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

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

Solution

We have two cases to consider: $x$ is positive or $x$ is negative. If $x$ is positive, we have $x+y=3$ and $xy+x^3=0$

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.

We conclude that $x$ is negative. In this case $-x+y=3$ and $-xy+x^3=0$ . Negating the top equation gives us $x-y=-3$ . We seek $x-y$ , so the answer is $\boxed{(A) -3}$

-solution by jmania

1994 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Problem 26
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination

Page

Toolbox

Search

1994 AHSME Problems/Problem 25

Problem

Solution

See Also