Difference between revisions of "2014 AMC 10B Problems/Problem 9"
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==Solution== | ==Solution== | ||
− | 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 (left hand side) 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>. |
==Solution 2== | ==Solution 2== |
Revision as of 22:30, 14 September 2024
Contents
[hide]Problem
For real numbers and , What is ?
Solution
Multiply the numerator and denominator of the LHS (left hand side) 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 simple values for z and w ( and , respectively), and find , or answer choice .
Solution 3
Let and . To find values for a and b, we can try and . However, that leaves us with a fractional solution, so scaling it by 2, we get and . Solving by adding the equations together, we get and . Now, substituting back in, we get and . Now, putting this into the desired equation with (since it will cancel out), we get . Dividing, we get .
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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.