## Intermediate Algebra (6th Edition)

$y=-36$ and $y=24$
We are given that $|\frac{y}{3}+2|=10$. So, $\frac{y}{3}+2=10$ and $\frac{y}{3}+2=-10$. We can solve both of these equations for y. For $\frac{y}{3}+2=10$, subtract 2 from both sides of the equation. $\frac{y}{3}=8$ Multiply both sides of the equation by 3. y=24 For $\frac{y}{3}+2=-10$, subtract 2 from both sides of the equation. $\frac{y}{3}=-12$ Multiply both sides of the equation by 3. y=-36