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

Revision as of 13:15, 6 August 2018

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

Revision as of 09:38, 21 September 2017 (view source) Pega969 (talk \| contribs) (→‎Solution) ← Older edit		Revision as of 13:15, 6 August 2018 (view source) Jmania (talk \| contribs) m (→‎Solution) Newer edit →
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	Negating the top equation gives us <math>x-y=-3</math>. We seek <math>x-y</math>, so the answer is <math>\boxed{(A) -3}</math>		Negating the top equation gives us <math>x-y=-3</math>. We seek <math>x-y</math>, so the answer is <math>\boxed{(A) -3}</math>
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		+	-solution by jmania