Difference between revisions of "1979 AHSME Problems/Problem 2"
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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. | ||
| + | |||
| + | == See also == | ||
| + | {{AHSME box|year=1979|num-b=1|num-a=3}} | ||
| + | |||
| + | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Revision as of 13:22, 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.
See also
| 1979 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
