I know this is easy, but i was wondering what the method is for doing this..
thank you!
$\displaystyle = \ \frac{6(x+5)-8(x+3)}{6(x+3)} \times \frac{1}{(x-3)}$
$\displaystyle =\\ \frac{6x+30-8x-24}{6(x+3)(x-3)}$
$\displaystyle =\\ \frac{-2x+6}{6(x+3)(x-3)}$
$\displaystyle = \\ \frac{-2(x-3)}{6(x+3)(x-3)}$
$\displaystyle =\\ \frac{-1}{3(x+3)}$