How do you solve for x if $\frac{1}{x - 3} + \frac{1}{x + 3} = \frac{4}{x^{2} - 9}$ $1/(x-3)+1/(x+3)=4/(x^2-9)$ ?

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Answer 1 · 2015-02-03T18:48:11

Remembering that $x^{2} - 9 = (x + 3) (x - 3)$ $x^2-9=(x+3)(x-3)$
you have:
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How do you solve for x if 1x−3+1x+3=4x2−91/(x-3)+1/(x+3)=4/(x^2-9)?