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3 x 2 + 4 x + 5 ( x + 6 ) ( x 2 ) + 2 x 2 + x + 6 x 2 + 4 x 12 x 2 4 x 6 x 2 + 4 x 12 Factor the denominators to determine if they're the same . 3 x 2 + 4 x + 5 ( x + 6 ) ( x 2 ) + 2 x 2 + x + 6 ( x + 6 ) ( x 2 ) x 2 4 x 6 ( x + 6 ) ( x 2 ) The denominators are the same . Combine the numerators being careful to note the negative sign . 3 x 2 + 4 x + 5 + 2 x 2 + x + 6 ( x 2 4 x + 6 ) ( x + 6 ) ( x 2 ) 3 x 2 + 4 x + 5 + 2 x 2 + x + 6 x 2 + 4 x + 6 ( x + 6 ) ( x 2 ) 4 x 2 + 9 x + 17 ( x + 6 ) ( x 2 )

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Practice set a

Add or Subtract the following rational expressions.

5 x 2 y 2 3 x 2 y 2

x y 2

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x + y x y + 2 x + 3 y x y

3 x + 4 y x y

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4 x 2 x + 4 3 x + 10 x 2 + 2 x + 5 3 x + 10

3 x 2 3 x 1 3 x + 10

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x ( x + 1 ) x ( 2 x + 3 ) + 3 x 2 x + 7 2 x 2 + 3 x

4 x 2 + 7 x ( 2 x + 3 )

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4 x + 3 x 2 x 6 8 x 4 ( x + 2 ) ( x 3 )

4 x + 7 ( x + 2 ) ( x 3 )

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5 a 2 + a 4 2 a ( a 6 ) + 2 a 2 + 3 a + 4 2 a 2 12 a + a 2 + 2 2 a 2 12 a

4 a 2 + 2 a + 1 a ( a 6 )

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8 x 2 + x 1 x 2 6 x + 8 + 2 x 2 + 3 x x 2 6 x + 8 5 x 2 + 3 x 4 ( x 4 ) ( x 2 )

5 x 2 + x + 3 ( x 4 ) ( x 2 )

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Fractions with different denominators

Sample set b

Add or Subtract the following rational expressions.

4 a 3 y + 2 a 9 y 2 . The denominators are  n o t  the same . Find the LCD . By inspection, the LCD is 9 y 2 . 9 y 2 + 2 a 9 y 2 The denominator of the first rational expression has been multiplied by 3 y , so the numerator must be multiplied by 3 y . 4 a · 3 y = 12 a y 12 a y 9 y 2 + 2 a 9 y 2 The denominators are now the same . Add the numerators . 12 a y + 2 a 9 y 2

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3 b b + 2 + 5 b b 3 . The denominators are n o t the same . The LCD is ( b + 2 ) ( b 3 ) . ( b + 2 ) ( b 3 ) + ( b + 2 ) ( b 3 ) The denominator of the first rational expression has been multiplied by b 3 , so the numerator must be multiplied by b 3. 3 b ( b 3 ) 3 b ( b 3 ) ( b + 2 ) ( b 3 ) + ( b + 2 ) ( b 3 ) The denominator of the second rational expression has been multiplied by b + 2 , so the numerator must be multiplied by b + 2. 5 b ( b + 2 ) 3 b ( b 3 ) ( b + 2 ) ( b 3 ) + 5 b ( b + 2 ) ( b + 2 ) ( b 3 ) The denominators are now the same . Add the numerators . 3 b ( b 3 ) + 5 b ( b + 2 ) ( b 3 ) ( b + 2 ) = 3 b 2 9 b + 5 b 2 + 10 b ( b 3 ) ( b + 2 ) = 8 b 2 + b ( b 3 ) ( b 2 )

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x + 3 x 1 + x 2 4 x + 4 . The denominators are  n o t  the same . Find the LCD . x + 3 x 1 + x 2 4 ( x + 1 ) The LCD is  ( x + 1 ) ( x 1 ) 4 ( x + 1 ) ( x 1 ) + 4 ( x + 1 ) ( x 1 ) The denominator of the first rational expression has been multiplied by  4 ( x + 1 )  so  the numerator must be multiplied by  4 ( x + 1 ) . 4 ( x + 3 ) ( x + 1 ) 4 ( x + 3 ) ( x + 1 ) 4 ( x + 1 ) ( x 1 ) + 4 ( x + 1 ) ( x 1 ) The denominator of the second rational expression has been multiplied by  x 1 so the numerator must be multiplied by  x 1. ( x 1 ) ( x 2 ) 4 ( x + 3 ) ( x + 1 ) 4 ( x + 1 ) ( x 1 ) + ( x 1 ) ( x 2 ) 4 ( x + 1 ) ( x 1 ) The denominators are now the same. Add the numerators. 4 ( x + 3 ) ( x + 1 ) + ( x 1 ) ( x 2 ) 4 ( x + 1 ) ( x 1 ) 4 ( x 2 + 4 x + 3 ) + x 2 3 x + 2 4 ( x + 1 ) ( x 1 ) 4 x 2 + 16 x + 12 + x 2 3 x + 2 4 ( x + 1 ) ( x 1 ) = 5 x 2 + 13 x + 14 4 ( x + 1 ) ( x 1 )

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x + 5 x 2 7 x + 12 + 3 x 1 x 2 2 x 3 Determine the LCD . x + 5 ( x 4 ) ( x 3 ) + 3 x 1 ( x 3 ) ( x + 1 ) The LCD is ( x 4 ) ( x 3 ) ( x + 1 ) . ( x 4 ) ( x 3 ) ( x + 1 ) + ( x 4 ) ( x 3 ) ( x + 1 ) The first numerator must be multiplied by x + 1 and the second by x 4. ( x + 5 ) ( x + 1 ) ( x 4 ) ( x 3 ) ( x + 1 ) + ( 3 x 1 ) ( x 4 ) ( x 4 ) ( x 3 ) ( x + 1 ) The denominators are now the same . Add the numerators . ( x + 5 ) ( x + 1 ) + ( 3 x 1 ) ( x 4 ) ( x 4 ) ( x 3 ) ( x + 1 ) x 2 + 6 x + 5 + 3 x 2 13 x + 4 ( x 4 ) ( x 3 ) ( x + 1 ) 4 x 2 7 x + 9 ( x 4 ) ( x 3 ) ( x + 1 )

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a + 4 a 2 + 5 a + 6 a 4 a 2 5 a 24 Determine the LCD. a + 4 ( a + 3 ) ( a + 2 ) a 4 ( a + 3 ) ( a 8 ) The LCD is  ( a + 3 ) ( a + 2 ) ( a 8 ) . ( a + 3 ) ( a + 2 ) ( a 8 ) ( a + 3 ) ( a + 2 ) ( a 8 ) The first numerator must be multiplied by  a 8  and the second by  a + 2. ( a + 4 ) ( a 8 ) ( a + 3 ) ( a + 2 ) ( a 8 ) ( a 4 ) ( a + 2 ) ( a + 3 ) ( a + 2 ) ( a 8 ) The denominators are now the same . Subtract the numerators .  ( a + 4 ) ( a 8 ) ( a 4 ) ( a + 2 ) ( a + 3 ) ( a + 2 ) ( a 8 ) a 2 4 a 32 ( a 2 2 a 8 ) ( a + 3 ) ( a + 2 ) ( a 8 ) a 2 4 a 32 a 2 + 2 a + 8 ( a + 3 ) ( a + 2 ) ( a 8 ) 2 a 24 ( a + 3 ) ( a + 2 ) ( a 8 ) Factor  2  from the numerator .  2 ( a + 12 ) ( a + 3 ) ( a + 2 ) ( a 8 )

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3 x 7 x + 5 x x 7 . The denominators are n e a r l y the same.  They differ only in sign. Our technique is to factor  1  from one of them. 3 x 7 x = 3 x ( x 7 ) = 3 x x 7 Factor  1  from the first term .  3 x 7 x + 5 x x 7 = 3 x x 7 + 5 x x 7 = 3 x + 5 x x 7 = 2 x x 7

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Practice set b

Add or Subtract the following rational expressions.

3 x 4 a 2 + 5 x 12 a 3

9 a x + 5 x 12 a 3

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5 b b + 1 + 3 b b 2

8 b 2 7 b ( b + 1 ) ( b 2 )

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a 7 a + 2 + a 2 a + 3

2 a 2 4 a 25 ( a + 2 ) ( a + 3 )

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4 x + 1 x + 3 x + 5 x 3

3 x 2 19 x 18 ( x + 3 ) ( x 3 )

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2 y 3 y + 3 y + 1 y + 4

5 y 2 + 6 y 12 y ( y + 4 )

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a 7 a 2 3 a + 2 + a + 2 a 2 6 a + 8

2 a 2 10 a + 26 ( a 2 ) ( a 1 ) ( a 4 )

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6 b 2 + 6 b + 9 2 b 2 + 4 b + 4

4 b 2 + 12 b + 6 ( b + 3 ) 2 ( b + 2 ) 2

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x x + 4 x 2 3 x 3

2 x 2 5 x + 8 3 ( x + 4 ) ( x 1 )

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5 x 4 x + 7 x x 4

2 x x 4

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Sample set c

Combine the following rational expressions.

3 + 7 x 1 . Rewrite the expression. 3 1 + 7 x 1 The LCD is  x 1. 3 ( x 1 ) x 1 + 7 x 1 = 3 x 3 x 1 + 7 x 1 = 3 x 3 + 7 x 1 = 3 x + 4 x 1

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3 y + 4 y 2 y + 3 y 6 . Rewrite the expression. 3 y + 4 1 y 2 y + 3 y 6 The LCD is  y 6. ( 3 y + 4 ) ( y 6 ) y 6 y 2 y + 3 y 6 = ( 3 y + 4 ) ( y 6 ) ( y 2 y + 3 ) y 6 = 3 y 2 14 y 24 y 2 + y 3 y 6 = 2 y 2 13 y 27 y 6

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Practice set c

Simplify 8 + 3 x 6 .

8 x 45 x 6

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Simplify 2 a 5 a 2 + 2 a 1 a + 3 .

a 2 a 14 a + 3

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Exercises

For the following problems, add or subtract the rational expressions.

6 y 5 x + 8 y 5 x

14 y 5 x

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15 n 2 m 6 n 2 m

9 n 2 m

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y + 4 y 6 + y + 8 y 6

2 y + 12 y 6

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y 1 y + 4 + y + 7 y + 4

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a + 6 a 1 + 3 a + 5 a 1

4 a + 11 a 1

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5 a + 1 a + 7 + 2 a 6 a + 7

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x + 1 5 x + x + 3 5 x

2 x + 4 5 x

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a 6 a + 2 + a 2 a + 2

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b + 1 b 3 + b + 2 b 3

2 b + 3 b 3

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a + 2 a 5 a + 3 a 5

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b + 7 b 6 b 1 b 6

8 b 6

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2 b + 3 b + 1 b 4 b + 1

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3 y + 4 y + 8 2 y 5 y + 8

y + 9 y + 8

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2 a 7 a 9 + 3 a + 5 a 9

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8 x 1 x + 2 15 x + 7 x + 2

7 x 8 x + 2

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2 3 x + 4 6 x 2

2 ( x + 1 ) 3 x 2

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2 5 a 2 1 10 a 3

4 a 1 10 a 3

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4 x 6 + 1 x 1

5 ( x 2 ) ( x 6 ) ( x 1 )

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6 y y + 4 + 2 y y + 3

2 y ( 4 y + 13 ) ( y + 4 ) ( y + 3 )

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x 1 x 3 + x + 4 x 4

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x + 2 x 5 + x 1 x + 2

2 x 2 2 x + 9 ( x 5 ) ( x + 2 )

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a + 3 a 3 a + 2 a 2

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y + 1 y 1 y + 4 y 4

6 y ( y 1 ) ( y 4 )

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

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y + 2 ( y + 1 ) ( y + 6 ) + y 2 y + 6

y 2 ( y + 1 ) ( y + 6 )

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2 a + 1 ( a + 3 ) ( a 3 ) a + 2 a + 3

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3 a + 5 ( a + 4 ) ( a 1 ) 2 a 1 a 1

2 a 2 4 a + 9 ( a + 4 ) ( a 1 )

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2 x x 2 3 x + 2 + 3 x 2

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4 a a 2 2 a 3 + 3 a + 1

7 a 9 ( a + 1 ) ( a 3 )

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3 y y 2 7 y + 12 y 2 y 3

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x 1 x 2 + 6 x + 8 + x + 3 x 2 + 2 x 8

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

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a 4 a 2 + 2 a 3 + a + 2 a 2 + 3 a 4

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b 3 b 2 + 9 b + 20 + b + 4 b 2 + b 12

2 b 2 + 3 b + 29 ( b 3 ) ( b + 4 ) ( b + 5 )

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y 1 y 2 + 4 y 12 y + 3 y 2 + 6 y 16

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x + 3 x 2 + 9 x + 14 x 5 x 2 4

x + 29 ( x 2 ) ( x + 2 ) ( x + 7 )

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x 1 x 2 4 x + 3 + x + 3 x 2 5 x + 6 + 2 x x 2 3 x + 2

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4 x x 2 + 6 x + 8 + 3 x 2 + x 6 + x 1 x 2 + x 12

5 x 4 3 x 3 34 x 2 + 34 x 60 ( x 2 ) ( x + 2 ) ( x 3 ) ( x + 3 ) ( x + 4 )

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y + 2 y 2 1 + y 3 y 2 3 y 4 y + 3 y 2 5 y + 4

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a 2 a 2 9 a + 18 + a 2 a 2 4 a 12 a 2 a 2 a 6

( a + 5 ) ( a 2 ) ( a + 2 ) ( a 3 ) ( a 6 )

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y 2 y 2 + 6 y + y + 4 y 2 + 5 y 6

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a + 1 a 3 + 3 a 2 a + 6 a 2 a

a 3 8 a 2 18 a 1 a 2 ( a + 3 ) ( a 1 )

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4 3 b 2 12 b 2 6 b 2 6 b

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3 2 x 5 4 x 4 + 2 8 x 3 + 24 x 2

x 3 + 2 x 2 + 6 x + 18 4 x 4 ( x 2 ) ( x + 3 )

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x + 2 12 x 3 + x + 1 4 x 2 + 8 x 12 x + 3 16 x 2 32 x + 16

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2 x x 2 9 x + 1 4 x 2 12 x x 4 8 x 3

14 x 4 9 x 3 2 x 2 + 9 x 36 8 x 3 ( x + 3 ) ( x 3 )

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8 + 2 x + 6

8 x + 50 x + 6

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3 + 5 x 6

3 x 13 x 6

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1 + 3 a a 1

2 a + 1 a 1

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2 x + x 2 4 x + 1

3 x 2 + 2 x 4 x + 1

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3 y + 4 y 2 + 2 y 5 y + 3

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x + 2 + x 2 + 4 x 1

2 x 2 + x + 2 x 1

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3 x 1 x 4 8

5 x + 31 x 4

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2 y 2 + 11 y 1 y + 4 3 y

( y 2 + y + 1 ) y + 4

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5 y 2 2 y + 1 y 2 + y 6 2

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4 a 3 + 2 a 2 + a 1 a 2 + 11 a + 28 + 3 a

7 a 3 + 35 a 2 + 85 a 1 ( a + 7 ) ( a + 4 )

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2 x 1 x + 6 x x 1

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5 m 6 m + 3 m m 6

2 m m 6

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a + 7 8 3 a + 2 a + 1 3 a 8

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2 y + 4 4 5 y 9 5 y 4

2 y 13 5 y 4

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m 1 1 m 2 m 1

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Exercises for review

( [link] ) Simplify ( x 3 y 2 z 5 ) 6 ( x 2 y z ) 2 .

x 22 y 14 z 32

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( [link] ) Write 6 a 3 b 4 c 2 a 1 b 5 c 3 so that only positive exponents appear.

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( [link] ) Construct the graph of y = 2 x + 4.
An xy coordinate plane with gridlines, labeled negative five to five on both axes.

A graph of a line passing through three points with coordinates zero, four; one, two; and two, zero.

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( [link] ) Find the product: x 2 3 x 4 x 2 + 6 x + 5 · x 2 + 5 x + 6 x 2 2 x 8 .

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( [link] ) Replace N with the proper quantity: x + 3 x 5 = N x 2 7 x + 10 .

( x + 3 ) ( x 2 )

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Questions & Answers

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Show that the equal masses particles emarge from collision at right angle by making explicit used of fact that momentum is a vector quantity
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A wave is described by the function D(x,t)=(1.6cm) sin[(1.2cm^-1(x+6.8cm/st] what are:a.Amplitude b. wavelength c. wave number d. frequency e. period f. velocity of speed.
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A body is projected upward at an angle 45° 18minutes with the horizontal with an initial speed of 40km per second. In hoe many seconds will the body reach the ground then how far from the point of projection will it strike. At what angle will the horizontal will strike
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Suppose hydrogen and oxygen are diffusing through air. A small amount of each is released simultaneously. How much time passes before the hydrogen is 1.00 s ahead of the oxygen? Such differences in arrival times are used as an analytical tool in gas chromatography.
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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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