How do you solve $\frac{x}{x^{2} - 36} + \frac{1}{x - 6} = \frac{1}{x + 6}$ $\frac{x}{x^2-36}+\frac{1}{x-6}=\frac{1}{x+6}$ ?

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Answer 1 · 2014-11-13T09:47:56

$\frac{x}{x^{2} - 36} + \frac{1}{x - 6} = \frac{1}{x + 6}$ $x/{x^2-36}+1/{x-6}=1/{x+6}$

by factoring out the denominator of the first term,

$\Rightarrow \frac{x}{(x - 6) (x + 6)} + \frac{1}{x - 6} = \frac{1}{x + 6}$ $=>x/{(x-6)(x+6)}+1/{x-6}=1/{x+6}$

by multiplying by $(x - 6) (x + 6)$ $(x-6)(x+6)$ ,

$\Rightarrow x + (x + 6) = (x - 6)$ $=>x+(x+6)=(x-6)$

by subtracting $x$ $x$ ,

$\Rightarrow x + 6 = - 6$ $=>x+6=-6$

by subtracting $6$ $6$ ,

$\Rightarrow x = - 12$ $=> x=-12$

I hope that this was helpful.

How do you solve \frac{x}{x^2-36}+\frac{1}{x-6}=\frac{1}{x+6}xx2−36+1x−6=1x+6\frac{x}{x^2-36}+\frac{1}{x-6}=\frac{1}{x+6}?