Difference between revisions of "2014 AMC 10B Problems/Problem 9"
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Multiply the numerator and denominator of the LHS by <math>wz</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>. | Multiply the numerator and denominator of the LHS by <math>wz</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>. | ||
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+ | ==Solution 2== | ||
+ | Muliply both sides by <math>\left(\frac{1}{w}-\frac{1}{z}\right)</math> to get <math>\frac{1}{w}+\frac{1}{z}=2014\left(\frac{1}{w}-\frac{1}{z}\right)</math>. Then, add <math>2014\cdot\frac{1}{z}</math> to both sides and subtract <math>\frac{1}{w}</math> from both sides to get <math>2015\cdot\frac{1}{z}=2013\cdot\frac{1}{w}</math>. Then, we can plug in the most simply values for z and w (<math>2015</math> and <math>2013</math>, respectively, and find <math>\frac{2013+2015}{2013-2015}=\frac{2(2014)}{-2}=-2014</math>, oranswer choice <math>\boxed{A}</math>. | ||
==Video Solution== | ==Video Solution== |
Revision as of 19:15, 29 May 2022
Contents
[hide]Problem
For real numbers and , What is ?
Solution
Multiply the numerator and denominator of the LHS by to get . Then since and , , or choice .
Solution 2
Muliply both sides by to get . Then, add to both sides and subtract from both sides to get . Then, we can plug in the most simply values for z and w ( and , respectively, and find , oranswer choice .
Video Solution
~savannahsolver
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.