## Intermediate Algebra (6th Edition)

$x=3$
Multiplying both sides by the $LCD= x ,$ then the solution to the given equation, $\dfrac{x^2+8}{x}-1=\dfrac{2(x+4)}{x} ,$ is \begin{array}{l}\require{cancel} 1(x^2+8)-x(1)=1(2(x+4)) \\\\ x^2+8-x=2x+8 \\\\ x^2+(-x-2x)+(8-8)=0 \\\\ x^2-3x=0 \\\\ x(x-3)=0 \\\\ x=\{ 0,3 \} .\end{array} Upon checking, only $x=3$ satisfies the original equation.