imc

Rewrite $\frac{(x+y)}{x}$ as $\frac{x}{x}+\frac{y}{x}=1+\frac{y}{x}$. We also know that $\frac{y}{x}<0$ because $x$ and $y$ are of opposite sign. Therefore, $1+\frac{y}{x}$ is maximized when $|\frac{y}{x}|$ is minimized, which occurs when $|x|$ is the largest and $|y|$ is the smallest. This occurs at $(-4,2)$, so $\frac{x+y}{x}=1-\frac{1}{2}=\frac{1}{2}\Rightarrow \boxed{\frac{1}{2}}$.