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3.2 Equivalent Fractions: Simplifying and Building Two fractions are said to be equivalent if they have the same value. Naturally, one approach we could use to determine if two fractions are equivalent is to convert each fraction to a decimal. For
example, since 35= 0.6 and 15
25= 0.6 , the fractions 3
5 and 15
25 are equivalent, and we could
write 35=15
25. Alternatively, consider the following forms of the number 1:
1 =2
2=3
3=4
4= ... =
100
100=n
n
Clearly 2 parts out of 2 is equal to 1, as is 100 parts out of 100, or n parts out of n. Now consider the following property (the Fundamental Property of Fractions):
If a, b, and c are nonzero: ab=a • c
b • c
This statement is saying if both the numerator and denominator of a fraction have the same factor (called a common factor), then that factor can be eliminated resulting in an equivalent fraction.
It is true because ab=a
b•c
c=a • c
b • c, so we are multiplying the fraction a
b by c
c (a form of 1) to
result in the fraction a • cb • c
. Recall that multiplying a number by 1 does not change its value (the
Identity Property of Multiplication). Using our fractions 35
and 1525
, note that:
15
25=3 • 5
5 • 5=3
5
Thus 1525
and 35
are equivalent fractions, or 1525
=3
5.
146
Example 1 Determine whether the two fractions are equivalent by using the Fundamental Property of Fractions.
a. 9
10,45
50
b. !5
7,!60
84
c. 7
12,49
96
d. 4
9,4xy
9xy
Solution a. For the two fractions to be equivalent, there must be a form of 1 (or a common factor) which can be multiplied by one fraction to create the other. Note that:
9
10•5
5=45
50
Since 55
is a form of 1, the two fractions are equivalent.
b. We must find a form of 1 (or common factor) which can be multiplied by one fraction to create the other. Note that:
!5
7•12
12= !
60
84
Since 1212
is a form of 1, the two fractions are equivalent.
c. We must find a form of 1 (or common factor) which can be multiplied by one fraction to create the other. Note that:
7
12•7
8=49
96
Since 78
is not a form of 1, the two fractions are not equivalent. An
alternate way to verify this is to convert each fraction to decimal:
7
12= 0.583
49
96= 0.510416
Note that these two decimal forms are not the same.
147
d. We must find a form of 1 (or common factor) which can be multiplied by one fraction to create the other. Note that:
4
9•xy
xy=4xy
9xy
Since xyxy
is a form of 1, the two fractions are equivalent.
Given a fraction, could you find other fractions which are equivalent to it. For example, given
the fraction 37
, what would be some other fractions equivalent to it? We could multiply by
different forms of 1:
3
7•2
2=6
14
3
7•5
5=15
35
3
7•a
a=3a
7a
Note that we could list as many equivalent fractions as we can list forms of 1, which is infinite. Example 2 For each fraction, list three equivalent fractions. Use variables in at least one of your fractions.
a. 5
8
b. !7
12
c. 2x
5
148
Solution a. Using the fractions 22
, 55
, and abab
as forms of 1 (yours will probably be
different):
5
8•2
2=10
16
5
8•5
5=25
40
5
8•ab
ab=5ab
8ab
Three equivalent fractions are 1016
, 2540
, and 5ab8ab
.
b. Using the fractions 99
, 100100
, and x2y
x2y
as forms of 1 (yours will probably be
different):
!7
12•9
9= !
63
108
!7
12•100
100= !
700
1200
!7
12•x2y
x2y= !
7x2y
12x2y
Three equivalent fractions are ! 63108
, ! 7001200
, and ! 7x2y
12x2y
.
c. Using the fractions 88
, 1515
, and xyxy
as forms of 1 (again yours will
probably be different):
2x
5•8
8=16x
40
2x
5•15
15=30x
75
2x
5•xy
xy=2x
2y
5xy
Three equivalent fractions are 16x40
, 30x75
, and 2x2y
5xy. Note how we
multiplied the numbers, and how the exponent was used to represent x • x in the third fraction.
149
In addition and subtraction of fractions, it will be necessary to convert a fraction to a specified
denominator. For example, given the fraction 56
, how could this fraction be converted to one
with a denominator of 72? That is, what numerator x would result in 56=x
72 being equivalent?
Since 6 •12 = 72 (we can find 12 by dividing 6 into 72), the form of 1 to use is 1212
. Thus:
5
6•12
12=60
72
The missing numerator is x = 60 . This idea is often referred to as building fractions. Example 3 Find the variable such that the two given fractions are equivalent.
a. 9
14=x
70
b. !3
4= !
33
y
c. a
15=96
120
d. !5
b= !
200
320
Solution a. Since 70 ÷14 = 5 , the form of 1 to use is 55
. Therefore:
9
14•5
5=45
70
The missing numerator is x = 45.
b. Since 33 ÷ 3 = 11 , the form of 1 to use is 1111
. Therefore:
!3
4•11
11= !
33
44
The missing denominator is y = 44.
150
c. Since 120 ÷15 = 8 , the form of 1 to use is 88
. Instead of multiplying the
first fraction by 88
, we can alternatively divide the second fraction:
96
120÷8
8=12
15
The missing numerator is a = 12. Note how we used the idea that division is the inverse of multiplication to do this problem.
d. Since 200 ÷ 5 = 40 , the form of 1 to use is 4040
. Again, we do this problem
“backwards” by dividing the second fraction:
!200
320÷40
40= !
5
8
The missing denominator is b = 8. This last example leads to the idea of simplifying (or reducing) fractions. That is, given a
fraction such as 3240
, can we apply the Fundamental Property of Fractions to reduce the numbers
to a “simpler” form? Using the form of 1 as 88
, we can write:
32
40=4 • 8
5 • 8=4
5
We say that 3240
reduces to 45
. Note that 45
does not reduce further, since there is no other form
of 1 we can use in the Fundamental Property of Fractions. But where did 88
come from? Recall
from Chapter 1 that the greatest common factor (GCF) of 32 and 40 is the largest number that will divide into both 32 and 40, which is precisely the number 8. In other words, using the GCF of the numerator and denominator as the common factor will always result in the form of 1 to use. In the past, you may have learned to reduce fractions by dividing the numerator and denominator by the same number (this is the same as our form of 1). The big problem, however, is knowing when to stop.
151
For example, we can attempt to reduce 3240
as:
32
40=16 • 2
20 • 2=16
20
However, the result can be reduced further. Thus the GCF becomes the quickest (and safest, in terms of errors) approach to simplify fractions. Example 4 Use the greatest common factor to simplify each fraction.
a. 56
80
b. !25
150
c. !48
132
d. 5xy
10x
Solution a. The GCF of 56 and 80 is 8, so the form of 1 to use is 88
. Therefore:
56
80=7 • 8
10 • 8=7
10
b. The GCF of 25 and 150 is 25, so the form of 1 to use is 2525
. Therefore:
!25
150= !
1• 25
6 • 25= !
1
6
Note how our “invisible” factor of 1 is used in this fraction.
152
c. The GCF of 48 and 132 is 12, so the form of 1 to use is 1212
. Therefore:
!48
132= !
4 •12
11•12= !
4
11
d. The GCF of 5xy and 10x is5x , so the form of 1 to use is 5x5x
. Therefore:
5xy
10x=y • 5x
2 • 5x=y
2
This illustrates how we can simplify fractions with symbols also. Thus far, we have found the GCF by guessing at it, but recall our alternate approach using primes, which works particularly well for larger numbers. For example, to reduce the fraction 168
180, it would be difficult to guess at the GCF of 168 and 180. We first factor each number into
primes:
168 = 8 • 21 = 2 • 4( ) • 3 • 7( ) = 2 • 2 • 2( ) • 3 • 7( ) = 2 • 2 • 2 • 3 • 7
180 = 10 •18 = 2 • 5( ) • 3 • 6( ) = 2 • 5( ) • 3 • 2 • 3( ) = 2 • 2 • 3 • 3 • 5
Instead of finding the GCF, we will use the primes in our fraction, remembering that common factors of the numerator and denominator will cancel:
168
180=
2 • 2 • 2 • 3 • 7
2 • 2 • 3 • 3 • 5prime factorizations
=/2 • /2 • 2 • /3 • 7
/2 • /2 • /3 • 3 • 5cancelling common factors
=2 • 7
3 • 5writing the remaining factors
=14
15multiplying
For fractions with larger numbers, this is usually the most efficient, and more importantly the most accurate, approach.
153
Example 5 Use prime numbers to simplify each fraction.
a. 21
112
b. !90
198
c. !168
224
d. 5x2y3
4x4y
Solution a. First find the prime factorizations of 21 and 112:
21 = 3 • 7
112 = 4 • 28 = 2 • 2( ) • 2 • 2 • 7( ) = 2 • 2 • 2 • 2 • 7
Now rewrite the fraction using prime numbers, and simplify:
21
112=
3 • 7
2 • 2 • 2 • 2 • 7prime factorizations
=3 • /7
2 • 2 • 2 • 2 • /7cancelling common factors
=3
2 • 2 • 2 • 2writing the remaining factors
=3
16multiplying
b. First find the prime factorizations of 90 and 198:
90 = 9 •10 = 3 • 3( ) • 2 • 5( ) = 2 • 3 • 3 • 5
198 = 2 • 99 = 2 • 9 •11( ) = 2 • 3 • 3 •11
Now rewrite the fraction using prime numbers, and simplify:
!90
198= !
2 • 3 • 3 • 5
2 • 3 • 3 •11prime factorizations
= !/2 • /3 • /3 • 5
/2 • /3 • /3 •11cancelling common factors
= !5
11writing the remaining factors
154
c. First find the prime factorizations of 168 and 224:
168 = 4 • 42 = 2 • 2( ) • 2 • 21( ) = 2 • 2( ) • 2 • 3 • 7( ) = 2 • 2 • 2 • 3 • 7
224 = 4 • 56 = 2 • 2( ) • 4 •14( ) = 2 • 2( ) • 2 • 2 • 2 • 7( ) = 2 • 2 • 2 • 2 • 2 • 7
Now rewrite the fraction using prime numbers, and simplify:
!168
224= !
2 • 2 • 2 • 3 • 7
2 • 2 • 2 • 2 • 2 • 7prime factorizations
= !/2 • /2 • /2 • 3 • /7
/2 • /2 • /2 • 2 • 2 • /7cancelling common factors
= !3
2 • 2writing the remaining factors
= !3
4multiplying
d. This may not seem to “fit”, but look carefully at the fraction. Treating the variables as prime numbers, we simplify:
5x2y
3
4x4y=
5 • x • x • y • y • y
2 • 2 • x • x • x • x • yprime factorizations
=5 • /x • /x • y • y • /y
2 • 2 • /x • /x • x • x • /ycancelling common factors
=5 • y • y
2 • 2 • x • xwriting the remaining factors
=5y2
4x2multiplying
This example is a typical algebra problem, and it is included here to illustrate that the ideas we are developing extend directly to algebra. In the last section, we found that some fractions result in a terminating decimal, while others result in a repeating decimal. Is there a way to tell in advance which will occur? Consider the two fractions and their decimal forms:
17
40= 0.425
19
30= 0.63
155
Now look at the prime factorizations of the denominators:
17
40=
17
2 • 2 • 2 • 5= 0.425
19
30=
19
2 • 3 • 5= 0.63
Recall that the decimal system has denominators which are powers of 10 = 2 • 5. Suppose we
want to build the first fraction 1740
up to a power of 10 in the denominator. Since each 10 = 2 • 5,
we will need to multiply by two additional factors of 5:
17
40=
17
2 • 2 • 2 • 5•5 • 5
5 • 5=
17 • 25
10 •10 •10=425
1000= 0.425
However, with the second fraction 1930
, the prime factor of 3 will always be in the prime
factorization of the denominator. Thus we can never build its denominator to a power of 10, and thus it can’t be represented as a terminating decimal. In summary, only fractions whose denominators have prime factors of 2 and 5 can be converted to terminating decimals. If a denominator of a fraction has prime factors other that 2 or 5, it will result in a repeating decimal (assuming the fraction is simplified). Thus the vast majority of fractions have repeating, rather than terminating, decimal forms. Example 6 Determine whether the decimal form of each rational number will be terminating or repeating. Do not actually convert the fraction to decimal!
a. 29
85
b. !23
400
c. 49
320
d. !97
440
156
Solution a. Finding the prime factorization of 85: 85 = 5 •17 Since the denominator has a prime factor other than 2 or 5 (17), the decimal form is a repeating decimal. b. Finding the prime factorization of 400:
400 = 4 •100
= 2 • 2( ) • 10 •10( )
= 2 • 2( ) • 2 • 5 • 2 • 5( )
= 2 • 2 • 2 • 2 • 5 • 5
Since the denominator has only 2 and 5 prime factors, the decimal form is a terminating decimal. c. Finding the prime factorization of 320:
320 = 10 • 32
= 2 • 5( ) • 4 • 8( )
= 2 • 5( ) • 2 • 2 • 2 • 2 • 2( )
= 2 • 2 • 2 • 2 • 2 • 2 • 5
Since the denominator has only 2 and 5 prime factors, the decimal form is a terminating decimal. d. Finding the prime factorization of 440:
440 = 10 • 44
= 2 • 5( ) • 4 •11( )
= 2 • 5( ) • 2 • 2 •11( )
= 2 • 2 • 2 • 5 •11
Since the denominator has a prime factor other than 2 or 5 (11), the decimal form is a repeating decimal. Terminology equivalent fractions Fundamental Property of Fractions common factor building fractions simplifying (or reducing) fractions greatest common factor (GCF)
157
Exercise Set 3.2 Determine whether the two fractions are equivalent by using the Fundamental Property of Fractions.
1. 5
7,40
56 2. 6
7,48
56
3. !13
15,!52
75 4. !
5
16,!20
80
5. !23
25,!368
400 6. !
15
28,!525
980
7. 5x
4y,25ax
20ay 8. 7y
9a,35y
2
45ay
9. !3a
7b,!6ab
14ab 10. !
9w
10z,!27awz
30az
For each fraction, list three equivalent fractions. Use variables in at least one of your fractions.
11. 6
13 12. 7
11
13. !5
12 14. !
8
15
15. !3x
5 16. !
4a
7
17. 6a
7b 18. 5s
9t
Find the variable such that the two given fractions are equivalent.
19. 5
7=x
56 20. 6
13=x
91
21. !14
25= !
x
500 22. !
9
15= !
x
210
23. 13
17=195
y 24. 12
19=216
y
25. !7
15= !
112
y 26. !
9
16= !
162
y
158
27. a
7=60
84 28. a
12=121
132
29. !a
15= !
187
255 30. !
a
13= !
198
286
31. 9
b=117
143 32. 6
b=108
198
33. !12
b= !
300
425 34. !
16
b= !
432
567
Use the greatest common factor to simplify each fraction.
35. 12
20 36. 25
40
37. !70
75 38. !
32
84
39. 35
72 40. 49
93
41. !60
100 42. !
45
150
43. 35
77 44. 26
143
45. !36
60 46. !
56
78
47. 12xy
15ax 48. 25ab
40ax
49. !16abx
30axy 50. !
45axy
75aby
159
Use prime numbers to simplify each fraction.
51. 48
72 52. 70
170
53. !36
40 54. !
45
81
55. 216
234 56. 220
242
57. !270
320 58. !
275
286
59. 81
192 60. 245
246
61. !155
248 62. !
153
272
63. 384
432 64. 364
468
65. !261
290 66. !
265
424
67. 45x2y5
15x3y3
68. 30x3y7
15x6y4
69. 8a7b6
6a5b2
70. 12a6b5
18a4b3
71. !16x
8y12
20x9y7
72. !36x
6y11
24x9y8
160
Determine whether the decimal form of each rational number will be terminating or repeating. Do not actually convert the fraction to decimal!
73. 14
75 74. 19
90
75. !19
64 76. !
38
125
77. !67
448 78. !
53
480
79. 21
120(Hint: Simplify first) 80. 49
175(Hint: Simplify first)
81. !9
480 82. !
21
448
Answer each question as true or false. If it is false, give a specific example to show that it is false. If it is true, explain why. 83. Every fraction has either a terminating or repeating decimal form. 84. Every repeating decimal can be written as a fraction.
85. If the numbers a and b are relatively prime, then the fraction ab
is already simplified.
86. If the numbers a and b are not relatively prime, then the fraction ab
is not simplified.
87. There is only one decimal form for every fraction. (Hint: Refer back to Exercises 95 and 96 of the previous section.) 88. There is only one fraction for every terminating decimal.
