6: More Fractional Equations and Problem Solving
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7.6 • More Fractional Equations and Problem Solving
Classroom Example
9
1
Solve x ϩ
ϭ .
2x
4
EXAMPLE 3
Solve n ϩ
299
10
1
ϭ .
n
3
Solution
nϩ
10
1
ϭ ,
n
3
n϶ 0
10
1
3n a n ϩ b ϭ 3n a b
n
3
3n2 ϩ 3 ϭ 10n
3n Ϫ 10n ϩ 3 ϭ 0
13n Ϫ 121n Ϫ 32 ϭ 0
3n Ϫ 1 ϭ 0
or
nϪ3ϭ0
3n ϭ 1
or
nϭ3
1
n ϭ or
nϭ3
3
2
Remember when we used the factoring
techniques to help solve equations of this
type in Chapter 6?
1
The solution set is e , 3 f .
3
Problem Solving
2
3
and are called multiplicative inverses, or reciprocals, of each other because
3
2
1
their product is 1. In general, the reciprocal of any nonzero real number n is the number .
n
Let’s use this idea to solve a problem.
Recall that
Classroom Example
The sum of a number and its
17
reciprocal is . Find the number.
4
EXAMPLE 4
The sum of a number and its reciprocal is
26
. Find the number.
5
Solution
We let n represent the number. Then
1
represents its reciprocal.
n
Number
ϩ
Its reciprocal
ϭ
26
5
n
ϩ
1
n
ϭ
26
,
5
n϶ 0
1
26
5n a n ϩ b ϭ 5n a b
n
5
Multiply both sides by 5n, the LCD
5n2 ϩ 5 ϭ 26n
5n2 Ϫ 26n ϩ 5 ϭ 0
15n Ϫ 121n Ϫ 52 ϭ 0
5n Ϫ 1 ϭ 0
or
5n ϭ 1 or
1
n ϭ or
5
nϪ5ϭ0
nϭ5
nϭ5
1
1
1
If the number is , its reciprocal is ϭ 5. If the number is 5, its reciprocal is .
5
1
5
5
300
Chapter 7 • Algebraic Fractions
Now let’s consider another uniform motion problem, which is a slight variation of those
we studied in the previous section. Again, keep in mind that we always use the distance–rate–
time relationships in these problems.
Classroom Example
To travel 280 miles, it takes Gary
one hour less than it takes Wayne
to travel 250 miles. Gary travels
20 miles per hour faster than Wayne.
Find the times and rates of both
travelers.
EXAMPLE 5
To travel 60 miles, it takes Sue, riding a moped, 2 hours less than it takes LeAnn, riding a bicycle, to travel 50 miles (see Figure 7.2). Sue travels 10 miles per hour faster than LeAnn. Find
the times and rates of both girls.
Sue
M
O
PE
D
LeAnn
50 miles
60 miles
Figure 7.2
Solution
We let t represent LeAnn’s time. Then t Ϫ 2 represents Sue’s time. We can record the information from Example 5 in the table.
Distance
Time
LeAnn
50
t
Sue
60
tϪ2
Rate a r ϭ
d
b
t
50
t
60
tϪ2
We use the fact that Sue travels 10 miles per hour faster than LeAnn as a guideline to set up
an equation.
Sue’s Rate
ϭ LeAnn’s Rate ϩ 10
60
tϪ2
ϭ
50
ϩ 10,
t
t ϶ 2 and t ϶ 0
Solving this equation yields
60
50
ϭ
ϩ 10
tϪ2
t
t1t Ϫ 22 a
50
60
b ϭ t1t Ϫ 22 a
ϩ 10b
tϪ2
t
60t ϭ 501t Ϫ 22 ϩ 10t1t Ϫ 22
60t ϭ 50t Ϫ 100 ϩ 10t2 Ϫ 20t
0 ϭ 10t2 Ϫ 30t Ϫ 100
0 ϭ t2 Ϫ 3t Ϫ 10
0 ϭ 1t Ϫ 521t ϩ 22
t Ϫ 5 ϭ 0 or t ϩ 2 ϭ 0
t ϭ 5 or
t ϭ Ϫ2
Copyright 2009 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.
7.6 • More Fractional Equations and Problem Solving
301
We must disregard the negative solution, so LeAnn’s time is 5 hours, and Sue’s time is 5 Ϫ 2 ϭ
50
60
3 hours. LeAnn’s rate is
ϭ 10 miles per hour, and Sue’s rate is
ϭ 20 miles per hour.
5
3
(Be sure that all of these results check back into the original problem!)
There is another class of problems that we commonly refer to as work problems, or
sometimes as rate-time problems. For example, if a certain machine produces 120 items
120
in 10 minutes, then we say that it is producing at a rate of
ϭ 12 items per minute.
10
Likewise, if a person can do a certain job in 5 hours, then that person is working at a rate of
1
of the job per hour. In general, if Q is the quantity of something done in t units of time, then
5
Q
the rate r is given by r ϭ . The rate is stated in terms of so much quantity per unit of time.
t
The uniform-motion problems we discussed earlier are a special kind of rate-time problem in
which the quantity is distance. The use of tables to organize information, as we illustrated
with the uniform-motion problems, is a convenient aid for some rate-time problems. Let’s
consider some examples.
Classroom Example
Printing press A can produce
30 posters per minute, and press B
can produce 20 posters per minute.
Printing press A is set up and starts
a job, and then 20 minutes later
printing press B is started, and both
presses continue printing until
2850 posters are produced. How long
would printing press B be used?
EXAMPLE 6
Printing press A can produce 35 fliers per minute, and press B can produce 50 fliers per
minute. Printing press A is set up and starts a job, and then 15 minutes later printing press B
is started, and both presses continue printing until 2225 fliers are produced. How long would
printing press B be used?
Solution
We let m represent the number of minutes that printing press B is used. Then m ϩ 15
represents the number of minutes that press A is used. The information in the problem can be
organized in a table.
Press A
Press B
Rate
Time
Quantity ؍Rate ؋ Time
35
50
m ϩ 15
m
35(m ϩ 15)
50m
Since the total quantity (total number of fliers) is 2225 fliers, we can set up and solve the following equation:
351m ϩ 152 ϩ 50m ϭ 2225
35m ϩ 525 ϩ 50m ϭ 2225
85m ϭ 1700
m ϭ 20
Therefore, printing press B must be used for 20 minutes.
Classroom Example
Sandy can shovel the walk in
50 minutes, and Ashley can shovel
the same walk in 75 minutes. How
long would it take the two of them
working together to shovel the walk?
EXAMPLE 7
Bill can mow a lawn in 45 minutes, and Jennifer can mow the same lawn in 30 minutes. How
long would it take the two of them working together to mow the lawn? (See Figure 7.3.)
302
Chapter 7 • Algebraic Fractions
Figure 7.3
Remark: Before you look at the solution of this problem, estimate the answer. Remember
that Jennifer can mow the lawn by herself in 30 minutes.
Solution
1
1
of the lawn per minute, and Jennifer’s rate is
of the lawn per minute.
45
30
1
If we let m represent the number of minutes that they work together, then represents the rate
m
when working together. Therefore, since the sum of the individual rates must equal the rate
working together, we can set up and solve the following equation:
Bill’s rate is
1
1
1
ϩ
ϭ ,
m϶ 0
m
30
45
1
1
1
90m a
ϩ b ϭ 90m a b
m
30
45
3m ϩ 2m ϭ 90
5m ϭ 90
m ϭ 18
Multiply both sides by 90m, the LCD
It should take them 18 minutes to mow the lawn when working together. (How close was your
estimate?)
Classroom Example
It takes Jake three times as long to
mow the lawn as it does Zack. How
long would it take each boy by
himself if they can mow the lawn
together in 36 minutes?
EXAMPLE 8
It takes Amy twice as long to deliver papers as it does Nancy. How long would it take each
girl by herself if they can deliver the papers together in 40 minutes?
Solution
We let m represent the number of minutes that it takes Nancy by herself. Then 2m represents
1
1
Amy’s time by herself. Therefore, Nancy’s rate is , and Amy’s rate is
. Since the combined
m
2m
1
rate is , we can set up and solve the following equation:
40
Nancy’s
Amy’s Combined
rate ϩ rate ϭ rate
1
1
1
ϩ
ϭ ,
m
2m
40
40m a
m϶ 0
1
1
1
ϩ
b ϭ 40m a b
m
2m
40
7.6 • More Fractional Equations and Problem Solving
303
40 ϩ 20 ϭ m
60 ϭ m
Therefore, Nancy can deliver the papers by herself in 60 minutes, and Amy can deliver them
by herself in 2(60) ϭ 120 minutes.
One final example of this section outlines another approach that some people find meaningful for work problems. This approach represents the fractional parts of a job. For example,
if a person can do a certain job in 7 hours, then at the end of 3 hours, that person has finished
3
5
of the job. (Again, we assume a constant rate of work.) At the end of 5 hours, of the
7
7
h
job has been done—in general, at the end of h hours, of the job has been completed. Let’s
7
use this idea to solve a work problem.
Classroom Example
It takes Chris 7 hours to install a
wood railing. After working for
2 hours he is joined by Carlos, and
together they finish the railing in
3 hours. How long would it take
Carlos to install the railing by
himself?
EXAMPLE 9
It takes Pat 12 hours to install a wood floor. After he had been working for 3 hours, he was
joined by his brother Mike, and together they finished the floor in 5 hours. How long would
it take Mike to install the floor by himself?
Solution
Let h represent the number of hours that it would take Mike to install the floor by himself.
The fractional part of the job that Pat does equals his working rate times his time. Because
1
it takes Pat 12 hours to do the entire floor, his working rate is . He works for 8 hours (3 hours
12
1
8
before Mike and then 5 hours with Mike). Therefore, Pat’s part of the job is
(8) ϭ .
12
12
The fractional part of the job that Mike does equals his working rate times his time. Because h
1
represents Mike’s time to install the floor, his working rate is . He works for 5 hours.
h
5
1
Therefore, Mike’s part of the job is (5) ϭ . Adding the two fractional parts together results
h
h
in 1 entire job being done. Let’s also show this information in chart form and set up our guideline. Then we can set up and solve the equation.
Time to do
entire job
Pat
12
Mike
h
Fractional part of
the job that Pat does
Working
rate
Time
working
1
12
1
h
Fractional part of
the job that Mike does
8
5
ϩ ϭ1
12
h
8
5
12h a
ϩ b ϭ 12h112
12
h
8
5
12h a b ϩ 12h a b ϭ 12h
12
h
8h ϩ 60 ϭ 12h
8
5
Fractional part
of the job done
8
12
5
h
304
Chapter 7 • Algebraic Fractions
60 ϭ 4h
15 ϭ h
It would take Mike 15 hours to install the floor by himself.
We emphasize a point made earlier. Don’t become discouraged if solving word problems
is still giving you trouble. The development of problem-solving skills is a long-term objective. If you continue to work hard and give it your best shot, you will gradually become more
and more confident in your approach to solving problems. Don’t be afraid to try some different approaches on your own. Our problem-solving suggestions simply provide a framework
for you to build on.
Concept Quiz 7.6
For Problems 1–10, answer true or false.
1. Assuming uniform motion, the rate at which a car travels is equal to the time traveled
divided by the distance traveled.
2. If a worker can lay 640 square feet of tile in 8 hours, we can say his rate of work is
80 square feet per hour.
5
3. If a person can complete 2 jobs in 5 hours, then the person is working at the rate of
2
of the job per hour.
4. In a time-rate problem involving two workers, the sum of their individual rates must
equal the rate working together.
2
5. If a person works at the rate of
of the job per hour, then at the end of 3 hours the
15
6
job would be
completed.
15
1
6. The solution set for x ϩ ϭ 4 is {2, 4}.
x
1
ϩ
xϩ2
x
2
8. The solution set for
Ϫ
xϪ1
x
7. The solution set for
1
5 Ϫ 2x
3
is e f .
ϭ 2
Ϫ3
2
x ϪxϪ6
3
Ϫx
is л.
ϭ 2
ϩ1
x Ϫ1
9. If Kim can do a certain job in 5 hours, then at the end of h hours she will have
5
completed of the job.
h
x
4
1
10. The solution set for
ϩ 2
ϭ is 5Ϫ86.
3x Ϫ 6
3
x Ϫ4
Problem Set 7.6
For Problems 1–32, solve each equation. (Objective 1)
1.
4
7
2
1
ϩ ϭ ϩ
x
x
6
3x
2.
2
25
9
Ϫ ϭϪ
x
3x
9
5.
5
3
Ϫ
ϭ1
2n Ϫ 10
nϪ5
6.
7
2
Ϫ
ϭ2
3x ϩ 6
xϩ2
3.
3
4
11
ϩ
ϭ
2x ϩ 2
xϩ1
12
7.
3
5
7
Ϫ ϭ ϩ1
2t
t
5t
4.
1
7
5
ϩ
ϭ
2x Ϫ 6
xϪ3
2
9.
x
4
ϩ
ϭ1
xϪ2
xϩ2
8.
2
3
5
ϩ ϭ1Ϫ
3t
4t
2t
7.6 • More Fractional Equations and Problem Solving
10.
2x
3
Ϫ
ϭ2
xϩ1
xϪ1
11.
x
2x
Ϫ
ϭ Ϫ1
xϪ4
xϩ4
12.
2x
x
ϩ
ϭ3
xϩ2
xϪ2
13.
3n
n
Ϫ
ϭ2
nϩ3
nϪ3
14.
4n
2n
Ϫ
ϭ2
nϪ5
nϩ5
15.
3
5
2
ϩ
ϭ
tϩ2
tϪ2
t Ϫ4
16.
t
16
1
ϩ 2
ϭ
2t Ϫ 8
2
t Ϫ 16
32. 3 ϩ
34. The sum of a number and three times its reciprocal is 4.
Find the number.
35. A number is
37. Suppose that Celia rides her bicycle 60 miles in
2 hours less time than it takes Tom to ride his bicycle
85 miles. If Celia rides 3 miles per hour faster than Tom,
find their respective rates.
3x Ϫ 1
4
5
ϩ
ϭ
2
xϩ3
xϪ3
x Ϫ9
5
3
ϭ
yϩ2
y2 ϩ 2y
20. 2 ϩ
4
4
ϭ 2
yϪ1
y Ϫy
21. n ϩ
1
17
ϭ
n
4
23.
15
15
ϩ
ϭ1
4n
41n ϩ 42
24.
10
10
ϩ
ϭ1
7x
71x ϩ 32
25. x Ϫ
26.
21
larger than its reciprocal. Find the number.
10
36. Suppose that the reciprocal of a number subtracted from
5
the number yields . Find the number.
6
2
19. 8 ϩ
6
6
ϭ 2
tϪ3
t Ϫ 3t
For Problems 33–50, set up an equation and solve the
problem. (Objective 2)
9
33. The sum of a number and twice its reciprocal is .
2
Find the number.
4
2x Ϫ 3
6
17.
Ϫ 2
ϭ
xϪ1
xϩ1
x Ϫ1
18.
22. n ϩ
38. To travel 300 miles, it takes a freight train 2 hours
longer than it takes an express train to travel 280 miles.
The rate of the express train is 20 miles per hour faster
than the rate of the freight train. Find the rates of both
trains.
3
ϭ4
n
39. One day, Jeff rides his bicycle out into the country
40 miles (see Figure 7.4). On the way back, he takes a
different route that is 2 miles longer, and it takes him an
hour longer to return. If his rate on the way out to the
country is 4 miles per hour faster than his rate back, find
both rates.
5x
Ϫ10
ϭ
xϪ2
xϪ2
xϩ1
3
12
Ϫ ϭ 2
x
xϪ3
x Ϫ 3x
40
t
5
1
27.
ϩ 2
ϭ
4t Ϫ 4
4
t Ϫ1
28.
305
42
x
4
1
ϩ 2
ϭ
3x Ϫ 6
3
x Ϫ4
N
E
W
3
4
2n ϩ 11
29.
ϩ
ϭ 2
nϪ5
nϩ7
n ϩ 2n Ϫ 35
30.
2
3
2n Ϫ 1
ϩ
ϭ 2
nϩ3
nϪ4
n Ϫ n Ϫ 12
31.
a
3
14
ϩ
ϭ 2
aϩ2
aϩ4
a ϩ 6a ϩ 8
S
Figure 7.4
Copyright 2009 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.
306
Chapter 7 • Algebraic Fractions
40. Rita jogs for 8 miles and then walks an additional
12 miles. She jogs at a rate twice her walking rate, and
she covers the entire distance of 20 miles in 4 hours.
Find the rate she jogs and the rate she walks.
41. A water tank can be filled by an inlet pipe in 5 minutes.
A drain pipe will empty the tank in 6 minutes. If
by mistake the drain is left open as the tank is
being filled, how long will it take before the tank
overflows?
42. Betty can do a job in 10 minutes. Doug can do the same
job in 15 minutes. If they work together, how long will
it take them to complete the job?
43. It takes Barry twice as long to deliver papers as it does
Mike. How long would it take each if they can deliver
the papers together in 40 minutes?
44. Working together, Cindy and Sharon can address
envelopes in 12 minutes. Cindy could do the addressing
by herself in 20 minutes. How long would it take
Sharon to address the envelopes by herself?
45. Mark can overhaul an engine in 20 hours, and Phil can
do the same job by himself in 30 hours. If they both
work together for a time, and then Mark finishes the job
by himself in 5 hours, how long did they work
together?
46. Working together, Pam and Laura can complete a
1
job in 1 hours. When working alone, it takes Laura
2
4 hours longer than Pam to do the job. How long does it
take each of them working alone?
47. A copy center has two copiers. Copier A can produce
copies at a rate of 40 pages per minute, and copier B
does 30 pages per minute. How long will copier B need
to run if copier A has been copying by itself for 6 minutes, and then both copier A and B are used until 520
copies are made?
48. It takes two pipes 3 hours to fill a water tank. Pipe B
can fill the tank alone in 8 hours more than it takes pipe
A to fill the tank alone. How long would it take each
pipe to fill the tank by itself?
49. In a survivor competition, the Pachena tribe can shuck
300 oysters in 10 minutes less time than it takes the
Tchaika tribe. If the Pachena tribe shucks oysters at a
rate of 5 oysters per minute faster than the Tchaika
tribe, find the rate of each tribe.
50. Machine A can wrap 600 pieces of candy in 5 minutes
less time than it takes machine B to wrap 600 pieces
of candy. If the rate of machine A is 20 candies
per minute faster than machine B, find the rate of each
machine.
Additional word problems can be found in Appendix B.
All of the problems in the Appendix marked as (7.5)
or (7.6) are appropriate for your practice.
Thoughts Into Words
51. Write a paragraph or two summarizing the new ideas
about problem solving that you have acquired thus far
in this course.
Further Investigations
For Problems 52–54, solve each equation.
52.
3x Ϫ 1
4
7
ϩ
ϭ
2
xϩ3
xϪ3
x Ϫ9
53.
xϪ2
3
Ϫ5
ϩ
ϭ
xϩ1
xϪ1
x2 Ϫ 1
Answers to the Concept Quiz
1. False
2. True
3. False
9. False
10. True
54.
4. True
5. True
7x Ϫ 12
5
2
Ϫ
ϭ
2
x
ϩ
4
x
Ϫ
4
x Ϫ 16
6. False
7. True
8. True
Chapter 7 Summary
OBJECTIVE
SUMMARY
Simplify rational expressions
using factoring techniques.
The fundamental principle of fractions
ak
a
a ϭ b provides the basis for simplifybk
b
ing rational expressions. For many problems, the numerator and denominator will
have to be factored before you can apply
the fundamental principle of fractions.
(Section 7.1/Objective 1)
Multiply rational expressions.
(Section 7.2/Objective 1)
To multiply algebraic fractions, multiply
the numerators, multiply the denominators,
and express the product in simplified form.
EXAMPLE
Simplify
3x2 Ϫ 15x
.
x Ϫ 3x Ϫ 10
2
Solution
3x1x Ϫ 52
3x2 Ϫ 15x
ϭ
1x ϩ 221x Ϫ 52
x2 Ϫ 3x Ϫ 10
3x
ϭ
xϩ2
Multiply
y
y ϩ 4y ϩ 3
2
Solution
y
y ϩ 4y ϩ 3
2
Divide rational expressions.
(Section 7.2/Objective 2)
To divide algebraic fractions, invert the
divisor and multiply.
и
3y ϩ 3
.
8
3y ϩ 3
8
ϭ
y
(y ϩ 1)(y ϩ 3)
ϭ
3y
81y ϩ 32
Divide
и
и
3(y ϩ 1)
8
a2 Ϫ 36
6a ϩ 6
Ϭ 2
.
2
a ϩ 8a ϩ 12
a ϩ 3a ϩ 2
Solution
a2 Ϫ 36
6a ϩ 6
Ϭ 2
2
a ϩ 8a ϩ 12
a ϩ 3a ϩ 2
ϭ
ϭ
ϭ
Combine rational expressions
with common denominators.
(Section 7.3/Objective 1)
Addition and subtraction of algebraic
fractions are based on the following
definitions:
a
c
aϩc
ϩ ϭ
Addition
b
b
b
a
c
aϪc
Ϫ ϭ
Subtraction
b
b
b
The final answer should always be in
simplified form.
a2 Ϫ 36
a2 ϩ 8a ϩ 12
и
1a ϩ 621a Ϫ 62
1a ϩ 621a ϩ 22
a2 ϩ 3a ϩ 2
6a ϩ 6
и
1a ϩ 221a ϩ 12
61a ϩ 12
aϪ6
6
Subtract
8n ϩ 1
5n Ϫ 8
Ϫ
.
6
6
Solution
8n ϩ 1 5n Ϫ 8 8n ϩ 1 Ϫ (5n Ϫ 8)
Ϫ
ϭ
6
6
6
8n ϩ 1 Ϫ 5n ϩ 8
ϭ
6
3n ϩ 9
ϭ
6
31n ϩ 32
nϩ3
ϭ
ϭ
6
2
(continued)
307
308
Chapter 7 • Algebraic Fractions
OBJECTIVE
SUMMARY
EXAMPLE
Add and subtract rational
expressions with different
denominators.
Use the following procedure when adding
and subtracting fractions.
Subtract
(Section 7.3/Objectives 2 and
3; Section 7.4/Objective 1)
Simplify complex fractions.
(Section 7.4/Objective 2)
x
1
Ϫ
.
xϪ4
x Ϫ 16
2
1. Find the least common denominator.
2. Change each fraction to an equivalent
fraction that has the LCD as its denominator.
3. Add or subtract the numerators, and
place this result over the LCD.
4. Look for possibilities to simplify the
final fraction.
Solution
Fractional forms that contain fractions in
the numerator and/or the denominator are
complex fractions. To simplify a complex
fraction means to express it as a single
fraction. One method for simplifying is to
multiply the entire complex fraction by a
form of 1. Another method is to simplify
the numerator, simplify the denominator,
and then proceed as with a division of
fractions problem.
3
2
ϩ
x
y
Simplify
.
1
1
ϩ
x2
y2
x
1
Ϫ
xϪ4
x Ϫ 16
x
1
ϭ
Ϫ
1x ϩ 421x Ϫ 42
xϪ4
11x ϩ 42
x
ϭ
Ϫ
1x ϩ 421x Ϫ 42
1x Ϫ 421x ϩ 42
x Ϫ 11x ϩ 42
ϭ
1x ϩ 421x Ϫ 42
xϪxϪ4
Ϫ4
ϭ
ϭ
1x ϩ 421x Ϫ 42
1x ϩ 421x Ϫ 42
2
Solution
3
2
3
2
ϩ
ϩ
2 2
x
y
x y
x
y
ϭ 2 2±
≤
1
1
1
1
x y
ϩ
ϩ
x2
y2
x2
y2
3
2
x2y2 a b ϩ x2y2 a b
x
y
ϭ
1
1
x2y2 a 2 b ϩ x2y2 a 2 b
x
y
2
2
3xy ϩ 2x y
ϭ
y2 ϩ x2
Solve rational equations that
have constants in the
denominator.
(Section 7.5/Objective 1)
To solve rational equations that have
constants in the denominator, multiply
both sides of the equation by the LCD of
all the denominators in the equation.
Solve
xϩ3
xϪ4
21
ϩ
ϭ .
2
5
10
Solution
xϩ3
xϪ4
21
ϩ
ϭ
2
5
10
xϩ3
xϪ4
21
10 a
ϩ
b ϭ 10 a b
2
5
10
51x ϩ 32 ϩ 21x Ϫ 42 ϭ 21
5x ϩ 15 ϩ 2x Ϫ 8 ϭ 21
7x ϩ 7 ϭ 21
7x ϭ 14
xϭ2
The solution set is {2}.
(continued)
Chapter 7 • Summary
OBJECTIVE
SUMMARY
Solve rational equations
that have variables in the
denominator.
If an equation contains a variable in one or
more of the denominators, we must avoid
any value of the variable that makes the
denominator zero.
(Section 7.5/Objective 1;
Section 7.6/Objective 1)
EXAMPLE
Solve
3
5
4
ϩ
ϭ .
n
2n
3
Solution
First we need to realize that n can not
equal zero.
3
5
4
6n a ϩ b ϭ 6n a b
n
2n
3
3
5
4
6n a b ϩ 6n a b ϭ 6n a b
n
2n
3
18 ϩ 15 ϭ 8n
33 ϭ 8n
33
ϭn
8
The solution set is e
Solve rational equations
that are in the form of a
proportion.
(Section 7.5/Objective 1)
c
a
ϭ if
b
d
and only if ad ϭ bc where b ϶ 0
and d ϶ 0, can be applied to solve
proportions. This property of proportions
is often referred to as cross products
are equal.
The property of proportions,
309
Solve
33
f.
8
6
3
ϭ
.
xϩ1
xϪ2
Solution
3
6
ϭ
xϩ1
xϪ2
61x Ϫ 22 ϭ 31x ϩ 12
6x Ϫ 12 ϭ 3x ϩ 3
3x ϭ 15
xϭ5
The solution set is {5}.
Solve proportion word
problems.
(Section 7.5/Objective 2)
Some of the word problems in this chapter
translate into equations that are proportions.
The numerator of a fraction is 4 less than
the denominator. The fraction in its
9
simplest form is . Find the fraction.
10
Solution
Let x represent the denominator. Then
x Ϫ 4 represents the numerator.
xϪ4
9
ϭ
x
10
101x Ϫ 42 ϭ 9x
10x Ϫ 40 ϭ 9x
x ϭ 40
The fraction is
36
.
40
(continued)