Copyright © 2013, 2009, 2005 Pearson Education, Inc. Section 6.3 Addition and Subtraction of...
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Transcript of Copyright © 2013, 2009, 2005 Pearson Education, Inc. Section 6.3 Addition and Subtraction of...
Copyright © 2013, 2009, 2005 Pearson Education, Inc.
Section 6.3
Addition and Subtraction of
Rational Expressions
Copyright © 2013, 2009, 2005 Pearson Education, Inc.
Objectives
• Least Common Multiples
• Review of Addition and Subtraction of Fractions
• Addition of Rational Expressions
• Subtraction of Rational Expressions
Copyright © 2013, 2009, 2005 Pearson Education, Inc.
The least common multiple (LCM) of two or more polynomials can be found as follows.
Step 1: Factor each polynomial completely.
Step 2: List each factor the greatest number of times that it occurs in either
factorization.
Step 3: Find the product of this list of factors. The result is the LCM.
FINDING THE LEAST COMMON MULTIPLE
Copyright © 2013, 2009, 2005 Pearson Education, Inc.
Example
Find the least common multiple of each pair of expressions.a. 6x, 9x4 b. x2 + 7x + 12, x2 + 8x + 16 SolutionStep 1: Factor each polynomial completely.
6x = 3 ∙ 2 ∙ x 9x4 = 3 ∙ 3 ∙ x ∙ x ∙ x ∙ x
Step 2: List each factor the greatest number of times.3 ∙ 3 ∙ 2 ∙ x ∙ x ∙ x ∙ x
Step 3: The LCM is 18x4.
Copyright © 2013, 2009, 2005 Pearson Education, Inc.
Example (cont)
b. x2 + 7x + 12, x2 + 8x + 16
Step 1: Factor each polynomial completely.x2 + 7x + 12 = (x + 3)(x + 4)x2 + 8x + 16 = (x+ 4)(x + 4)
Step 2: List each factor the greatest number of times.(x + 3), (x + 4), and (x + 4)
Step 3: The LCM is (x + 3)(x + 4)2.
Copyright © 2013, 2009, 2005 Pearson Education, Inc.
Example
Find the sum.a. b.
Solutiona. The LCD is 42.
b. The LCD is 18.
4 1
7 6
1 5
9 6
4 1
7 6
4 6 1 7
7 6 6 7 24 7
42 42
31
42
1 5
9 6
1 2 5 3
9 2 6 3
2 15
18 18
17
18
Copyright © 2013, 2009, 2005 Pearson Education, Inc.
Example
Find the difference.a. b.
Solutiona. The LCD is 36.
b. The LCD is 60.
13 7
18 12
5 11
12 30
13 7
18 12
13 2 7 3
18 2 12 3
26 21
36 36
5
36
5 11
12 30
5 5 11 2
12 5 30 2
3
60
25 22
60 60
1
20
Copyright © 2013, 2009, 2005 Pearson Education, Inc.
To add two rational expressions with like denominators, add their numerators. The denominator does not change.
C not zero
SUMS OF RATIONAL EXPRESSIONS
A B A B
C C C
Copyright © 2013, 2009, 2005 Pearson Education, Inc.
Example
Add and simplify.a. b.
Solutiona.
b.
4 1 2
3 3
x x
x x
2 2
5
7 10 7 10
x
x x x x
3
4
3
1 2
x x
x x
4 1 2
3
x x
x
3
5 1x
x
2 2
5
7 10 7 10
x
x x x x
2
5
7 10
x
x x
5
5 2
x
x x
1
2x
Copyright © 2013, 2009, 2005 Pearson Education, Inc.
Example
Add and simplify.a. b.
Solutiona. b.
2
2 5
x x
4 3
1 1x x
2
2 5
x x
2
2 5x
x x x
2
2 5x
x x x
2 2
2 5x
x x
2
2 5x
x
4 3
1 1x x
4 3 1
1 1 1x x
4 3
1 1x x
1
1x
Copyright © 2013, 2009, 2005 Pearson Education, Inc.
To subtract two rational expressions with like denominators, subtract their numerators. The denominator does not change.
C not zero
DIFFERENCES OF RATIONAL
EXPRESSIONS
A B A B
C C C
Copyright © 2013, 2009, 2005 Pearson Education, Inc.
Example
Subtract and simplify.a. b.
Solutiona. b.
2 2
6 6x
x x
2 2
2 3 4
1 1
x x
x x
2 2
6 6x
x x
2
6 6x
x
2
6 6x
x
2
x
x
1
x
2 2
2 3 4
1 1
x x
x x
2
2 3 4
1
x x
x
1
1 1
x
x x
2
1
1
x
x
1
1x
Copyright © 2013, 2009, 2005 Pearson Education, Inc.
Example
Subtract and simplify.
SolutionThe LCD is x(x + 7).
3 5
7
x
x x
3 5
7
x
x x
3 7 5
7 7
x x x
x x x x
3 7 5
7 7
x x x
x x x x
3 7 5
7
x x x
x x
2 4 21 5
7
x x x
x x
2 21
7
x x
x x
Copyright © 2013, 2009, 2005 Pearson Education, Inc.
Example
Simplify the expression. Write your answer in lowest terms and leave it in factored form.Solution
2 2
6 5
6 9 9x x x
6 5
3 3 3 3x x x x
2 2
6 5
6 9 9x x x
6 5
3 3 3 33
33
3
x
x x xx x
x
x
6 3 5 3
3 3 3 3 3 3
x x
x x x x x x
6 18 5 15
3 3 3
x x
x x x
33
3 3 3
x
x x x
Copyright © 2013, 2009, 2005 Pearson Education, Inc.
Step 1: If the denominators are not common, multiply each expression by 1
written in the appropriate form to obtain the LCD.
Step 2: Add or subtract the numerators. Combine like terms.
Step 3: If possible, simplify the final expression.
STEPS FOR FINDING SUMS AND
DIFFERENCES OF RATIONAL EXPRESSIONS
Copyright © 2013, 2009, 2005 Pearson Education, Inc.
Example
A 75-watt light bulb with a resistance of R1 = 160 ohms and a 60-watt light bulb with a resistance of R2 = 240 ohms are placed in an electrical circuit. Find the combined resistance. Solution
1 2
1 1 1
R R R
1 1
160 240
3 2
3
1
0 2
1
16 240
480
3
80
2
4
1 5
480
R
R = 96 ohms