Difference between revisions of "1994 AHSME Problems/Problem 25"

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==See Also==
==See Also==
{{AHSME box|year=1994|num-b=24|num-a=25}}
{{AHSME box|year=1994|num-b=24|num-a=26}}
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{{MAA Notice}}

Latest revision as of 03:40, 28 May 2021


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 $x+y=3$ 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

See Also

1994 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
Problem 26
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