Let's start by multiplying the first inequality by two. This gives us $6x+4y\le14$. That means that $6x+4y=14$ and $2x+4y=8$. Subtracting the second equation from the first gives us $4x=6$, which means $x=\frac{3}{2}$. Substituting into the second equation gives us $2\cdot\frac{3}{2}+4y=8$. Which reduces to $3+4y=8$. So $4y=5$ and $y=\frac{5}{4}$. So $x+y=\frac{3}{2}+\frac{5}{4}=\frac{11}{4}$.


Let $x$ and $y$ be real numbers such that $3x + 2y \le 7$ and $2x + 4y \le 8.$  Find the largest possible value of $x + y.$