Answer
$a.\quad \left\{\begin{array}{l}
x\neq 3\\
x\neq-1
\end{array}\right. $
$ b.\quad$ Solution set = $\emptyset$
Work Step by Step
Factor the denominators:
For $a^{2}+bx+c$, we search for
two factors ob whose sum is c.
Here, we find $-3$ and $+1$.
$\displaystyle \frac{1}{x-3}-\frac{2}{x+1}=\frac{8}{(x-3)(x+1)}$
$a.$
The denominators in the equation must not be zero:
$\left\{\begin{array}{l}
x-3\neq 0\\
x+1\neq 0\\
(x-3)(x+1)\neq 0
\end{array}\right\}\Rightarrow\left\{\begin{array}{l}
x\neq 3\\
x\neq-1
\end{array}\right. \qquad(*)$
$b.$
Multiply both sides with the LCD,$\quad (x-3)(x+1)$
$(x-3)(x+1) \displaystyle \left[ \frac{1}{x-3}-\frac{2}{x+1} \right]= (x-3)(x+1) \displaystyle \left[\frac{8}{(x-3)(x+1)} \right]\quad$ ... distribute and simplify
$ 1(x+1)-2(x-3)=8\quad$ ... distribute and simplify
$x+1-2x+6=8$
$-x+7=8\quad$ ... add $-7$ to both sides
$-x=1$
$ x=-1 \quad$ ... does not satisfy (*), not a valid solution.
Solution set = $\emptyset$
