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8.7 Rational Expressions with Unlike Denominators
CORD Math
Mrs. Spitz
Fall 2006
Objective
• Add or subtract rational expressions with unlike denominators.
Upcoming
• 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:
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. 339
326
224
Ex. 1: Find the LCM of 12x2y and 15x2y2.
12x2y = 2 2 3 x x y15x2y2 = 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.
Ex. 3: Find .10
7
5
62xx
22 10
7
2
2
5
6
10
7
5
6
xx
x
xxx
22 10
7
10
12
xx
x
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.
210
712
x
x
← Multiply by a representation of 1 2x/2x.
← Simplify
← Add
Ex. 4: Find .2
4
42
aa
a
)2(
)2(
)2(
4
)2)(2(2
4
42
a
a
aaa
a
aa
a
a2 – 4 = (a + 2)(a – 2) ← Use each factor the greater number of times it appears.
← Multiply by a representation of 1.
← Simplify
← Add a and -4a to get -3a.
)2)(2(
)2(4
aa
aa
4
842
a
aa
4
832
a
a
Ex. 5: Find .)2(
5
)3)(2(
42
x
x
xx
x
22 )2(
5
)3)(2(
)4(
1
1
)2(
5
)3)(2(
4
x
x
xx
x
x
x
xx
x
2)2(
5
)3)(2(
)4(
x
x
xx
x
Multiply the first fraction by -1/-1 to change (2 – x) to (x – 2).
)3(
)3(
)2(
5
)2(
)2(
)3)(2(
)4(2
x
x
x
x
x
x
xx
x
← Change signs.
← LCD is (x – 2)(x – 2)(x + 3)
← First term needs another (x – 2) and the second terms is missing only (x + 3).
Ex. 5: Find .)2(
5
)3)(2(
42
x
x
xx
x
)3()2(
82
)2)(3)(2(
)82()2)(4(2
22
xx
xx
xxx
xxxx
)3()2(
)152
)3()2(
)3)(5(2
2
2
xx
xx
xx
xx
)3(
)3(
)2(
5
)2(
)2(
)3)(2(
)4(2
x
x
x
x
x
x
xx
x← First term needs another (x – 2) and the second terms is missing only (x + 3).
← First term needs another (x – 2) and distribute the negative sign correctly.
← Second term needs a (x + 3).
)3)(2)(2(
15282 22
xxx
xxxx← Combine terms.
Ex. 5: Find .)2(
5
)3)(2(
42
x
x
xx
x
)3)(2)(2(
15282 22
xxx
xxxx
)3()2(
742
xx
x
← 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).
