Difference between revisions of "1978 AHSME Problems/Problem 3"
(Created page with "== Problem 3 == For all non-zero numbers <math>x</math> and <math>y</math> such that <math>x = 1/y</math>, <math>\left(x-\frac{1}{x}\right)\left(y+\frac{1}{y}\right)</math> e...") |
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Simplifying, we get <math>(x-y)(y+x)</math>. Multiplying, we get <math>\boxed{\textbf{(D) }x^2-y^2}</math> | Simplifying, we get <math>(x-y)(y+x)</math>. Multiplying, we get <math>\boxed{\textbf{(D) }x^2-y^2}</math> | ||
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~awin | ~awin | ||
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| + | ==See Also== | ||
| + | {{AHSME box|year=1978|num-b=2|num-a=4}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 10:59, 13 February 2021
Problem 3
For all non-zero numbers 
and 
such that 
, 
equals
Solution 1
Using substitution, we can substitute y into the equation in the first parentheses.
Therefore, we'll get ![\[\left(x-\frac{1}{\frac{1}{y}}\right)\left(y+\frac{1}{y}\right)\]](http://latex.artofproblemsolving.com/f/1/e/f1edb160fe876c934f42669cd4e6f20ab3f0eb36.png)
Because 
, we can also see that 
. Using substitution again, we can substitute x into the second equation getting
![\[\left(x-\frac{1}{\frac{1}{y}}\right)\left(y+\frac{1}{\frac{1}{x}}\right)\]](http://latex.artofproblemsolving.com/f/c/d/fcdfcbf7e311050f8da32449b234f7ca1f355042.png)
Simplifying, we get 
. Multiplying, we get 
~awin
See Also
| 1978 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 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. 
