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 $\boxed{(A) -3}$

-solution by jmania