Difference between revisions of "1979 AHSME Problems/Problem 2"
(→Solution) |
(→Solution 2) |
||
| Line 14: | Line 14: | ||
==Solution 2== | ==Solution 2== | ||
| − | Notice that we can do <math>\frac{x | + | Notice that we can do <math>\frac{x-y}{xy} = \frac{xy}{xy}</math>. We are left with <math>\frac{1}{y} - \frac{1}{x} = 1</math>. Multiply by <math>-1</math> to achieve <math>\frac{1}{x} - \frac{1}{y} = \boxed{-1}</math>. |
== See also == | == See also == | ||
Latest revision as of 22:29, 6 February 2018
Contents
Problem 2
For all non-zero real numbers 
and 
such that 
equals
Solution 1
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.
Solution 2
Notice that we can do 
. We are left with 
. Multiply by to achieve 
.
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. 
