8.7 Rational Expressions with
Unlike Denominators
CORD Math
Mrs. Spitz
Fall 2006
Objective
• Add or subtract rational expressions with
unlike denominators.
Upcoming
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Midchapter Quiz Thursday
8.7 Thursday 10/26
8.8 Friday 10/27
8.9 Monday 10/30
8.10 Tuesday/Wed
Chapter 8 Review Wed/Thur
Chapter 8 Test Friday
Assignment
• Pp. 331-332 #3-38 all
Multiples – LCM
• What is the LCM of 4, 6 and 9? (Refer to
the dancer example in your book if you
want to read how it pertains to dancing
and choreography).
4 – 8 – 12 – 16 – 20 – 24 – 28 – 32 – 36 – 40
6 – 12 – 18 – 24 – 30 – 36 – 42 – 48
9 – 18 – 27 – 36 – 45 – 54 So, the LCM of 4, 6
and 9 is 36.
What else could you do?
• You could use prime
factorization to find
LCM.
1. Find the prime
factorization of each
number:
4  22
6  23
9  3 3
2. Use each prime
factor the greatest
number of times it
appears in any of the
factorizations.
2 appears two times in
4 and 3 appears two
times in 9. Thus the
LCM of 4, 6, and 9 is
2  2  3  3, or 36.
Ex. 1: Find the LCM of 12x2y and 15x2y2.
12x2y = 2  2  3  x  x  y
15x2y2 = 5  3  x  x  y  y
LCM = 2  2  3  5  x  x  y  y = 60x2y2 –Use
each factor the greatest number of times it
appears in either factorization.
Ex. 2: Find the LCM of x2 + x - 2 and x2 + 5x - 6.
x2 +x - 2 = (x - 1)(x + 2)
X2 + 5x - 6 = (x + 6)(x – 1)
LCM = (x – 1)(x + 2)(x + 6)–Use each factor the
greatest number of times it appears in either
factorization.
Take notes on this . . .
• To add or subtract fractions with unlike
denominators, first rename the fractions so
the denominators are alike. Any common
denominator could be used. However, the
computation is usually easier if you use
the least common denominator (LCD).
Recall that the least common denominator
is the LCM of the denominators.
Use the following steps to add or
subtract rational expressions.
1. Find the LCD.
2. Change each rational expression into an
equivalent expression with LCD as the
denominator.
3. Add or subtract as with rational
expressions with like denominators.
4. Simplify if necessary.
6
7

.
2
5 x 10 x
Ex. 3: Find
5x = 5  x
10x2 = 2  5  x  x
← Use each factor the greater number of
times it appears.
The LCD for the expression above is 2  5  x  x or 10x2. Since the
denominator of the second term is already 10x2, only the first term
needs to be renamed.
6
7
6 2x
7




2
5 x 10 x
5 x 2 x 10 x 2
12 x
7


2
10 x 10 x 2
12 x  7

10 x 2
← Multiply by a representation of 1 2x/2x.
← Simplify
← Add
Ex. 4: Find
a
4

.
2
a 4 a2
← Use each factor the greater number of
times it appears.
a2 – 4 = (a + 2)(a – 2)
a
4
a
4
(a  2)




2
a  4 a  2 (a  2)( a  2) (a  2) (a  2)
a  4(a  2)

(a  2)( a  2)
a  4a  8

a2  4
 3a  8
 2
a 4
← Multiply by a representation of 1.
← Simplify
← Add a and -4a to get -3a.
Ex. 5: Find
x4
x5

.
2
(2  x)( x  3) ( x  2)
Multiply the first fraction by -1/-1 to change (2 – x) to (x – 2).
x4
x5
1
( x  4)
x 5




2
(2  x)( x  3) ( x  2)
 1 (2  x)( x  3) ( x  2) 2
 ( x  4)
x 5


( x  2)( x  3) ( x  2) 2
← Change signs.
← LCD is (x – 2)(x – 2)(x + 3)
 ( x  4) ( x  2) x  5 ( x  3)




2
( x  2)( x  3) ( x  2) ( x  2) ( x  3)
← First term needs another (x – 2) and
the second terms is missing only (x + 3).
x4
x5

.
Ex. 5: Find
2
(2  x)( x  3) ( x  2)
 ( x  4) ( x  2) x  5 ( x  3)




2
( x  2)( x  3) ( x  2) ( x  2) ( x  3)
← First term needs another (x – 2) and
the second terms is missing only (x + 3).
 ( x  4)( x  2)  ( x 2  2 x  8)  x 2  2 x  8


( x  2)( x  3)( x  2)
( x  2)2 ( x  3)
← First term needs another (x – 2) and
distribute the negative sign correctly.
( x  5)( x  3)
x 2  2 x  15)


2
( x  2) ( x  3) ( x  2) 2 ( x  3)
 x 2  2 x  8  x 2  2 x  15

( x  2)( x  2)( x  3)
← Second term needs a (x + 3).
← Combine terms.
Ex. 5: Find
x4
x5

.
2
(2  x)( x  3) ( x  2)
 x 2  2 x  8  x 2  2 x  15

( x  2)( x  2)( x  3)
 4x  7

2
( x  2) ( x  3)
← Combine terms.
← x2 drop out. -2x and -2x
add to give you -4x and +8
and -15 give you -7 in the
numerator. In the
denominator, we have what
we have had throughout as
the LCD (x - 2)2(x + 3).