Answer
The solution set of the equation is $\left\{ 7 \right\}$.
Work Step by Step
Consider the equation $\frac{4}{x+2}+\frac{2}{x-4}=\frac{30}{{{x}^{2}}-2x-8}$.
The equation is not defined for $x=-2\text{ and }x=4$.
First of all, simplify the equation.
That is,
$\begin{align}
& \frac{4}{x+2}+\frac{2}{x-4}=\frac{30}{{{x}^{2}}-2x-8} \\
& \frac{4\left( x-4 \right)+2\left( x+2 \right)}{\left( x+2 \right)\left( x-4 \right)}=\frac{30}{{{x}^{2}}-2x-8} \\
& \frac{4x-16+2x+4}{{{x}^{2}}-2x-8}=\frac{30}{{{x}^{2}}-2x-8}
\end{align}$
Now, combine the like terms as follows
$\frac{6x-12}{{{x}^{2}}-2x-8}=\frac{30}{{{x}^{2}}-2x-8}$
Now, multiply both sides by ${{x}^{2}}-2x-8$ and get,
$\begin{align}
& \frac{6x-12}{{{x}^{2}}-2x-8}\cdot \left( {{x}^{2}}-2x-8 \right)=\frac{30}{{{x}^{2}}-2x-8}\cdot \left( {{x}^{2}}-2x-8 \right) \\
& 6x-12=30 \\
& 6x=30+12 \\
& 6x=42
\end{align}$
Now, divide both sides by $6$ and get,
$\begin{align}
& \frac{6x}{6}=\frac{42}{6} \\
& x=7
\end{align}$
Check:
Substitute the value of $x=7$ in $\frac{4}{x+2}+\frac{2}{x-4}=\frac{30}{{{x}^{2}}-2x-8}$ to verify the solution.
$\begin{align}
& \frac{4}{x+2}+\frac{2}{x-4}=\frac{30}{{{x}^{2}}-2x-8} \\
& \frac{4}{7+2}+\frac{2}{7-4}\overset{?}{\mathop{=}}\,\frac{30}{{{7}^{2}}-2\left( 7 \right)-8} \\
& \text{ }\frac{4}{9}+\frac{2}{3}\overset{?}{\mathop{=}}\,\frac{30}{49-14-8} \\
& \text{ }\frac{12+18}{27}\overset{?}{\mathop{=}}\,\frac{30}{27} \\
\end{align}$
Further solve and get,
$\frac{30}{27}=\frac{30}{27}$
That is true.
Hence, the solution set of the equation is $\left\{ 7 \right\}$.
