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Observe the following examples; then try to speculate on the method.

3 4 = ? 20 .

The original denominator 4 was multiplied by 5 to yield 20. What arithmetic process will yield 5 using 4 and 20?

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9 10 = ? 10 y .

The original denominator 10 was multiplied by y to yield 10 y .

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6 x y 2 a 3 b = ? 16 a 5 b 3 .

The original denominator 2 a 3 b was multiplied by 8 a 2 b 2 to yield 16 a 5 b 3 .

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5 a x ( a + 1 ) 2 = ? 4 ( a + 1 ) 2 ( a 2 ) .

The original denominator ( a + 1 ) 2 was multiplied by 4 ( a 2 ) to yield 4 ( a + 1 ) 2 ( a 2 ) .

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To determine the quantity that the original denominator was multiplied by to yield the new denominator, we ask, "What did I multiply the original denominator by to get the new denominator?" We find this factor by dividing the original denominator into the new denominator.

It is precisely this quantity that we multiply the numerator by to build the rational expression.

Sample set a

Determine N in each of the following problems.

8 3 = N 15 . The original denominator is 3 and the new  denominato is 15 . Divide the original  denominator into the new denominator and  multiply the numerator 8 by this result . 15 ÷ 3 = 5 Then, 8 5 = 40. So, 8 3 = 40 15 and N = 40. Check by reducing 40 15 .

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2 x 5 b 2 y = N 20 b 5 y 4 . The original denominator is  5 b 2 y  and the new  denominator is  20 b 5 y 4 . Divide the original  denominator into the new denominator and  multiply the numerator  2 x  by this result . 20 b 5 y 4 5 b 2 y = 4 b 3 y 3 So, 2 x · 4 b 3 y 3 = 8 b 3 x y 3 . Thus, 2 x 5 b 2 y = 8 b 3 x y 3 20 b 5 y 4 and N = 8 b 3 x y 3 .

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6 a a + 2 = N ( a + 2 ) ( a 7 ) . The new denominator divided by the original denominator is  ( a + 2 ) ( a 7 ) a + 2 = a 7 Multiply  6 a by a 7. 6 a ( a 7 ) = 6 a 2 + 42 a 6 a a + 2 = 6 a 2 + 42 a ( a + 2 ) ( a 7 )  and  N = 6 a 2 + 42 a .

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3 ( a 1 ) a 4 = N a 2 16 . The new denominator divided by the original denominator is a 2 16 a 4    = ( a + 4 ) ( a 4 ) a 4 = a + 4 Multiply  3 ( a 1 ) by a + 4. 3 ( a 1 ) ( a + 4 ) = 3 ( a 2 + 3 a 4 ) = 3 a 2 9 a + 12 3 ( a 1 ) a 4 = 3 a 2 9 a + 12 a 2 16 and N = 3 a 2 9 a + 12

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7 x = N x 2 y 3 . Write  7 x  as  7 x 1 . 7 x 1 = N x 2 y 3 Now we can see clearly that the original denominator  1 was multiplied by  x 2 y 3 .  We need to multiply the  numerator  7 x  by  x 2 y 3 . 7 x = 7 x · x 2 y 3 x 2 y 3 7 x = 7 x 3 y 3 x 2 y 3  and  N = 7 x 3 y 3 .

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5 x x + 3 = 5 x 2 20 x N . The same process works in this case .  Divide the original  numerator  5 x  into the new numerator  5 x 2 20 x . 5 x 2 20 x 5 x = 5 x ( x 4 ) 5 x = x 4 Multiply the denominator by  x 4. ( x + 3 ) ( x 4 ) 5 x x + 3 = 5 x 2 20 ( x + 3 ) ( x 4 ) and N = 5 x 2 20.

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4 x 3 x = N x 3 . The two denominators have nearly the same terms ;  each has  the opposite sign .  Factor  1  from the original denominator . 3 x = 1 ( 3 + x ) = ( x 3 ) 4 x 3 x = 4 x ( x 3 ) = 4 x x 3  and  N = 4 x .

It is important to note that we factored 1 from the original denominator. We did not multiply it by 1 . Had we multiplied only the denominator by 1 we would have had to multiply the numerator by 1 also.

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Practice set a

Determine N .

9 a 5 b = N 35 b 2 x 3

N = 63 a b x 3

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2 y y 1 = N y 2 1

N = 2 y 2 2 y

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a + 7 a 5 = N a 2 3 a 10

N = a 2 + 9 a + 14

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4 a = N 6 a 3 ( a 1 )

N = 24 a 4 ( a 1 )

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2 x = N 8 x 3 y 3 z 5

N = 16 x 4 y 3 z 5

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6 a b b + 3 = N b 2 + 6 b + 9

N = 6 a b 2 + 18 a b

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3 m m + 5 = 3 m 2 18 m N

N = m 2 m 30

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2 r 2 r 3 = 2 r 3 + 8 r 2 N

N = r 2 7 r + 12

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8 a b 2 a 4 = N 4 a

N = 8 a b 2

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The reason for building rational expressions

Building rational expressions

Normally, when we write a rational expression, we write it in reduced form. The reason for building rational expressions is to make addition and subtraction of rational expressions convenient (simpler).

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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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