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== | ||
| − | We have two cases to consider: x is positive or x is negative. If x 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: |
| − | x+y=3 | + | <math>x+y=3</math> |
| − | xy+x^3=0 | + | <math>xy+x^3=0</math> |
| − | Solving for y in the top equation gives us 3-x. Plugging this in gives us: | + | Solving for <math>y</math> in the top equation gives us <math>3-x</math>. Plugging this in gives us: |
| − | x^3-x^2+3x=0 | + | <math>x^3-x^2+3x=0</math> |
| − | Since we're told x is not zero, we can divide by x, giving us: | + | Since we're told <math>x</math> is not zero, we can divide by <math>x</math>, giving us: |
| − | x^2-x+3=0 | + | <math>x^2-x+3=0</math> |
| − | 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: | + | 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: |
| − | -x+y=3 | + | <math>-x+y=3</math> |
| − | -xy+x^3=0 | + | <math>-xy+x^3=0</math> |
| − | Negating the top equation gives us x-y=-3. We seek x-y, so the answer is A) -3 | + | 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> |
Revision as of 10:38, 21 September 2017
Problem
If 
and 
are non-zero real numbers such that
![\[|x|+y=3 \qquad \text{and} \qquad |x|y+x^3=0,\]](http://latex.artofproblemsolving.com/1/e/6/1e64d1bb45926f49276d0f83ec302a83f25f2f45.png)
then the integer nearest to 
is
Solution
We have two cases to consider: is positive or is negative. If is positive, we have:
Solving for in the top equation gives us 
. Plugging this in gives us:
Since we're told is not zero, we can divide by , giving us:
The discriminant of this is 
, which means the equation has no real solutions. Therefore, is negative. Now we have:
Negating the top equation gives us 
. We seek , so the answer is 
