2004 AMC 10A Problem 15

1. The fraction is easier to think about if you split it:
$$\frac{x+y}{x} = \frac{x}{x} + \frac{y}{x} = 1 + \frac{y}{x}.$$
2. Here $y$ is always positive and $x$ is always negative, so $\dfrac{y}{x}$ is always negative. The whole expression is $1$ minus something.
3. To make $1$ minus something as big as possible, make that something as small as possible. The size of $\dfrac{y}{x}$ is $y$ divided by the size of $x$. That is smallest when $y$ is as small as possible and $x$ is as far from $0$ as possible.
4. So take $y = 2$ and $x = -4$:
$$1 + \frac{2}{-4} = 1 - \frac12 = \frac12.$$

The answer is $\boxed{\textbf{(D)}\ \frac12}$.

Rewriting the fraction as $1 + \frac{y}{x}$ separates the fixed part from the part that changes, and makes it clear which way each variable should go. For a fraction like this, the largest and smallest values happen when each variable is at one end of its allowed range, so the corners are the places to look.

Try all four corners, where each variable is at one end of its range:

The largest is $\tfrac12$. Choice (E), $1$, cannot happen, because it would need $y = 0$.

Split the fraction:
$$\frac{x+y}{x} = 1 + \frac{y}{x}.$$
Since $y > 0$ and $x < 0$, the term $\frac{y}{x}$ is always negative, so the expression is always less than $1$. To make it as large as possible we want $\frac{y}{x}$ as close to $0$ as possible, i.e. $\left|\frac{y}{x}\right| = \frac{y}{|x|}$ as small as possible.

That means the smallest $y$ and the largest $|x|$: $y = 2$ and $x = -4$. Then
$$1 + \frac{2}{-4} = 1 - \frac12 = \frac12.$$

The answer is $\boxed{\textbf{(D)}\ \frac12}$.