Difference between revisions of "1979 AHSME Problems/Problem 2"
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\textbf{(E) }y-x </math> | \textbf{(E) }y-x </math> | ||
− | ==Solution== | + | ==Solution 1== |
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. | ||
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+ | ==Solution 2== | ||
+ | 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 23:29, 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 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.