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 12:19, 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 Plugging in and as the and sides respectively, we get and . Plugging this in to gives us as our final answer.