## Intermediate Algebra for College Students (7th Edition)

Solution set = $\{3\}$.
The LHS denominator $x^{2}+2x-8$ is factored by searching for two factors of $c=-8$ whose sum is $b=2$ ... we find $+4$ and $-2$ $x^{2}+2x-8=(x-2)(x+4)$ First, we exclude those values of x that yield a zero in any of the denominators. $x\not\in\{-4,2 \}\qquad (*)$ Multiply the equation with the LCD=$(x-2)(x+4)$ $(2x-1)+2(x-2)=x+4\qquad$ ... simplify (distribute) $2x-1+2x-4=x+4$ $4x-5=x+4\qquad$ ... add $5-x$ to both sides $3x=9$ $x=3 \qquad$ ...Checking (*), this is a valid solution Solution set = $\{3\}$.