Answer
Solution set = $\displaystyle \{-\frac{4}{3}\}$.
Work Step by Step
Factor each denominator
$(x^{2}+bx+c$ is factored by searching for factors of c whose sum is b).
$x^{2}+3x-10$ = $\left[\begin{array}{lll}
factors & of & -10\\
+5, & -2, & 5-2=3
\end{array}\right]=(x-2)(x+5)$
$x^{2}+9x+20$ = $\left[\begin{array}{lll}
factors & of & 20\\
+5, & +4, & 5+4=9
\end{array}\right]=(x+4)(x+5)$
$x^{2}+2x-8$ = $\left[\begin{array}{lll}
factors & of & -8\\
-2, & +4, & -2+4=2
\end{array}\right]=(x+4)(x-2)$
Rewrite the equation
$\displaystyle \frac{4}{(x-2)(x+5)}+\frac{1}{(x+4)(x+5)}=\frac{2}{(x+4)(x-2)}$
First, we exclude those values of x that yield a zero in any of the denominators.
$x\not\in\{-5,-4,2 \}\qquad (*)$
Multiply the equation with the LCD=$(x-2)(x+5)(x+4)$
$ 4(x+4)+1(x-2)=2(x+5)\qquad$ ... simplify (distribute)
$4x+16+x-2=2x+10$
$ 5x+14=2x+10 \qquad$ ... add $-14-2x$ to both sides
$3x=-4$
$ x=-\displaystyle \frac{4}{3}\qquad$ ...Checking (*), this is a valid solution
Solution set = $\displaystyle \{-\frac{4}{3}\}$.
