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

m (Solution)
(Solution)
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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==
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:
+
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>
  
<math>x+y=3</math>
+
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>
<math>xy+x^3=0</math>
 
  
Solving for <math>y</math> in the top equation gives us <math>3-x</math>. Plugging this in gives us:
+
The discriminant of this is <math>(-1)^2-4(1)(3)=-11</math>, which means the equation has no real solutions.  
  
<math>x^3-x^2+3x=0</math>
+
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>
  
Since we're told <math>x</math> is not zero, we can divide by <math>x</math>, giving us:
+
-solution by jmania
  
<math>x^2-x+3=0</math>
+
==See Also==
  
The discriminant of this is <math>(-1)^2-4(1)(3)=-11</math>, which means the equation has no real solutions. Therefore, <math>x</math> is negative. Now we have:
+
{{AHSME box|year=1994|num-b=24|num-a=25}}
 
+
{{MAA Notice}}
<math>-x+y=3</math>
 
<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>
 
 
 
-solution by jmania
 

Revision as of 03:39, 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 (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
Problem 25
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