Difference between revisions of "2017 AMC 10A Problems/Problem 5"
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| − | Let the two real numbers be <math>x,y</math>. We are given that | + | Let the two real numbers be <math>x,y</math>. We are given that <math>x+y=4xy,</math> and dividing both sides by <math>xy</math>, <math>\frac{x}{xy}+\frac{y}{xy}=4.</math> |
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| − | and dividing both sides by <math>xy</math>, | ||
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<cmath>\frac{1}{y}+\frac{1}{x}=\boxed{\textbf{(C) } 4}.</cmath> | <cmath>\frac{1}{y}+\frac{1}{x}=\boxed{\textbf{(C) } 4}.</cmath> | ||
Revision as of 14:15, 9 February 2017
Problem
The sum of two nonzero real numbers is 4 times their product. What is the sum of the reciprocals of the two numbers?
Solution
Let the two real numbers be 
. We are given that 
and dividing both sides by 
, 
![\[\frac{1}{y}+\frac{1}{x}=\boxed{\textbf{(C) } 4}.\]](http://latex.artofproblemsolving.com/e/a/1/ea1dadbf53587b5968e46e59c6f61f87125eeed4.png)
Note: we can easily verify that this is the correct answer; for example, 1/2 and 1/2 work, and the sum of their reciprocals is 4.
See Also
| 2017 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 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 | ||
| All AMC 10 Problems and Solutions | ||
| 2017 AMC 12A (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 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
