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

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==Problem==
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==Problem 9==
 
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>?  
 
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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Substitute the new values into the first equation  
 
Substitute the new values into the first equation  
  
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<math>1/2 + 1 = 3/2</math>,
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<math>\frac{1}{2} + 1 = \frac{3}{2}</math>,
  
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<math>1/2 - 1 = -1/2</math>,
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<math>\frac{1}{2} - 1 = -\frac{1}{2}</math>,
  
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<math>(3/2) / (-1/2) = -3</math>
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<math>\frac{\frac{3}{2} / \frac{-1}{2}} = -3</math>
  
 
Substitute in the second equation with new values of \( w \) and \( z \):
 
Substitute in the second equation with new values of \( w \) and \( z \):
  
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(2 + 1) / (2 - 1) = 3.
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<math>/frac{/frac{2 + 1} / \frac{2 - 1}} = 3. </math>
  
 
Answers of each equation (where X is the quotient): <math>x</math> and <math>-x</math>  
 
Answers of each equation (where X is the quotient): <math>x</math> and <math>-x</math>  
  
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Therefore, the answers to the equations are the negatives of each other. Thus the answer is (A)
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Therefore, the answers to the equations are the negatives of each other. Thus the answer is <math>/boxed{(A)}</math>
  
  
 
~WalkEmDownTrey
 
~WalkEmDownTrey
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~minor <math>latex</math> edits by SwordAxe
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==Video Solution (CREATIVE THINKING)==
 
==Video Solution (CREATIVE THINKING)==

Latest revision as of 19:39, 5 June 2025

Problem 9

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 (left hand side) 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 (SIGMA)

Basic algebra at the end of the day, so simplify the numerator and the denominator. The numerator simplifies out to $\frac{w+z}{wz}$ and the denominator simplifies out to $\frac{z-w}{wz}$.

This results in $\cfrac{\frac{w+z}{zw}}{\frac{z-w}{zw}} = 2014$.

Division results in the elimination of $zw$, so we get $\frac{w+z}{z-w} = 2014$.

$z-w$ is just $-(w-z)$ so the equation above is $-(\frac{w+z}{w-z} = 2014$.

Solving this results in $\frac{w+z}{w-z} = \boxed{\textbf{(A)}\ -2014}$.

~AkCANdo ~minor edit by SwordAxe

Solution 3

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 $\boxed{A}$.

Solution 4

Let $a = \frac{1}{w}$ and $b = \frac{1}{z}$. To find values for a and b, we can try $a+b = 2014$ and $a-b=1$. However, that leaves us with a fractional solution, so scaling it by 2, we get $a+b = 4028$ and $a-b=2$. Solving by adding the equations together, we get $b = 2015$ and $a = 2013$. Now, substituting back in, we get $w = \frac{1}{2015}$ and $z = \frac{1}{2013}$. Now, putting this into the desired equation with $n = 2015 \cdot 2013$ (since it will cancel out), we get $\frac{\frac{2013+2015}{n}}{\frac{2013-2015}{n}}$. Dividing, we get $\frac{4028}{-2} = \boxed{\textbf{(A)}\ -2014}$.

~idk12345678

Solution 5

Set \( w = 2 \) and \( z = 1 \).

Substitute the new values into the first equation

$\frac{1}{2} + 1 = \frac{3}{2}$,

$\frac{1}{2} - 1 = -\frac{1}{2}$,

$\frac{\frac{3}{2} / \frac{-1}{2}} = -3$

Substitute in the second equation with new values of \( w \) and \( z \):

$/frac{/frac{2 + 1} / \frac{2 - 1}} = 3.$ (Error compiling LaTeX. Unknown error_msg)

Answers of each equation (where X is the quotient): $x$ and $-x$

Therefore, the answers to the equations are the negatives of each other. Thus the answer is $/boxed{(A)}$


~WalkEmDownTrey ~minor $latex$ edits by SwordAxe

Video Solution (CREATIVE THINKING)

https://youtu.be/Y37KozgBEXg

~Education, the Study of Everything


Video Solution

https://youtu.be/6Uh77bue0bE

~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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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