Difference between revisions of "2014 AMC 10B Problems/Problem 9"
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==Solution 2== | ==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 simple 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>, or answer choice <math>\boxed{A}</math>. | 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 simple 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>, or answer choice <math>\boxed{A}</math>. | ||
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+ | ==Solution 3== | ||
+ | Let <math>a = \frac{1}{w}</math> and <math>b = \frac{1}{z}</math>. To find values for a and b, we can try <math>a+b = 2014</math> and <math>a-b=1</math>. However, that leaves us with a fractional solution, so scaling it by 2, we get <math>a+b = 4028</math> and <math>a-b=2</math>. Solving by adding the equations together, we get <math>b = 2015</math> and <math>a = 2013</math>. Now, substituting back in, we get <math>w = \frac{1}{2015}</math> and <math>z = \frac{1}{2013}</math>. Now, putting this into the desired equation with <math>n = 2015 \cdot 2013</math> (since it will cancel out), we get <math>\frac{\frac{2013+2015}{n}}{\frac{2013-2015}{n}}</math>. Dividing, we get <math>\frac{4028}{-2} = \boxed{\textbf{(A)}\ -2014}</math>. | ||
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+ | ~idk12345678 | ||
==Video Solution (CREATIVE THINKING)== | ==Video Solution (CREATIVE THINKING)== |
Revision as of 19:47, 15 April 2024
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 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 .
~idk12345678
Video Solution (CREATIVE THINKING)
~Education, the Study of Everything
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.