-mm telephoto lens.
Photo for Exercise 64
6.7
In This Section
429
Applications of Ratios and Proportions
In this section, we will use the ideas of rational expressions in ratio and proportion
problems. We will solve proportions in the same way we solved equations in
Section 6.6.
U1V Ratios
In Chapter 1 we defined a rational number as the ratio of two integers. We will now
give a more general definition of ratio. If a and b are any real numbers (not just integers),
with b � 0, then the expression a is called the ratio of a and b or the ratio of a to b.
b
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Chapter 6 Rational Expressions
The ratio of a to b is also written as a: b. A ratio is a comparison of two numbers. Some
examples of ratios are
1
3 4.2
3.6
100
4
,
, —,
, and
.
1
4 2.1
5
1
2
Ratios are treated just like fractions. We can reduce ratios, and we can build them up.
We generally express ratios as ratios of integers. When possible, we will convert a
ratio into an equivalent ratio of integers in lowest terms.
E X A M P L E
1
Finding equivalent ratios
Find an equivalent ratio of integers in lowest terms for each ratio.
1
4
b) —
1
2
4.2
a)
2.1
c)
3.6
5
Solution
a) Because both the numerator and the denominator have one decimal place, we will
multiply the numerator and denominator by 10 to eliminate the decimals:
4.2
2.1
4.2(10)
2.1(10)
42
21
21 2
21 1
2
Do not omit the 1 in a ratio.
1
So the ratio of 4.2 to 2.1 is equivalent to the ratio 2 to 1.
b) This ratio is a complex fraction. We can simplify this expression using the LCD
method as shown in Section 6.5. Multiply the numerator and denominator of this
ratio by 4:
1
1
4
4
4
1
1
1
2
4
2
2
c) We can get a ratio of integers if we multiply the numerator and denominator by 10.
3.6
5
3.6(10)
5(10)
18
25
36
50
Reduce to lowest terms.
Now do Exercises 1–16
In Example 2, a ratio is used to compare quantities.
E X A M P L E
2
Nitrogen to potash
In a 50-pound bag of lawn fertilizer there are 8 pounds of nitrogen and 12 pounds of
potash. What is the ratio of nitrogen to potash?
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6.7
Applications of Ratios and Proportions
431
Solution
The nitrogen and potash occur in this fertilizer in the ratio of 8 pounds to 12 pounds:
8
12
2 4�
3 4�
2
3
So the ratio of nitrogen to potash is 2 to 3.
Now do Exercises 17–18
E X A M P L E
3
Males to females
In a class of 50 students, there were exactly 20 male students. What was the ratio of males
to females in this class?
Solution
Because there were 20 males in the class of 50, there were 30 females. The ratio of males
to females was 20 to 30, or 2 to 3.
Now do Exercises 19–20
Ratios give us a means of comparing the size of two quantities. For this reason the
numbers compared in a ratio should be expressed in the same units. For example, if
one dog is 24 inches high and another is 1 foot high, then the ratio of their heights is 2
to 1, not 24 to 1.
E X A M P L E
4
Quantities with different units
What is the ratio of length to width for a poster with a length of 30 inches and a width of
2 feet?
Solution
Because the width is 2 feet, or 24 inches, the ratio of length to width is 30 to 24. Reduce
as follows:
30 5 6 5
24 4 6 4
So the ratio of length to width is 5 to 4.
Now do Exercises 21–24
U2V Proportions
A proportion is any statement expressing the equality of two ratios. The statement
a c
or
a:b c:d
b d
is a proportion. In any proportion the numbers in the positions of a and d shown here
are called the extremes. The numbers in the positions of b and c as shown are called
the means. In the proportion
30 5 ,
24 4
the means are 24 and 5, and the extremes are 30 and 4.
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Chapter 6 Rational Expressions
If we multiply each side of the proportion
a c
b d
by the LCD, bd, we get
U Helpful Hint V
The extremes-means property or
cross-multiplying is nothing new.
You can accomplish the same thing
by multiplying each side of the equa
tion by the LCD.
a
bd
b
c
bd
d
a d
b c.
or
We can express this result by saying that the product of the extremes is equal to
the product of the means. We call this fact the extremes-means property or crossmultiplying.
Extremes-Means Property (Cross-Multiplying)
Suppose a, b, c, and d are real numbers with b � 0 and d � 0. If
a
b
c
, then
d
ad
bc.
We use the extremes-means property to solve proportions.
E X A M P L E
5
Using the extremes-means property
Solve the proportion 3
x
5
x+5
for x.
Solution
Instead of multiplying each side by the LCD, we use the extremes-means property:
3
x
3(x + 5)
5
x+5
5x
3x + 15
5x
15
2x
15
2
x
Original proportion
Extremes-means property
Distributive property
Check:
3
15
2
3
2
15
2
5
5
5
2
2
5
15
25 5
25
+5
2
2
So 15 is the solution to the equation or the solution to the proportion.
2
Now do Exercises 25–38
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E X A M P L E
6.7
6
Applications of Ratios and Proportions
433
Solving a proportion
The ratio of men to women at Brighton City College is 2 to 3. If there are 894 men, then
how many women are there?
Solution
Because the ratio of men to women is 2 to 3, we have
Number of men
Number of women
2
.
3
If x represents the number of women, then we have the following proportion:
894
x
2
3
2x
2682 Extremes-means property
x
1341
The number of women is 1341.
Now do Exercises 39–42
Note that any proportion can be solved by multiplying each side by the LCD
as we did when we solved other equations involving rational expressions. The
extremes-means property gives us a shortcut for solving proportions.
E X A M P L E
7
Solving a proportion
In a conservative portfolio the ratio of the amount invested in bonds to the amount invested
in stocks should be 3 to 1. A conservative investor invested $2850 more in bonds than she
did in stocks. How much did she invest in each category?
Solution
Because the ratio of the amount invested in bonds to the amount invested in stocks is 3 to 1,
we have
Amount invested in bonds
Amount invested in stocks
3
.
1
If x represents the amount invested in stocks and x + 2850 represents the amount invested
in bonds, then we can write and solve the following proportion:
x + 2850
x
3
1
3x
x + 2850 Extremes-means property
2x
2850
x
1425
x + 2850
4275
So she invested $4275 in bonds and $1425 in stocks. As a check, note that these amounts
are in the ratio of 3 to 1.
Now do Exercises 43–46
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Chapter 6 Rational Expressions
Example 8 shows how conversions from one unit of measurement to another can
be done by using proportions.
E X A M P L E
8
Converting measurements
There are 3 feet in 1 yard. How many feet are there in 12 yards?
Solution
Let x represent the number of feet in 12 yards. There are two proportions that we can
write to solve the problem:
3 feet
x feet
1 yard
12 yards
3 feet
1 yard
x feet
12 yards
The ratios in the second proportion violate the rule of comparing only measurements that
are expressed in the same units. Note that each side of the second proportion is actually
the ratio 1 to 1, since 3 feet 1 yard and x feet 12 yards. For doing conversions we can
use ratios like this to compare measurements in different units. Applying the extremesmeans property to either proportion gives
3 12
x 1,
or
x
36.
So there are 36 feet in 12 yards.
Now do Exercises 47–50
Warm-Ups
▼
Fill in the blank.
1. A
is a comparison of two numbers.
2. A
is an equation that expresses the equality
of two ratios.
a c
3. In
, b and c are the
.
b d
a c
4. In
, a and d are the
.
b d
a c
5. The
property says that if
, then
b d
ad bc.
True or false?
6. The ratio of 40 men to 30 women can be expressed as
the ratio 4 to 3.
7. The ratio of 3 feet to 2 yards can be expressed as the
ratio 3 to 2.
8. The ratio of 1.5 to 2 is equivalent to the ratio 3 to 4.
9. The product of the extremes is equal to the product of
the means.
2 3
10. If
, then 5x 6.
x 5
11. If 4 of the 12 members of the supreme council are
women, then the ratio of men to women is 1 to 3.
Exercises
U Study Tips V
• Get an early start studying for your final exams.
• If you have several final exams, it can be difficult to find the time to prepare for all of them in the last couple of days.
U1V Ratios
For each ratio, find an equivalent ratio of integers in lowest
terms. See Example 1.
4
1.
6
10
2.
20
200
3.
150
1000
4.
200
2.5
5.
3.5
4.8
6.
1.2
0.32
7.
0.6
0.05
8.
0.8
35
9.
10
3
88
4.5
10.
11.
12.
33
7
2.5
1
2
2
3
5
13.
14.
15.
1
3
1
5
4
3
4
16.
1
4
Find a ratio for each of the following, and write it as a ratio of
integers in lowest terms. See Examples 2–4.
17. Men and women. Find the ratio of men to women in a
bowling league containing 12 men and 8 women.
19. Smokers. A life insurance company found that among
its last 200 claims, there were six dozen smokers. What
is the ratio of smokers to nonsmokers in this group of
claimants?
20. Hits and misses. A woman threw 60 darts and hit the
target a dozen times. What is her ratio of hits to misses?
21. Violence and kindness. While watching television for one
week, a consumer group counted 1240 acts of violence and
40 acts of kindness. What is the violence to kindness ratio
for television, according to this group?
22. Length to width. What is the ratio of length to width for
the rectangle shown?
L
W
2.5 ft
48 in.
Figure for Exercise 22
23. Rise to run. What is the ratio of rise to run for the stairway shown in the figure?
18. Coffee drinkers. Among 100 coffee drinkers, 36 said
that they preferred their coffee black and the rest did
not prefer their coffee black. Find the ratio of those
who prefer black coffee to those who prefer nonblack
coffee.
Rise
Run
8 in.
1 ft
Figure for Exercise 23
24. Rise and run. If the rise is 3 and the run is 5, then what is
2
the ratio of the rise to the run?
U2V Proportions
Solve each proportion. See Example 5.
Photo for Exercise 18
4 2
25.
x 3
9 3
26.
x 2
6.7
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dug84356_ch06b.qxd
436
27.
a
2
Page 436
1
5
28.
3
x
30.
5
9
x
2
x
7
5
33.
10
x
35.
a
a+1
m
m
37.
12:45 PM
6-56
Chapter 6 Rational Expressions
29.
31.
9/14/10
43. Basketball blowout. As the final buzzer signaled the end
of the basketball game, the Lions were 34 points ahead of
the Tigers. If the Lions scored 5 points for every 3 scored
by the Tigers, then what was the final score?
3
4
3
4
5
x
4
32.
x+1
x
3
a+3
a
36.
c+3
c 1
m 3
m+4
38.
c+2
c 3
h
h
44. The golden ratio. The ancient Greeks thought that the
most pleasing shape for a rectangle was one for which the
ratio of the length to the width was approximately 8 to 5,
the golden ratio. If the length of a rectangular painting is
2 ft longer than its width, then for what dimensions would
the length and width have the golden ratio?
2
x
x+1
2
34.
34
x + 12
1
2
b
3
h
3
h
9
Use a proportion to solve each problem. See Examples 6–8.
39. New shows and reruns. The ratio of new shows to reruns
on cable TV is 2 to 27. If Frank counted only eight new
shows one evening, then how many reruns were there?
45. Automobile sales. The ratio of sports cars to luxury cars
sold in Wentworth one month was 3 to 2. If 20 more sports
cars were sold than luxury cars, then how many of each
were sold that month?
46. Foxes and rabbits. The ratio of foxes to rabbits in the
Deerfield Forest Preserve is 2 to 9. If there are 35 fewer
foxes than rabbits, then how many of each are there?
47. Inches and feet. If there are 12 inches in 1 foot, then how
many inches are there in 7 feet?
40. Fast food. If four out of five doctors prefer fast food,
then at a convention of 445 doctors, how many prefer fast
food?
48. Feet and yards. If there are 3 feet in 1 yard, then how
41. Voting. If 220 out of 500 voters surveyed said that they
would vote for the incumbent, then how many votes could
the incumbent expect out of the 400,000 voters in the
state?
49. Minutes and hours. If there are 60 minutes in 1 hour, then
how many minutes are there in 0.25 hour?
many yards are there in 28 feet?
50. Meters and kilometers. If there are 1000 meters in
1 kilometer, then how many meters are there in
2.33 kilometers?
51. Miles and hours. If Alonzo travels 230 miles in 3 hours,
then how many miles does he travel in 7 hours?
52. Hiking time. If Evangelica can hike 19 miles in 2 days on
the Appalachian Trail, then how many days will it take her
to hike 63 miles?
Photo for Exercise 41
42. New product. A taste test with 200 randomly selected
people found that only three of them said that they would
buy a box of new Sweet Wheats cereal. How many boxes
could the manufacturer expect to sell in a country of
280 million people?
53. Force on basketball shoes. The force exerted on shoe soles
in a jump shot is proportional to the weight of the person
jumping. If a 70-pound boy exerts a force of 980 pounds on
his shoe soles when he returns to the court after a jump,
then what force does a 6 ft 8 in. professional ball player
weighing 280 pounds exert on the soles of his shoes when
he returns to the court after a jump? Use the accompanying
graph to estimate the force for a 150-pound player.
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12:45 PM
Page 437
6-57
6.7
Applications of Ratios and Proportions
437
Force (thousands of pounds)
WASTE GENERATION AT A FAST-FOOD RESTAURANT
5
34% Corrugated shipping boxes
4
8%
3%
4%
7%
4%
6%
3
2
1
Liquids, office paper, misc.
Plastic wraps, syrup containers
Uncoated paper (napkins)
Coated paper (sandwich wrap)
Polystyrene (hot cups, lids, etc.)
Customer’s waste (Diapers, etc.)
34% Food waste
0
50 100 150 200 250 300
Weight (pounds)
Figure for Exercise 53
54. Force on running shoes. The ratio of the force on the
shoe soles to the weight of a runner is 3 to 1. What force
does a 130-pound jogger exert on the soles of her shoes?
55. Capture-recapture. To estimate the number of trout in
Trout Lake, rangers used the capture-recapture method.
They caught, tagged, and released 200 trout. One week
later, they caught a sample of 150 trout and found that
5 of them were tagged. Assuming that the ratio of tagged
trout to the total number of trout in the lake is the same
as the ratio of tagged trout in the sample to the number
of trout in the sample, find the number of trout in
the lake.
56. Bear population. To estimate the size of the bear
population on the Keweenaw Peninsula, conservationists
captured, tagged, and released 50 bears. One year later, a
random sample of 100 bears included only 2 tagged bears.
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