Difference between revisions of "1979 AHSME Problems/Problem 2"
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| + | == Problem 2 == | ||
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| + | For all non-zero real numbers <math>x</math> and <math>y</math> such that <math>x-y=xy, \frac{1}{x}-\frac{1}{y}</math> equals | ||
| + | |||
| + | <math>\textbf{(A) }\frac{1}{xy}\qquad | ||
| + | \textbf{(B) }\frac{1}{x-y}\qquad | ||
| + | \textbf{(C) }0\qquad | ||
| + | \textbf{(D) }-1\qquad | ||
| + | \textbf{(E) }y-x </math> | ||
| + | |||
==Solution== | ==Solution== | ||
Moving all variables to one side of the equation, we can use Simon's Favorite Factoring Trick to factor the equation into <cmath>(x+1)(y-1) = -1</cmath> Plugging in <math>-1</math> and <math>1</math> as the <math>x</math> and <math>y</math> sides respectively, we get <math>x = -2</math> and <math>y = 2</math>. Plugging this in to <math>\frac{1}{x}-\frac{1}{y}</math> gives us <math>\boxed{-1}</math> as our final answer. | Moving all variables to one side of the equation, we can use Simon's Favorite Factoring Trick to factor the equation into <cmath>(x+1)(y-1) = -1</cmath> Plugging in <math>-1</math> and <math>1</math> as the <math>x</math> and <math>y</math> sides respectively, we get <math>x = -2</math> and <math>y = 2</math>. Plugging this in to <math>\frac{1}{x}-\frac{1}{y}</math> gives us <math>\boxed{-1}</math> as our final answer. | ||
Revision as of 13:19, 3 January 2017
Problem 2
For all non-zero real numbers 
and 
such that 
equals
Solution
Moving all variables to one side of the equation, we can use Simon's Favorite Factoring Trick to factor the equation into ![\[(x+1)(y-1) = -1\]](http://latex.artofproblemsolving.com/1/a/c/1ac3e870b41f0001119009c86cbdaec0d369a6e3.png)
Plugging in 
and 
as the and sides respectively, we get 
and 
. Plugging this in to 
gives us 
as our final answer.
