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

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<math> \textbf{(A)}\ -3 \qquad\textbf{(B)}\ -1 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 5 </math>
 
<math> \textbf{(A)}\ -3 \qquad\textbf{(B)}\ -1 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 5 </math>
 
==Solution==
 
==Solution==
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We have two cases to consider: x is positive or x is negative. If x is positive, we have:
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We have two cases to consider: <math>x</math> is positive or <math>x</math> is negative. If <math>x</math> is positive, we have <math>x+y=3</math> and <math>xy+x^3=0</math>
  
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x+y=3
+
Solving for <math>y</math> in the top equation gives us <math>3-x</math>. Plugging this in gives us: <math>x^3-x^2+3x=0</math>.  Since we're told <math>x</math> is not zero, we can divide by <math>x</math>, giving us: <math>x^2-x+3=0</math>
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xy+x^3=0
 
  
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Solving for y in the top equation gives us 3-x. Plugging this in gives us:
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The discriminant of this is <math>(-1)^2-4(1)(3)=-11</math>, which means the equation has no real solutions.  
  
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x^3-x^2+3x=0
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We conclude that <math>x</math> is negative. In this case <math>-x+y=3</math> and <math>-xy+x^3=0</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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Since we're told x is not zero, we can divide by x, giving us:
+
-solution by jmania
  
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x^2-x+3=0
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==See Also==
  
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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:
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{{AHSME box|year=1994|num-b=24|num-a=26}}
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{{MAA Notice}}
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-x+y=3
 
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-xy+x^3=0
 
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Negating the top equation gives us x-y=-3. We seek x-y, so the answer is A) -3
 

Latest revision as of 02:40, 28 May 2021

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

See Also

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

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