Answer
the set of all real numbers except $2$ and $-1$.
Work Step by Step
Expressing the terms of the given equation, $
\dfrac{x-8}{x^2-x-2}+\dfrac{2}{x-2}=\dfrac{3}{x+1}
,$ in factored form results to
\begin{array}{l}\require{cancel}
\dfrac{x-8}{(x-2)(x+1)}+\dfrac{2}{x-2}=\dfrac{3}{x+1}
.\end{array}
Multiplying both sides by the $LCD=
(x-2)(x+1)
$, then the solution to the equation above is
\begin{array}{l}
1(x-8)+(x+1)(2)=(x-2)(3)
\\\\
x-8+2x+2=3x-6
\\\\
(x+2x-3x)=-6+8-2
\\\\
0=0
\text{ (TRUE)}
.\end{array}
Since the solution above ended with a TRUE statement, then the values of $x$ that satisfy the original equation is the set of all real numbers. But when $x=\{2,-1\}$, the denominator becomes $0$. Hence, the solution is the $\text{
the set of all real numbers except $2$ and $-1$
}.$
