Answer
$\{-\frac{5}{2},1\}$
Work Step by Step
Multiply both sides by the LCD $(x+4)(x-2)$.
$(x+4)(x-2)\left (\frac{5}{x+4} \right )= (x+4)(x-2)\left (4+\frac{3}{x-2}\right )$
Use distributive property.
$(x+4)(x-2)\left (\frac{5}{x+4} \right )= (x+4)(x-2)\left (4\right )+ (x+4)(x-2)\left (\frac{3}{x-2}\right )$
Cancel the common factors.
$\require{cancel}
\begin{align*}
\cancel{(x+4)}(x-2)\left (\frac{5}{\cancel{x+4}} \right )&= (x+4)(x-2)\left (4\right )+ (x+4)\cancel{(x-2)}\left (\frac{3}{\cancel{x-2}}\right )\\
\\5(x-2)&=4(x+4)(x-2) +3(x+4)
\end{align*}$
Use distributive property and FOIL method.
$5x-10=4(x^2-2x+4x-8) +3x+12$
$5x-10=4(x^2+2x-8) +3x+12$
$5x-10=4x^2+8x-32 +3x+12$
Simplify by combining like terms.
$5x-10=4x^2+11x-20$
Add $-5x+10$ to both sides.
$5x-10-5x+10=4x^2+11x-20-5x+10$
Simplify.
$0=4x^2+6x-10$
Divide the equation by $2$.
$0=2x^2+3x-5$
Rewrite $3x$ as $5x-2x$.
$0=2x^2+5x-2x-5$
Group the terms.
$0=(2x^2+5x)+(-2x-5)$
Factor out the GCF in each group.
$0=x(2x+5)-1(2x+5)$
Factor out $(2x+5)$.
$0=(2x+5)(x-1)$
Use Zero-Product Property.
$2x+5=0\quad $ or $\quad x-1 =0$
Solve for $x$.
$x=-\frac{5}{2}\quad $ or $\quad x =1$
Hence, the solution set of the given equation is $\{-\frac{5}{2},1\}$.
