OpenSAT

To solve for $x + y$, we can add the two equations together. The $y$ terms will cancel out, leaving us with $3x = 16$. Solving for $x$, we get $x = \frac{16}{3}$. Substituting this value back into either of the original equations, we can solve for $y$. Substituting into the first equation, we get $2 \cdot \frac{16}{3} + 3y = 12$, or $\frac{32}{3} + 3y = 12$. Subtracting $\frac{32}{3}$ from both sides, we get $3y = 12 - \frac{32}{3}$, or $3y = \frac{4}{3}$. Dividing both sides by 3, we get $y = \frac{4}{9}$. Therefore, $x + y = \frac{16}{3} + \frac{4}{9} = 6$.