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4 b b 1 , 2 b b + 3 . By inspection, the LCD is  ( b 1 ) ( b + 3 ) . Rewrite each fraction with new denominator  ( b 1 ) ( b + 3 ) . ( b 1 ) ( b + 3 ) , ( b 1 ) ( b + 3 ) The denominator of the first rational expression has been multiplied  by  b + 3 , so the numerator  4 b  must be multiplied by  b + 3. 4 b ( b + 3 ) = 4 b 2 + 12 b 4 b 2 + 12 b ( b 1 ) ( b + 3 ) , ( b 1 ) ( b + 3 ) The denominator of the second rational expression has been multiplied  by  b 1 , so the numerator  2 b  must be multiplied by  b 1. 2 b ( b 1 ) = 2 b 2 + 2 b 4 b 2 + 12 b ( b 1 ) ( b + 3 ) , 2 b 2 + 2 b ( b 1 ) ( b + 3 )

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6 x x 2 8 x + 15 , 2 x 2 x 2 7 x + 12 . We first find the LCD .  Factor . 6 x ( x 3 ) ( x 5 ) , 2 x 2 ( x 3 ) ( x 4 ) The LCD is  ( x 3 ) ( x 5 ) ( x 4 ) . Rewrite each of these  fractions with new denominator  ( x 3 ) ( x 5 ) ( x 4 ) . ( x 3 ) ( x 5 ) ( x 4 ) , ( x 3 ) ( x 5 ) ( x 4 ) By comparing the denominator of the first fraction with the LCD  we see that we must multiply the numerator  6 x  by  x 4. 6 x ( x 4 ) = 6 x 2 24 x 6 x 2 24 x ( x 3 ) ( x 5 ) ( x 4 ) , ( x 3 ) ( x 5 ) ( x 4 ) By comparing the denominator of the second fraction with the LCD,  we see that we must multiply the numerator  2 x 2  by  x 5. 2 x 2 ( x 5 ) = 2 x 3 + 10 x 2 6 x 2 24 x ( x 3 ) ( x 5 ) ( x 4 ) , 2 x 3 + 10 x 2 ( x 3 ) ( x 5 ) ( x 4 )

These examples have been done step-by-step and include explanations. This makes the process seem fairly long. In practice, however, the process is much quicker.

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6 a b a 2 5 a + 4 , a + b a 2 8 a + 16 6 a b ( a 1 ) ( a 4 ) , a + b ( a 4 ) 2 LCD = ( a 1 ) ( a 4 ) 2 . 6 a b ( a 4 ) ( a 1 ) ( a 4 ) 2 , ( a + b ) ( a 1 ) ( a 1 ) ( a 4 ) 2

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x + 1 x 3 + 3 x 2 , 2 x x 3 4 x , x 4 x 2 4 x + 4 x + 1 x 2 ( x + 3 ) , 2 x x ( x + 2 ) ( x 2 ) , x 4 ( x 2 ) 2 LCD = x 2 ( x + 3 ) ( x + 2 ) ( x 2 ) 2 . ( x + 1 ) ( x + 2 ) ( x 2 ) 2 x 2 ( x + 3 ) ( x + 2 ) ( x 2 ) 2 , 2 x 2 ( x + 3 ) ( x 2 ) x 2 ( x + 3 ) ( x + 2 ) ( x 2 ) 2 , x 2 ( x + 3 ) ( x + 2 ) ( x 4 ) x 2 ( x + 3 ) ( x + 2 ) ( x 2 ) 2

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Practice set c

Change the given rational expressions into rational expressions with the same denominators.

4 x 3 , 7 x 5

4 x 2 x 5 , 7 x 5

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2 x x + 6 , x x 1

2 x ( x 1 ) ( x + 6 ) ( x 1 ) , x ( x + 6 ) ( x + 6 ) ( x 1 )

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3 b 2 b , 4 b b 2 1

3 ( b + 1 ) b ( b 1 ) ( b + 1 ) , 4 b 2 b ( b 1 ) ( b + 1 )

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8 x 2 x 6 , 1 x 2 + x 2

8 ( x 1 ) ( x 3 ) ( x + 2 ) ( x 1 ) , 1 ( x 3 ) ( x 3 ) ( x + 2 ) ( x 1 )

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10 x x 2 + 8 x + 16 , 5 x x 2 16

10 x ( x 4 ) ( x + 4 ) 2 ( x 4 ) , 5 x ( x + 4 ) ( x + 4 ) 2 ( x 4 )

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2 a b 2 a 3 6 a 2 , 6 b a 4 2 a 3 , 2 a a 2 4 a + 4

2 a 2 b 2 ( a 2 ) 2 a 3 ( a 6 ) ( a 2 ) 2 , 6 b ( a 6 ) ( a 2 ) a 3 ( a 6 ) ( a 2 ) 2 , 2 a 4 ( a 6 ) a 3 ( a 6 ) ( a 2 ) 2

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Exercises

For the following problems, replace N with the proper quantity.

x 2 4 y 2 = N 20 y 4

5 x 2 y 2

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4 a 5 x 2 y = N 15 x 3 y 3

12 a x y 2

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10 z 7 a 3 b = N 21 a 4 b 5

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8 x 2 y 5 a 3 = N 25 a 3 x 2

40 x 4 y

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2 a 2 = N a 2 ( a 1 )

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5 x 3 = N x 3 ( x 2 )

5 ( x 2 )

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4 x a = N a 4 4 a 2

4 a x ( a + 2 ) ( a 2 )

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6 b 3 5 a = N 10 a 2 30 a

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4 x 3 b = N 3 b 5 15 b

4 x ( b 4 5 )

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2 m m 1 = N ( m 1 ) ( m + 2 )

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3 s s + 12 = N ( s + 12 ) ( s 7 )

3 s ( s 7 )

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a + 1 a 3 = N ( a 3 ) ( a 4 )

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a + 2 a 2 = N ( a 2 ) ( a 4 )

( a + 2 ) ( a 4 )

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b + 7 b 6 = N ( b 6 ) ( b + 6 )

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5 m 2 m + 1 = N ( 2 m + 1 ) ( m 2 )

5 m ( m 2 )

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4 a + 6 = N a 2 + 5 a 6

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9 b 2 = N b 2 6 b + 8

9 ( b 4 )

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3 b b 3 = N b 2 11 b + 24

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2 x x 7 = N x 2 4 x 21

2 x ( x + 3 )

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6 m m + 6 = N m 2 + 10 m + 24

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4 y y + 1 = N y 2 + 9 y + 8

4 y ( y + 8 )

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x + 2 x 2 = N x 2 4

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y 3 y + 3 = N y 2 9

( y 3 ) 2

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a + 5 a 5 = N a 2 25

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z 4 z + 4 = N z 2 16

( z 4 ) 2

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4 2 a + 1 = N 2 a 2 5 a 3

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1 3 b 1 = N 3 b 2 + 11 b 4

b + 4

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a + 2 2 a 1 = N 2 a 2 + 9 a 5

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3 4 x + 3 = N 4 x 2 13 x 12

3 ( x 4 )

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b + 2 3 b 1 = N 6 b 2 + 7 b 3

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x 1 4 x 5 = N 12 x 2 11 x 5

( x 1 ) ( 3 x + 1 )

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4 y + 6 = 4 y + 8 N

( y + 6 ) ( y + 2 )

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6 a 1 = 6 a 18 N

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8 a a + 3 = 8 a 2 40 a N

( a + 3 ) ( a + 5 )

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y + 1 y 8 = y 2 2 y 3 N

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x 4 x + 9 = x 2 + x 20 N

( x + 9 ) ( x + 5 )

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7 a 5 a = N a 5

7 a

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m + 1 3 m = N m 3

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k + 6 10 k = N k 10

k 6

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For the following problems, convert the given rational expressions to rational expressions having the same denominators.

5 b 2 , 4 b 3

5 b b 3 , 4 b 3

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9 x 2 , 1 4 x

36 4 x 2 , x 4 x 2

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2 x + 5 , 4 x 5

2 ( x 5 ) ( x + 5 ) ( x 5 ) , 4 ( x + 5 ) ( x + 5 ) ( x 5 )

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10 y + 2 , 1 y + 8

10 ( y + 8 ) ( y + 2 ) ( y + 8 ) , y + 2 ( y + 2 ) ( y + 8 )

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3 b 2 , b 2 b + 5

3 ( b + 5 ) b 2 ( b + 5 ) , b 4 b 2 ( b + 5 )

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10 a a 6 , 2 a 2 6 a

10 a 2 a ( a 6 ) , 2 a ( a 6 )

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4 x 2 + 2 x , 1 x 2 4

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x + 1 x 2 x 6 , x + 4 x 2 + x 2

( x + 1 ) ( x 1 ) ( x 1 ) ( x + 2 ) ( x 3 ) , ( x + 4 ) ( x 3 ) ( x 1 ) ( x + 2 ) ( x 3 )

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x 5 x 2 9 x + 20 , 4 x 2 3 x 10

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4 b 2 + 5 b 6 , b + 6 b 2 1

4 ( b + 1 ) ( b + 1 ) ( b 1 ) ( b + 6 ) , ( b + 6 ) 2 ( b + 1 ) ( b 1 ) ( b + 6 )

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b + 2 b 2 + 6 b + 8 , b 1 b 2 + 8 b + 12

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x + 7 x 2 2 x 3 , x + 3 x 2 6 x 7

( x + 7 ) ( x 7 ) ( x + 1 ) ( x 3 ) ( x 7 ) , ( x + 3 ) ( x 3 ) ( x + 1 ) ( x 3 ) ( x 7 )

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2 a 2 + a , a + 3 a 2 1

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x 2 x 2 + 7 x + 6 , 2 x x 2 + 4 x 12

( x 2 ) 2 ( x + 1 ) ( x 2 ) ( x + 6 ) , 2 x ( x + 1 ) ( x + 1 ) ( x 2 ) ( x + 6 )

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x 2 2 x 2 + 5 x 3 , x . 1 5 x 2 + 16 x + 3

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2 x 5 , 3 5 x

2 x 5 , 3 x 5

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4 a 6 , 5 6 a

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6 2 x , 5 x 2

6 x 2 , 5 x 2

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2 m m 8 , 7 8 m

2 m m 8 , 7 m 8

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Excercises for review

( [link] ) Factor m 2 x 3 + m x 2 + m x .

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( [link] ) Factor y 2 10 y + 21.

( y 7 ) ( y 3 )

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( [link] ) Write the equation of the line that passes through the points ( 1 , 1 ) and ( 4 , 2 ) . Express the equation in slope-intercept form.

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( [link] ) Reduce y 2 y 6 y 3 .

y + 2

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( [link] ) Find the quotient: x 2 6 x + 9 x 2 x 6 ÷ x 2 + 2 x 15 x 2 + 2 x .

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Source:  OpenStax, Elementary algebra. OpenStax CNX. May 08, 2009 Download for free at http://cnx.org/content/col10614/1.3
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