Difference between revisions of "2014 AMC 10B Problems/Problem 9"
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==Solution== | ==Solution== | ||
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| + | Multiply numerator and denominator by <math>w+z</math> to get <math>\frac{z+w}{z-w}=2014</math>. Then since <math>z+w=w+z</math> and <math>w-z=-(z-w)</math>, <math>\frac{w+z}{w-z}=-\frac{z+w}{z-w}=-2014</math>, or choice <math>\boxed{A}</math>. | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2014|ab=B|num-b=8|num-a=10}} | {{AMC10 box|year=2014|ab=B|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 15:49, 20 February 2014
Problem
For real numbers 
and 
, ![\[\cfrac{\frac{1}{w} + \frac{1}{z}}{\frac{1}{w} - \frac{1}{z}} = 2014.\]](http://latex.artofproblemsolving.com/6/9/a/69a1b108beb9e11391b1e636b5839f88ac0d9af1.png)
What is 
?
Solution
Multiply numerator and denominator by 
to get 
. Then since 
and 
, 
, or choice 
.
See Also
| 2014 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 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. 
