Transcript Slide 1

Chapter 7
Section 3
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
7.3
1
2
Least Common Denominators
Find the least common denominator for a
group of fractions.
Rewrite rational expressions with given
denominators.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Objective 1
Find the least common
denominator for a group of
fractions.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 7.3 - 3
Find the least common denominator for a
group of fractions.
Adding or subtracting rational expressions often requires
a least common denominator (LCD), the simplest
expression that is divisible by all of the denominators in all
of the expressions. For example, the least common
5
2
denominator for the fractions 9 and 12 is 36, because 36 is
the smallest positive number divisible by both 9 and 12.
We can often find least common denominators by
1
2
inspection. For example, the LCD for 6 and 3m is 6m. In
other cases, we find the LCD by a procedure similar to that
used in Section 6.1 for finding the greatest common factor.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 7.3 - 4
Find the least common denominator for a
group of fractions. (cont’d)
To find the least common denominator, use the following
steps.
Step 1: Factor each denominator into prime factors.
Step 2: List each different denominator factor the
greatest number of times it appears in any of
the denominators.
Step 3: Multiply the denominator factors from Step 2 to
get the LCD.
When each denominator is factored into prime factors, every
prime factor must be a factor of the least common denominator.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 7.3 - 5
EXAMPLE 1
Finding the LCD
Find the LCD for each pair of fractions.
Solution:
7 1
10  2  5
,
25  5  5
10 25
LCD  52  2
4
11
,
4
6
8m 12m
 25
2
5
 50
8m4  2  2  2  m4  23  m4
6
6
2
6
12m  2  2  3  m  2  3  m
LCD  23  3  m6
 24m6
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 7.3 - 6
EXAMPLE 2
Finding the LCD
Find the LCD for
4
5
and
.
3
5
16m n
9m
Solution:
16m n  2  2  2  2  m  n
3
3
9m  3  3  m
5
5
LCD  24  32  m5  n
 2 m n
4
3
 3 m
2
5
 144m5n
When finding the LCD, use each factor the greatest number of times
it appears in any single denominator, not the total number of times it
appears.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 7.3 - 7
EXAMPLE 3
Finding the LCD
Find the LCD for the fractions in each list.
Solution:
6
3x  1
, 2
2
x  4 x x  16
x  x  4  x  x  4
 x  4 x  4
  x  4 x  4
LCD  x  x  4 x  4
4
1
,
x 1 1 x
Either x − 1 or 1 − x, since they are
opposite expressions.
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Slide 7.3 - 8
Objective 2
Rewrite rational expressions with
given denominators.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 7.3 - 9
Rewrite rational expressions with given
denominators.
Once the LCD has been found, the next step in preparing
to add or subtract two rational expressions is to use the
fundamental property to write equivalent rational expressions.
Step 1: Factor both denominators.
Step 2: Decide what factor(s) the denominator must
be multiplied by in order to equal the specified
denominator.
Step 3: Multiply the rational expression by the factor
divided by itself. (That is, multiply by 1.)
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 7.3 - 10
EXAMPLE 4
Writing Rational Expressions
with Given Denominotors
Rewrite each rational expression with the indicated
denominator.
Solution:
3 ?

4 36
7k
?

5 30k
3
?

4 49
3 3 9
 
4 4 9
7k
?

5 5  6k
7 k 7 k 6k


5
5 6k
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
27

36
2
42k

30k
Slide 7.3 - 11
EXAMPLE 5
Writing Rational Expressions
with Given Denominators
Rewrite each rational expression with the indicated
denominator.
Solution:
9
?

2a  5 6a  15
9
?

2a  5 3  2 a  5 
9
9
3


2a  5  2a  5 3
27

6a  15
5k  1
?
5k  1
?


2
k  2k k  k  2  k  1 k  k  2  k  k  2  k  1
5k  1
5k  1  k  1  5k  1 k  1



k  k  2 k  k  2  k  1 k  k  2  k  1
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 7.3 - 12