Answer
$a.\quad \left\{\begin{array}{l}
x\neq 4\\
x\neq-2
\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 $-4$ and $+2$.
$\displaystyle \frac{1}{x-4}-\frac{5}{x+2}=\frac{6}{(x-4)(x+2)}$
$a.$
The denominators in the equation must not be zero:
$\left\{\begin{array}{l}
x-4\neq 0\\
x+2\neq 0\\
(x-4)(x+2)\neq 0
\end{array}\right\}\Rightarrow\left\{\begin{array}{l}
x\neq 4\\
x\neq-2
\end{array}\right. \qquad(*)$
$b.$
Multiply both sides with the LCD,$\quad (x-4)(x+2)$
$(x-4)(x+2) \left[ \displaystyle \frac{1}{x-4}-\frac{5}{x+2} \right]= (x-4)(x+2) \displaystyle \left[\frac{6}{(x-4)(x+2)} \right]\quad$ ... distribute and simplify
$ 1(x+2)-5(x-4)=6\quad$ ... distribute and simplify
$x+2-5x+20=6$
$-4x+22=6\quad$ ... add $-22$ to both sides
$-4x=-16$
$ x=4 \quad$ ... does not satisfy (*), not a valid solution.
Solution set = $\emptyset$
