Download Section 7.2

Chapter 7
Rational
Expressions
and
Equations
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CHAPTER
7
7.1
7.2
7.3
7.4
Rational Expressions and
Equations
Simplifying, Multiplying, and Dividing
Rational Expressions
Adding and Subtracting Rational
Expressions
Simplifying Complex Rational Expressions
Solving Equations Containing Rational
Expressions
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7.2
Adding and Subtracting Rational
Expressions
1. Add or subtract rational expressions with the same
denominator.
2. Find the least common denominator (LCD).
3. Write equivalent rational expressions with the LCD as
the denominator.
4. Add or subtract rational expressions with unlike
denominators.
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Adding or Subtracting Rational Expressions with
the Same Denominator
To add or subtract rational expressions with the same
denominator,
1. Add or subtract the numerators and keep the same
denominator.
2. Simplify to lowest terms.
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Example
x2
16

Subtract.
x4
x4
Solution
2
x
16

x4
x4
x 2  16

x4
Note: The numerator
can be factored, so we
may be able to simplify.
x  4  x  4 


x4
 x4
Divide out the common
factor, x – 4.
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Example
b 2  7b
b 2  4b
 2
Add. 2
b  5b
b  5b
Solution
b 2  7b
b 2  4b
 2
2
b  5b
b  5b

2
2
b

7
b

b

   4b 
b 2  5b
2b 2  3b
Combine like terms in
 2
the numerator.
b  5b
b  2b  3 Factor the numerator

b  b  5  and the denominator.
2b  3

b5
Divide out the common
factor, b.
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Example
Subtract.
3x 2  2 x  7
x2  x  8

3x  1
3x  1
Solution
2
2
3
x

2
x

7

x
 x  8
3x 2  2 x  7
x2  x  8





3x  1
3x  1
3x  1
Note: To write an equivalent
addition, change the operation
symbol from a minus sign to a plus
sign and change all the signs in the
subtrahend (second) polynomial.

2
2
3
x

2
x

7


x

   x  8
3x  1
2 x2  x  1

3x  1
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Example
x2  5x  6
x 2  3x  2

Add.
3
2x  4x
2 x3  4 x
Solution
x  5x  6
x  3x  2


3
3
2x  4x
2x  4x
2
2
2
2
x

5
x

6

x

   3x  2 
2 x3  4 x
Combine like terms in
the numerator.
2 x2  8x  8

2 x3  4 x
Factor the numerator
and the denominator.
2  x  2  x  2 

2 x  x  2  x  2 
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continued
2  x  2  x  2 

2 x  x  2  x  2 
x2
x2

or 2
x  x  2
x  2x
Divide out the common
factors, 2 and x + 2.
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Remember that when adding or subtracting fractions
with different denominators, we must first find a
common denominator. It is helpful to use the least
common denominator (LCD), which is the smallest
number that is evenly divisible by all the
denominators.
8  23
LCD  23
12  22
3
3 = 24
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Finding the LCD
To find the LCD of two or more rational expressions,
1. Find the prime factorization of each denominator.
2. Write the product that contains each unique prime
factor the greatest number of times that factor
occurs in any factorization. Or, if you prefer to
use exponents, write the product that contains
each unique prime factor raised to the greatest
exponent that occurs on that factor in any
factorization.
3. Simplify the product found in step 2.
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Example
5
2
and
Find the LCD.
2
12 y
8 y3
Solution We first factor the denominators 12y2 and
8y3 by writing their prime factorizations.
12 y 2  22
8 y3  23
3 y2
y3
The unique factors are 2, 3, and y. To generate the
LCD, include 2, 3, and y the greatest number of times
each appears in any of the factorizations.
LCD = 23
3 y3  24 y3
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Example
x
x7
and 2
Find the LCD.
x 3
x  6x
Solution Find the factors.
x  3 is prime and x2  6 x  x  x  6
The unique factors are x – 3, x, and x + 6, and the
highest power of each is 1.
LCD = x  x  3 x  6
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Example
8
3
Find the LCD. 2
and
x  25
2 x  10
Solution Factor the denominators x2 – 25 and 2x – 10.
x2  25 
 x  5 x  5
2 x 10  2  x  5
The unique factors are 2, (x + 5), and (x – 5).
The greatest number of times that 2 appears is once.
The greatest number of times that (x + 5) appears is
once.
The greatest number of times that (x – 5) appears is
once.
LCD = 2  x  5 x  5
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Example
x5
x 2  4 x  32
and
Find the LCD.
4
3
2
4 x  24 x  5 x
3x 2  6 x  3
Solution Factor the denominators.
4 x4  24 x3  5x2  4 x2  x  1 x  5
3x  6 x  3  3  x  1
2
2
The unique factors are 3, 4, x, (x + 1), and (x + 5).
The highest power of 3, 4, and x + 5 is 1, and the
highest power of x and x + 1 is 2.
LCD = 3 4 x
2
 x  1
2
 x  5
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Example
Write the fractions with the LCD as the denominator.
5
2
and
2
12 y
8 y3
Solution In example 2(a), we found that the LCD
was 24y3
5
2
12 y
2
and
8 y3
Convert each to a fraction whose denominator is 24y3.
5
5 2 y 10 y
=

2
2
12 y
12 y 2 y 24 y 3
2
2 3
6
= 3 
3
8 y 8 y 3 24 y 3
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Example
Write the fractions with the LCD as the denominator.
8
3
and
2
x  25
2 x  10
Solution In example 2(c), we found that the LCD
was 2(x + 5)(x − 5).
Convert each to a fraction whose denominator is 2(x +
5)(x − 5).
8
8
2
16
=
=
 x  5 x  5  x  5 x  5 2 2  x  5 x  5
x  5

3
3
3 x  15


2  x  5  2  x  5   x  5  2  x  5  x  5 
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Adding or Subtracting Rational Expressions with
Different Denominators
To add or subtract rational expressions with different
denominators,
1. Find the LCD.
2. Write each rational expression as an equivalent
expression with the LCD.
3. Add or subtract the numerators and keep the LCD
as the denominator.
4. Simplify.
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Example
Add.
x  3 5x 1

8x
6 x2
Solution The LCD is 24x2.
x  3 5x 1
x  3 3x   5 x  1 4 




2
8x
6x
8x  3x   6 x 2   4 
3x 2  9 x 20 x  4


2
24 x
24 x 2
3x 2  9 x  20 x  4

24 x 2
3x 2  11x  4

24 x 2
Write equivalent rational
expressions with the
LCD, 24x2.
Subtract numerators.
Note: Remember that to add
polynomials, we combine like
terms.
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Example
7
3x
+
Add.
x6
6 x
Solution
3 x  1
7
7
3x

+
+
6  x  1
x6
6 x x6
7
3 x

+
x6
x6
3x  7

x6
Since x – 6 and 6 – x
are additive inverses,
we obtain the LCD by
multiplying the
numerator and
denominator of one of
the rational expressions
by –1. We chose the
second rational
expression.
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Example
x
28
 2
x  4 x  x  12
Solution Find the LCD.
x  4 is prime and x2  x  12   x  4 x  3
LCD=  x  4 x  3
x
28
x
28
 2


x  4 x  x  12 x  4  x  4  x  3
x  x  3
28


x  4  x  3  x  4  x  3
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continued
x  x  3
28


 x  4  x  3  x  4  x  3
x 2  3x
28


 x  4  x  3  x  4  x  3
x 2  3 x  28

 x  4  x  3
x  4  x  7 


 x  4  x  3
x7

x3
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Example
Solution
2y  5 7y  2
Subtract: 2

.
2
3 y 1 1  3 y
2 y  5 (7 y  2)(1)
 2

3 y  1 (1  3 y 2 )(1)
2 y  5 7 y  2
 2

3 y 1 3 y2 1
2 y  5  (7 y  2)

3 y2 1
2y  5  7y  2
9y  7

 2
2
3 y 1
3 y 1
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