7.4
Complex Fractions
7.4
OBJECTIVES
1. Use the fundamental principle to simplify a
complex fraction
2. Use division to simplify a complex fraction
Our work in this section deals with two methods for simplifying complex fractions. We
begin with a definition. A complex fraction is a fraction that has a fraction in its numerator or denominator (or both). Some examples are
5
6
3
4
NOTE Fundamental principle:
P
PR
Q QR
when Q 0 and R 0
NOTE Again, we are
multiplying by
10
or 1.
10
4
x
3
x1
1
x
1
1
x
1
and
Two methods can be used to simplify complex fractions. Method 1 involves the fundamental principle, and Method 2 involves inverting and multiplying.
Recall that by the fundamental principle we can always multiply the numerator and
denominator of a fraction by the same nonzero quantity. In simplifying a complex fraction,
we multiply the numerator and denominator by the LCD of all fractions that appear within
the complex fraction.
Here the denominators are 5 and 10, so we can write
3
3
10
5
5
6
7
7
7
10
10
10
Our second approach interprets the complex fraction as indicating division and applies our
earlier work in dividing fractions in which we invert and multiply.
3
5
3
7
3 10
6
7
5
10
5 7
7
10
Invert and multiply.
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Which method is better? The answer depends on the expression you are trying to simplify. Both approaches are effective, and you should be familiar with both. With practice
you will be able to tell which method may be easier to use in a particular situation.
Let’s look at the same two methods applied to the simplification of an algebraic complex
fraction.
Example 1
Simplifying Complex Fractions
Simplify.
2x
y
x
2
y
1
523
524
CHAPTER 7
RATIONAL EXPRESSIONS AND FUNCTIONS
2x
x
Method 1 The LCD of 1, , 2, and is y. So we multiply the numerator and denomiy
y
nator by y.
2x
2x
1
y
y
y
x
x
2
2
y
y
y
1
2x
y
y
x
2y y
y
Distribute y over the numerator and denominator.
1y
Simplify.
y 2x
2y x
Method 2 In this approach, we must first work separately in the numerator and denominator to form single fractions.
understand the steps in forming
a single fraction in the
numerator and denominator.
2x
y
2x
y 2x
y
y
y
y
x
2y
x
2y x
2
y
y
y
y
1
y 2x
y
y
2y x
y 2x
2y x
Invert the divisor and multiply.
CHECK YOURSELF 1
Simplify.
x
1
y
2x
2
y
Again, simplifying a complex fraction means writing an equivalent simple fraction in
lowest terms, as Example 2 illustrates.
Example 2
Simplifying Complex Fractions
Simplify.
2y
y2
2
x
x
y2
1 2
x
1
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NOTE Make sure you
COMPLEX FRACTIONS
SECTION 7.4
525
We choose the first method of simplification in this case. The LCD of all the fractions
that appear is x2. So we multiply the numerator and denominator by x2.
2y
y2
2y
y2
2
1
2 x2
x
x
x
x
2
2
y
y
1 2
1 2 x2
x
x
1
Distribute x2 over the
numerator and denominator,
and simplify.
x2 2xy y2
x2 y2
Factor the numerator and
denominator.
(x y)(x y)
xy
(x y)(x y)
xy
Divide by the common factor
x y.
CHECK YOURSELF 2
Simplify.
5
6
2
x
x
9
1 2
x
1
In Example 3, we will illustrate the second method of simplification for purposes of
comparison.
Example 3
Simplifying Complex Fractions
Simplify.
1
x2
2
x
x1
1
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NOTE Again, take time to
make sure you understand how
the numerator and
denominator are rewritten as
single fractions.
NOTE Method 2 is probably
the more efficient in this case.
The LCD of the denominators
would be (x 2)(x 1), leading
to a somewhat more
complicated process if method 1
were used.
1
x2
1
x
x2
x2
x2
x
2
2
x(x 1)
2
x x
x
x1
x1
x1
x
1
1
2
2
1
x1
x1
2
x2 x x2
x1
x1
x 2 (x 2)(x 1)
x1
(x 2)(x 2)
CHAPTER 7
RATIONAL EXPRESSIONS AND FUNCTIONS
CHECK YOURSELF 3
Simplify.
5
x3
1
x
2x 1
2
The following algorithm summarizes our work with complex fractions.
Step by Step: Simplifying Complex Fractions
Method 1
Step 1 Multiply the numerator and denominator of the complex fraction by
the LCD of all the fractions that appear within the numerator and
denominator.
Step 2 Simplify the resulting rational expression, writing the expression in
lowest terms.
Method 2
Step 1
Step 2
Write the numerator and denominator of the complex fraction as
single fractions, if necessary.
Invert the denominator and multiply as before, writing the result in
lowest terms.
CHECK YOURSELF ANSWERS
1.
xy
2x 2y
2.
x2
x3
3.
2x 1
(x 3)(x 1)
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526
Name
7.4
Exercises
Section
Date
In exercises 1 to 39, simplify each complex fraction.
2
3
1.
6
8
5
6
2.
10
15
ANSWERS
1.
2.
2
1
3
2
3.
3
1
4
3
3
1
4
2
4.
7
1
8
4
1
2
3
5.
1
3
5
3
1
4
6.
1
2
8
3.
4.
5.
6.
7.
8.
9.
x
8
7. 2
x
4
2
a
10
8. 3
a
15
10.
11.
12.
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3
m
9.
6
m2
15
x2
10.
20
x3
y1
y
11.
y1
2y
x3
4x
12.
x3
2x
a 2b
3a
13. 2
a 2ab
9b
m 3n
4m
14. 2
m 3mn
8n
13.
14.
527
ANSWERS
15.
16.
x2
x2 9
15. 2
x 4
x2 3x
x5
x2 6x
16. 2
x 25
x2 36
1
x
17.
1
2
x
1
b
18.
1
3
b
17.
2
18.
19.
3
20.
21.
22.
1
1
x
y
19.
1
xy
1
ab
20.
x2
1
y2
21.
x
1
y
m
2
n
22. 2
m
4
n2
3
4
2
a
a
23.
2
3
1 2
a
a
2
8
2
x
x
24.
1
6
1 2
x
x
x2
2x y
y
25.
1
1
2
y2
x
a
2b
1
b
a
26.
1
4
2
b2
a
1
x1
27.
1
1
x1
1
m2
28.
1
2
m2
1
y1
29.
8
y
y2
1
x2
30.
18
x
x3
1
1
a
b
23.
24.
25.
26.
1
27.
28.
1
29.
1
1
528
2
1
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30.
ANSWERS
1
x3
x
31.
1
x3
x
1
3
1
3
2
m2
m
32.
2
m2
m
1
3
1
3
31.
32.
33.
x
x1
x
33.
x
x1
x
1
1
1
1
y
y4
y
34.
4
y4
y
1
2
1
2
34.
35.
36.
a
a
35.
a
a
a1
1
1
a1
1
a1
1
a1
x
x
36.
x
x
x2
2
2
x2
2
x2
2
x2
37.
38.
39.
40.
37. 1 1
38. 1 1
1
x
(a)
1
1
1
y
(b)
41.
1
39. 1 1
1
1
1
x
42.
40. (a) Extend the “continued fraction” patterns in exercises 37 and 39 to write the next
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complex fraction. (b) Simplify the complex fraction obtained in (a).
41. Compare your results in exercises 37, 39, and 40. Could you have predicted the
result?
42. Outline the two different methods used to simplify a complex fraction. What are the
advantages of each method?
529
ANSWERS
43.
43. Can the expression
x1 y1
x2 y2
? If not, what is the correct
2
2 be written as
x y
xy
simplified form?
44.
44. Write and simplify a complex fraction that is the reciprocal of x 45.
6
.
x1
3
. Write and simplify a complex fraction whose numerator is
x
f (3 h) f (3) and whose denominator is h.
45. Let f(x) 46.
47.
46. Write and simplify a complex fraction that is the arithmetic mean of
47. Write and simplify a complex fraction that is the reciprocal of
1
1
.
and
x
x1
1
1
.
x
y
Suppose you drive at 40 mi/h from city A to city B. You then return along the same route
from city B to city A at 50 mi/h. What is your average rate for the round trip? Your obvious
guess would be 45 mi/h, but you are in for a surprise.
Suppose that the cities are 200 mi apart. Your time from city A to city B is the distance
divided by the rate, or
200 mi
5h
40 mi/h
Similarly, your time from city B to city A is
200 mi
4h
50 mi/h
The total time is then 9 h, and now using rate equals distance divided by time, we have
Note that the rate for the round trip is independent of the distance involved. For instance,
try the same computations above if cities A and B are 400 mi apart.
The answer to the problem above is the complex fraction
R
2
1
1
R1
R2
in which R1 rate going
R2 rate returning
R rate for round trip
530
© 2001 McGraw-Hill Companies
400 mi
400
4
mi/h 44 mi/h
9h
9
9
ANSWERS
Use this information to solve exercises 48 to 51.
4
9
48.
48. Verify that if R1 40 mi/h and R2 50 mi/h, then R 44 mi/h, by simplifying the
complex fraction after substituting those values.
49.
50.
49. Simplify the given complex fraction first. Then substitute 40 for R1 and 50 for R2 to
calculate R.
51.
50. Repeat exercise 48, with R1 50 mi/h and R2 60 mi/h.
52.
51. Use the procedure in exercise 49 with the above values for R1 and R2.
52. The following inequality is used when the U.S. House of Representatives seats are
apportioned (see the chapter opener for more information).
E
A
A
E
e
a1
a
e1
A
E
a1
e1
Show that this inequality can be simplified to
A
E
.
1a(a 1)
1e(e 1)
Here, A and E represent the populations of two states of the United States, and a and
e are the number of representatives each of these two states have in the U.S. House of
Representatives.
53. Mathematicians have shown that there are situations in which the method for
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apportionment described in the chapter’s introduction does not work, and a state may
not even get its basic quota of representatives. They give the table below of a
hypothetical seven states and their populations as an example.
State
Population
Exact Quota
Number of Reps.
A
B
C
D
E
F
G
Total
325
788
548
562
4263
3219
295
10,000
1.625
3.940
2.740
2.810
21.315
16.095
1.475
50
2
4
3
3
21
15
2
50
531
ANSWERS
In this case, the total population of all states is 10,000, and there are 50
representatives in all, so there should be no more than 10,00050 or 200 people
per representative. The quotas are found by dividing the population by 200. Whether
a state, A, should get an additional representative before another state, E, should
get one is decided in this method by using the simplified inequality below. If the ratio
53.
54.
A
E
1a(a 1)
1e(e 1)
is true, then A gets an extra representative before E does.
(a) If you go through the process of comparing the inequality above for each pair of
states, state F loses a representative to state G. Do you see how this happens? Will
state F complain?
(b) Alexander Hamilton, one of the signers of the Constitution, proposed that the
extra representative positions be given one at a time to states with the largest
remainder until all the “extra” positions were filled. How would this affect the
table? Do you agree or disagree?
54. In Italy in the 1500s, Pietro Antonio Cataldi expressed square roots as infinite,
continued fractions. It is not a difficult process to follow. For instance, if you want the
square root of 5, then let
x 1 15
Squaring both sides gives
(x 1)2 5
x2 2x 1 5
or
which can be written
x(x 2) 4
x
4
x2
One can continue replacing x with
4
:
x2
4
x
4
4
2
2
4
2 ...
to obtain
15 1
(a) Evaluate this complex fraction and then add 1 and see how close it is to the
square root of 5. What should you put where the ellipses (. . .) are? Try a number
you feel is close to 15. How far would you have to go to get the square root
correct to the nearest hundredth?
(b) Develop an infinite complex fraction for 110.
532
© 2001 McGraw-Hill Companies
2
ANSWERS
In exercises 55 and 56, use the table utility on your graphing calculator to complete the
table. Comment on the equivalence of the two expressions.
55.
56.
x
55.
3
2
1
0
1
2
3
2
x
4
1 2
x
x
x2
57.
1
x
56.
58.
3
2
1
0
1
2
3
20
x
25
4 2
x
4x
2x 5
8 57. Here is yet another method for simplifying a complex fraction. Suppose we want to
simplify
3
5
7
10
Multiply the numerator and denominator of the complex fraction by
10
.
7
© 2001 McGraw-Hill Companies
(a) What principle allows you to do this?
10
(b) Why was
chosen?
7
(c) When learning to divide fractions, you may have heard the saying “Yours is not to
reason why . . . just invert and multiply.” How does this method serve to explain
the “reason why” we invert and multiply?
1
2
58. Suppose someone wrote a fraction as follows . Give two ways that this fraction can
3
be interpreted, and simplify each. Do you see why it is important to clearly indicate
the “main fraction line”? On your graphing calculator, type 1/2/3 and press ENTER.
Which way is your calculator interpreting the fraction?
533
Answers
1
m
2(y 1)
9.
11.
2x
2
y1
x
2x 1
xy
21.
15.
17.
19. y x
(x 2)(x 3)
2x 1
y
x2y(x y)
x
y2
x
25.
27.
29.
31.
xy
x2
(y 1)(y 4)
3
2a
2x 1
3x 2
35. 2
37.
39.
41.
a 1
x1
2x 1
1.
8
9
3.
43. No;
14
5
5.
xy(x y)
y2 x2
5
6
45.
3b
a2
a4
23.
a3
7.
1
3h
47.
xy
xy
13.
33. 1
4
9
49. 44 mi/h
51. 54
6
mi/h
11
53.
55.
x
3
2
1
0
1
2
3
3
error
1
error
1
3
error
3
5
3
error
1
0
1
3
1
2
3
5
2
x
4
1 2
x
x
x2
1
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57.
534