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- Nov 13th 2013, 05:47 AM #1

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- Nov 13th 2013, 06:18 AM #2

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- Nov 13th 2013, 07:42 AM #3

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## Re: please help

You can do that simply by

**adding**the fractions as indicated: $\displaystyle \dfrac{C}{x- 3}+ \dfrac{D}{x+ 8}= \dfrac{C(x+8)}{(x- 3)(x+ 8)}+ \dfrac{D(x- 3)}{(x-3)(x+ 8)}$$\displaystyle = \dfrac{(C+D)x+ (8C- 3D)}{(x- 3)(x+ 8)}= \dfrac{4x- 23}{(x- 3)(x+ 8)}$

Do you see that you must have C+ D= 4 and 8C- 3D= -23?

Surely you see that $\displaystyle (x- 3)(x+ 8)= x^2+ 5x- 24$?

- Nov 13th 2013, 07:45 AM #4

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- Nov 13th 2013, 07:52 AM #5

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