Difference between revisions of "2014 AMC 10B Problems/Problem 9"

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==Problem==
 
==Problem==
For real numbers <math> w </math> and <math> z </math>, <cmath> \cfrac{\frac{1}{w} + \frac{1}{z}}{\frac{1}{w} - \frac{1}{z}} = 2014. </cmath> What is <math> \frac{w+z}{w-z} </math>? \\
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For real numbers <math> w </math> and <math> z </math>, <cmath> \cfrac{\frac{1}{w} + \frac{1}{z}}{\frac{1}{w} - \frac{1}{z}} = 2014. </cmath> What is <math> \frac{w+z}{w-z} </math>?  
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<math> \textbf{(A) }-2014\qquad\textbf{(B) }\frac{-1}{2014}\qquad\textbf{(C) }\frac{1}{2014}\qquad\textbf{(D) }1\qquad\textbf{(E) }2014 </math>
 
<math> \textbf{(A) }-2014\qquad\textbf{(B) }\frac{-1}{2014}\qquad\textbf{(C) }\frac{1}{2014}\qquad\textbf{(D) }1\qquad\textbf{(E) }2014 </math>
  
 
==Solution==
 
==Solution==
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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>.
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==Solution 2==
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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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==Video Solution==
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https://youtu.be/6Uh77bue0bE
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~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}}

Latest revision as of 09:14, 30 May 2022

Problem

For real numbers $w$ and $z$, \[\cfrac{\frac{1}{w} + \frac{1}{z}}{\frac{1}{w} - \frac{1}{z}} = 2014.\] What is $\frac{w+z}{w-z}$?

$\textbf{(A) }-2014\qquad\textbf{(B) }\frac{-1}{2014}\qquad\textbf{(C) }\frac{1}{2014}\qquad\textbf{(D) }1\qquad\textbf{(E) }2014$

Solution

Multiply the numerator and denominator of the LHS by $wz$ to get $\frac{z+w}{z-w}=2014$. Then since $z+w=w+z$ and $w-z=-(z-w)$, $\frac{w+z}{w-z}=-\frac{z+w}{z-w}=-2014$, or choice $\boxed{A}$.

Solution 2

Muliply both sides by $\left(\frac{1}{w}-\frac{1}{z}\right)$ to get $\frac{1}{w}+\frac{1}{z}=2014\left(\frac{1}{w}-\frac{1}{z}\right)$. Then, add $2014\cdot\frac{1}{z}$ to both sides and subtract $\frac{1}{w}$ from both sides to get $2015\cdot\frac{1}{z}=2013\cdot\frac{1}{w}$. Then, we can plug in the most simple values for z and w ($2015$ and $2013$, respectively), and find $\frac{2013+2015}{2013-2015}=\frac{2(2014)}{-2}=-2014$, or answer choice $\boxed{A}$.

Video Solution

https://youtu.be/6Uh77bue0bE

~savannahsolver

See Also

2014 AMC 10B (ProblemsAnswer KeyResources)
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

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