Answer
Solution set = $\displaystyle \{\frac{1}{7}\}$.
Work Step by Step
Factor each denominator
$(x^{2}+bx+c$ is factored by searching for factors of c whose sum is b).
$x^{2}+3x-10$ = $\left[\begin{array}{lll}
factors & of & -10\\
+5,-2 & , & 5-2=3
\end{array}\right]=(x-2)(x+5)$
$x^{2}+x-6$ = $\left[\begin{array}{lll}
factors & of & -6\\
+3,-2, & , & 3-2=1
\end{array}\right]=(x+3)(x-2)$
$x^{2}-x-12$ = $\left[\begin{array}{lll}
factors & of & -12\\
-4,+3, & & -4+3=-1
\end{array}\right]=(x-4)(x+3)$
Rewrite the equation
$\displaystyle \frac{4}{(x-2)(x+5)}-\frac{1}{(x+3)(x-2)}=\frac{3}{(x-4)(x+3)}$
First, we exclude those values of x that yield a zero in any of the denominators.
$x\not\in\{-5,-3,2,4 \}\qquad (*)$
Multiply the equation with the LCD=$(x+5)(x+3)(x-2)(x-4)$
$ 4(x+3)(x-4)-1(x+5)(x-4)=3(x+5)(x-2)\qquad$ ... simplify (distribute)
$4(x^{2}-x-12)-(x^{2}+x-20)=3(x^{2}+3x-10)$
$4x^{2}-4x-48-x^{2}-x+20=3x^{2}+9x-30$
$ 3x^{2}-5x-28=3x^{2}+9x-30 \qquad$ ... add $28-3x^{2}-9x$ to both sides
$-14x=-2$
$ x=\displaystyle \frac{-2}{-14}=\frac{1}{7}\qquad$ ...Checking (*), this is a valid solution
Solution set = $\displaystyle \{\frac{1}{7}\}$.
