Distributing multiplication of rational functions I am having trouble distributing with fractions. This
$$ \left(\frac{1}{(x + 3)} + \frac{(x + 3)}{(x - 3)}\right)\, (9 - x^2) = -\frac{(x^2-3)}{9-x^2}(9-x^2) $$
has the answer $\left\{\left[x = - \frac{9}{7}\right]\right\}$.
I start solving by cancelling $(9-x^2)$ on the right hand side, and multiplying $(9-x^2)$ with the terms on the left hand side by foiling (multiplying term by term). So the first step is
$$9\times\frac{1}{x+3}+\frac{9 (x+3)}{x-3}-x^2\times\frac{1}{x+3}-\frac{x^2 (x+3) }{x-3}=-x^2-3$$
This is apparently wrong because it has another answer, $\left\{\left[x = - \frac{3}{7}\right]\right\}$ What am I doing wrong?

Cancelling $(9-x^2)$ on the RHS leaves you with $-(x^2-3)$. Next I'd recommend finding a common denominator and combining the fractions inside the parentheses on the LHS. You should get $$ \frac{1}{\left(x + 3\right)} + \frac{\left(x + 3\right)}{\left(x - 3\right)} =  \frac{x-3+(x+3)^2}{x^2 - 9}$$ which means the LHS will simplify to $-(x-3+(x+3)^2)$, leaving you with $$-(x-3+(x+3)^2) = -(x^2-3)$$ Can you take it from here? If I carry out the rest of this algebra I find that $x = \frac{-9}{7}$.
