Answer
$y=\frac{21}{11}$
Work Step by Step
Given:
$\frac{2}{3}y+\frac{1}{2}(y-3)=\frac{y+1}{4}$
Distribute $\frac{1}{2}$ and $\frac{1}{4}$ to simply the equation to:
$\frac{2}{3}y+\frac{1}{2}y-\frac{3}{2}=\frac{1}{4}y+\frac{1}{4}$
Subtract $\frac{1}{4}y$ from both sides:
$\frac{2}{3}y+\frac{1}{2}y-\frac{3}{2}-\frac{1}{4}y=\frac{1}{4}$
Add $\frac{3}{2}$ to both sides:
$\frac{2}{3}y+\frac{1}{2}y-\frac{1}{4}y=\frac{1}{4}+\frac{3}{2}$
Group like terms:
$\frac{8}{12}y+\frac{6}{12}y-\frac{3}{12}y=\frac{1}{4}+\frac{6}{4}$
$\frac{11}{12}y=\frac{7}{4}$
Multiply both sides by $\frac{12}{11}$:
$y=\frac{7}{4}\cdot\frac{12}{11}=\frac{21}{11}$
