Answer
$\dfrac{\dfrac{6}{x^{2}+2x-15}-\dfrac{1}{x-3}}{\dfrac{1}{x+5}+1}=-\dfrac{x-1}{(x+6)(x-3)}$
Work Step by Step
$\dfrac{\dfrac{6}{x^{2}+2x-15}-\dfrac{1}{x-3}}{\dfrac{1}{x+5}+1}$
Factor the rational expression completely:
$\dfrac{\dfrac{6}{x^{2}+2x-15}-\dfrac{1}{x-3}}{\dfrac{1}{x+5}+1}=\dfrac{\dfrac{6}{(x+5)(x-3)}-\dfrac{1}{x-3}}{\dfrac{1}{x+5}+1}=...$
Evaluate the operations indicated in the numerator and in the denominator:
$...=\dfrac{\dfrac{6-(x+5)}{(x+5)(x-3)}}{\dfrac{1+x+5}{x+5}}=\dfrac{\dfrac{6-x-5}{(x+5)(x-3)}}{\dfrac{x+6}{x+5}}=\dfrac{\dfrac{-x+1}{(x+5)(x-3)}}{\dfrac{x+6}{x+5}}=...$
Simplify:
$...=\dfrac{(-x+1)(x+5)}{(x+6)(x+5)(x-3)}=\dfrac{-x+1}{(x+6)(x-3)}=-\dfrac{x-1}{(x+6)(x-3)}$
