Intermediate Algebra for College Students (7th Edition)

Solution set = $\emptyset$. (no solution)
First, we exclude those values of x that yield a zero in any of the denominators. $x\not\in\{-2,+2 \}\qquad (*)$ The last denominator $x^{2}-4$ is factored as $(x-2)(x+2)$ Multiply the equation with the LCD=$(x-2)(x+2)$ $1(x+2)+1(x-2)=4\qquad$ ... simplify (distribute) $x+2+x-2=4$ $2x=4$ $x=2 \qquad$ ...Checking (*), this is not a valid solution Solution set = $\emptyset$. (no solution)