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

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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>.
 
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>.
  
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==Solution 2==
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==Solution 2 (SIGMA)==
  
 
Basic algebra at the end of the day, so simplify the numerator and the denominator. The numerator simplifies out to  
 
Basic algebra at the end of the day, so simplify the numerator and the denominator. The numerator simplifies out to  
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~AkCANdo
 
~AkCANdo
 +
~minor edit by SwordAxe
  
 
==Solution 3==
 
==Solution 3==

Revision as of 19:33, 5 June 2025

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 (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

$1/2 + 1 = 3/2$,

$1/2 - 1 = -1/2$,

$(3/2) / (-1/2) = -3$

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

(2 + 1) / (2 - 1) = 3.

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 (A)


~WalkEmDownTrey


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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