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Solve \(\dfrac{3}{x^2 + 7x + 12} + \dfrac{2}{x^2 - 9} = \dfrac{4}{x^2 + x - 12}\)

 May 7, 2020
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\(\frac{3}{(x+4)(x+3)}+\frac{2}{(x-3)(x+3)}=\frac{4}{(x+4)(x-3)}\)

Multiply by \((x+3)(x-3) \)

\(\frac{3(x-3)}{(x+4)}+2=\frac{4(x+3)}{(x+4)}\)

Subtract \(\frac{3(x-3)}{(x+4)}\)

\(2=\frac{4(x+3)}{(x+4)}-\frac{3(x-3)}{(x+4)}=\frac{4x+12-3x+9}{x+4}=\frac{x+21}{x+4}\)

Thus we have \(\frac{x+21}{x+4}=2\)

Multiply by \((x+4)\) and expand

\(x+21=2x+8\) Solve for x

\(21=x+8\)

\(x=13\)

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 May 7, 2020
edited by Guest  May 7, 2020

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