Answer
$\{-23,2\}$.
Work Step by Step
Determine the Least Common Denominator ($LCD$) so that we clear the fractions.
$LCD=4(x-1)(x+2)$.
Multiply the given equation by $4(x-1)(x+2)$.
$\Rightarrow 4(x-1)(x+2) \left (\frac{2}{x-1}\right )=4(x-1)(x+2)\left (\frac{1}{4}+\frac{7}{x+2}\right )$
Use distributive property and cancel common terms.
$\Rightarrow 4(x-1)(x+2) \left (\frac{2}{x-1}\right )=4(x-1)(x+2)\left (\frac{1}{4}\right )+4(x-1)(x+2)\left (\frac{7}{x+2}\right )$
Cancel common factors.
$\Rightarrow 8(x+2) =(x-1)(x+2)+28(x-1)$
Use the FOIL method and the distributive property.
$\Rightarrow 8x+16 =x^2+2x-1x-2+28x-28$
Add like terms.
$\Rightarrow 8x+16 =x^2+29x-30$
Add $-8x-16$ to both sides.
$\Rightarrow 8x+16-8x-16 =x^2+29x-30-8x-16$
Simplify.
$\Rightarrow 0=x^2+21x-46$
Rewrite the middle term as $21x$ as $23x-2x$.
$\Rightarrow 0=x^2+23x-2x-46$
Group the terms.
$\Rightarrow 0=(x^2+23x)+(-2x-46)$
Factor each group.
$\Rightarrow 0=x(x+23)-2(x+23)$
Factor out $(x+23)$.
$\Rightarrow 0=(x+23)(x-2)$
By using zero product rule set each factor equal to zero.
$\Rightarrow x+23=0$ or $x-2=0$
Isolate $x$.
$\Rightarrow x=-23$ or $x=2$
The solution set is $\{-23,2\}$.
Note: The equation is defined for all real values of $x$ except the zeros of the denominators which are: $-2$ and $1$, therefore the solution set we found is correct.
