Answer
$\displaystyle \frac{1-x}{(x+6)(x-3)}, \qquad x\neq-6,-5,3$
Work Step by Step
Factor all denominators of the complex rational equation.
(For $x^{2}+2x-15$, we find factors of $-15$ that add to $+2$.
These are $+5$ and $-3.$)
$...=\displaystyle \frac{\frac{6}{(x+5)(x-3)}-\frac{1}{x-3}}{\frac{1}{x+5}+1}=...$
Complex rational expressions have rational expressions in the numerator or/and in the denominator.
Here, multiplying both the numerator and denominator with $(x+5)(x-3)$ would get rid of these fractions.
We have to exclude any value that yields 0 in any denominator.
Excluded: $x=-5,3$.
The denominator of the expression itself can not be zero,
so we will exclude any values for which it is 0:
$\displaystyle \frac{1}{x+5}+1=0\quad/\times(x+5)$
$1+(x+5)=0$
$x+6=0\qquad -6$ is also excluded.
$...=\displaystyle \frac{\frac{6}{(x+5)(x-3)}-\frac{1}{x-3}}{\frac{1}{x+5}+1}\times\frac{(x+5)(x-3)}{(x+5)(x-3)}$
$=\displaystyle \frac{6-1(x+5)}{(x-3)+(x+5)(x-3)}$
$=\displaystyle \frac{1-x}{x-3+x^{2}+2x-15}$
$=\displaystyle \frac{1-x}{x^{2}+3x-18}\qquad$ factors of -18 with sum 3 are $+6$ and $-3$
$=\displaystyle \frac{1-x}{(x+6)(x-3)}, \qquad x\neq-6,-5,3$
