For the first term, we have $x^{2}-9 = (x+3)(x-3)$. For the second term, we have $x^{2}+3x = x(x+3)$. For the third term, we have $x^{2}-3x = x(x-3)$. So, the equation becomes:
$$
\frac{x}{(x+3)(x-3)}-\frac{x-4}{x(x+3)}=\frac{10}{x(x-3)}
$$
Show more…