Difference between revisions of "2017 AMC 10A Problems/Problem 5"
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<cmath>\frac{x}{xy}+\frac{y}{xy}=4</cmath> | <cmath>\frac{x}{xy}+\frac{y}{xy}=4</cmath> | ||
| − | <cmath>\frac{1}{y}+\frac{1}{x}=\boxed{4 (C)}.</cmath> | + | <cmath>\frac{1}{y}+\frac{1}{x}=\boxed{4 (\textbf{C})}.</cmath> |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2017|ab=A|num-b=4|num-a=6}} | {{AMC10 box|year=2017|ab=A|num-b=4|num-a=6}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 17:28, 8 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
![\[x+y=4xy,\]](http://latex.artofproblemsolving.com/3/7/2/3725abeb6b7352c14ac4f65351d5610aa2796ae1.png)
and dividing both sides by 
,
![\[\frac{x}{xy}+\frac{y}{xy}=4\]](http://latex.artofproblemsolving.com/4/6/6/466dc3f07a1168b0568096aef338d7c267efb651.png)
![\[\frac{1}{y}+\frac{1}{x}=\boxed{4 (\textbf{C})}.\]](http://latex.artofproblemsolving.com/d/c/d/dcd9a01e3749bd98af810ed96c8fb7e759fd0f11.png)
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
