Question
Solve the rational equation. Be sure to check for extraneous solutions.$$\frac{1}{x+3}+\frac{1}{x-3}=\frac{x^{2}-3}{x^{2}-9}$$
Step 1
The common denominator is $(x+3)(x-3)$, which is the factored form of $x^{2}-9$. So, we rewrite the equation as follows: $$\frac{1(x-3)}{(x+3)(x-3)}+\frac{1(x+3)}{(x+3)(x-3)}=\frac{x^{2}-3}{x^{2}-9}$$ Show more…
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