Distributing multiplication of rational functions

Question

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?

Answer 1

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 $x = \frac{-9}{7}$.

Answer 2

$\left(\frac{1}{x+3} + \frac{x+3}{x-3} \right) ( 9 -x^2) -x^2 -3$

It will be

$\frac{9-x^2}{x+3} + \frac{(x+3)(9-x^2)}{x-3} = -x^2 - 3 $

after that

$\frac{(9-x^2)(x-3)}{(x+3)(x-3)} + \frac{(x+3)(x+3)(9-x^2)}{(x-3)(x+3)} = -x^2 -3 $

Note that $9-x^2 = -(x^2 - 9) = -(x+3)(x-3)$, i.e., you can cancel it with the denominator