Answer
$\{-3,2\}$.
Work Step by Step
Factor the denominators in order to determine the Least Common Denominator ($LCD$) so that we clear the fractions.
Factor $x^2+x-20$.
Rewrite the middle term $x$ as $5x-4x$.
$\Rightarrow x^2+5x-4x-20$.
Group the terms.
$\Rightarrow (x^2+5x)+(-4x-20)$.
Factor each group.
$\Rightarrow x(x+5)-4(x+5)$.
Factor out $(x+5)$.
$\Rightarrow (x+5)(x-4)$.
Back substitute into the given equation.
$\Rightarrow \frac{2x+7}{x+5}-\frac{x-8}{x-4}=\frac{x+18}{(x+5)(x-4)}$
Multiply the equation by $LCD=(x+5)(x-4)$.
$\Rightarrow (x+5)(x-4)\left (\frac{2x+7}{x+5}-\frac{x-8}{x-4}\right )=(x+5)(x-4)\left (\frac{x+18}{(x+5)(x-4)}\right)$
Use the distributive property.
$\Rightarrow (x+5)(x-4)\left (\frac{2x+7}{x+5}\right )-(x+5)(x-4)\left (\frac{x-8}{x-4}\right )=(x+5)(x-4)\left (\frac{x+18}{(x+5)(x-4)}\right)$
Cancel the common terms.
$\Rightarrow (x-4)(2x+7)-(x+5)(x-8)=x+18$
Use the FOIL method.
$\Rightarrow 2x^2+7x-8x-28-x^2+8x-5x+40=x+18$
Add like terms.
$\Rightarrow x^2+2x+12=x+18$
Add $-x-18$ to both sides.
$\Rightarrow x^2+2x+12-x-18=x+18-x-18$
Simplify.
$\Rightarrow x^2+x-6=0$
Rewrite the middle term $x$ as $3x-2x$.
$\Rightarrow x^2+3x-2x-6=0$
Group the terms.
$\Rightarrow (x^2+3x)+(-2x-6)=0$
Factor each group.
$\Rightarrow x(x+3)-2(x+3)=0$
Factor out $(x+3)$.
$\Rightarrow (x+3)(x-2)=0$
By using the Zero Product Rule set each factor equal to zero.
$\Rightarrow x+3=0$ or $x-2=0$
Isolate $x$.
$\Rightarrow x=-3$ or $x=2$
The solution set is $\{-3,2\}$.
Note: The equation is defined for all real values of $x$ except the zeros of the denominators which are $-5$ and $4$. Since the solution set does not contain any of these values, it is correct.
