Subtracting Rational Expressions

Question

$$ \frac{3a+2}{36-a^2} - \frac{a-4}{a^2-8a+12} $$

Got stuck on this equation. $$ \frac{3a+2}{(6-a)(6+a)} - \frac{a-4}{(a-6)(a-2)} $$

For this part, what do I do about the $(6-a)(6+a)$? Do I take away a $-1$ or something else?

Edit: I think I figured out the answer, thanks to all of your help.

Answer 1

Hint: $$ -\frac{a-4}{(a-6)(a-2)}=+\frac{a-4}{(6-a)(a-2)} $$

Answer 2

HINT Note that $-(a-6) = 6-a$, so $$ \frac{3a+2}{(6-a)(6+a)} - \frac{a-4}{(a-6)(a-2)} = \frac{1}{6-a} \left[ \frac{3a+2}{6+a} + \frac{a-4}{a-2}\right] $$ and now just bring the fractions inside the bracket to a common denominator.