## Algebra 1

$\frac{-1}{x+3}$ ; x $\neq$ 3, x $\neq$ 5
(x-3) (x-5). Begin by factoring the denominator $\frac{x-5}{(x+3)(x-5)}$ plug the factored denominator in. The (x-5) cancels out in numerator and denominator. Equating with zero, we consider the two equations. First, x-5=0, so x=5 Second, x-3=0, so x=3 Hence, $\frac{-1}{x+3}$ ; x $\neq$ 3, x $\neq$ 5 is the final answer.