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>. | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/6Uh77bue0bE | ||
+ | |||
+ | ~savannahsolver | ||
==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 14:17, 24 July 2020
Contents
Problem
For real numbers and
,
What is
?
Solution
Multiply the numerator and denominator of the LHS by to get
. Then since
and
,
, or 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.