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

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Answer 1 · 2014-11-13T09:47:56

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

by factoring out the denominator of the first term,

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

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

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

by subtracting $x$ ,

$=>x+6=-6$

by subtracting $6$ ,

$=> x=-12$

I hope that this was helpful.

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