Suppose that $x$ and $y$ are real numbers with $-4 \leq x \leq-2$ and $2 \leq y \leq 4$. The greatest possible value of $\frac{x+y}{x}$ is
(A) 1
(B) -1
(C) $-\frac{1}{2}$
(D) 0
(E) $\frac{1}{2}$
We note that $\frac{x+y}{x}=\frac{x}{x}+\frac{y}{x}=1+\frac{y}{x}$.

The greatest possible value of $\frac{x+y}{x}=1+\frac{y}{x}$ thus occurs when $\frac{y}{x}$ is as great as possible.

Since $x$ is always negative and $y$ is always positive, then $\frac{y}{x}$ is negative.

Therefore, for $\frac{y}{x}$ to be as great as possible, it is as least negative as possible (i.e. closest to 0 as possible).

Since $x$ is negative and $y$ is positive, this happens when $x$ is as negative as possible and $y$ is as small as possible - that is, when $x=-4$ and $y=2$.

Therefore, the greatest possible value of $\frac{x+y}{x}$ is $1+\frac{2}{-4}=\frac{1}{2}$.

ANSwer: (E)