## Intermediate Algebra (6th Edition)

$y=\left\{ 5,-3 \right\}$
Multiplying both sides by the $LCD= 30$, then the solution to the given equation, $\dfrac{y^2}{30}=\dfrac{y}{15}+\dfrac{1}{2}$, is \begin{array}{} 1(y^2)=2(y)+15(1) \\\\ y^2=2y+15 \\\\ y^2-2y-15=0 \\\\ (y-5)(y+3)=0 .\end{array} Equating each factor to zero, then $y=\left\{ 5,-3 \right\}$.