1994 AHSME Problems/Problem 25

Problem

If $x$ and $y$ are non-zero real numbers such that $|x|+y=3 \qquad \text{and} \qquad |x|y+x^3=0,$ then the integer nearest to $x-y$ is

$\textbf{(A)}\ -3 \qquad\textbf{(B)}\ -1 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 5$

We have two cases to consider: x is positive or x is negative. If x is positive, we have:

x+y=3 xy+x^3=0

Solving for y in the top equation gives us 3-x. Plugging this in gives us:

x^3-x^2+3x=0

Since we're told x is not zero, we can divide by x, giving us:

x^2-x+3=0

The discriminant of this is (-1)^2-4(1)(3)=-11, which means the equation has no real solutions. Therefore, x is negative. Now we have:

-x+y=3 -xy+x^3=0

Negating the top equation gives us x-y=-3. We seek x-y, so the answer is A) -3