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

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==Solution==
 
==Solution==
  
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Multiply the numerator and denominator of the LHS by <math>w+z</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==
 +
 
 +
Basic algebra at the end of the day, so simplify the numerator and the denominator. The numerator simplifies out to
 +
<math>\frac{w+z}{wz}</math> and the denominator simplifies out to <math>\frac{z-w}{wz}</math>.
 +
 
 +
This results in <math>\cfrac{\frac{w+z}{zw}}{\frac{z-w}{zw}} = 2014</math>.
 +
 
 +
Division results in the elimination of <math>zw</math>, so we get <math>\frac{w+z}{z-w} = 2014</math>.
 +
 
 +
<math>z-w</math> is just <math>-(w-z)</math> so the equation above is <math>-(\frac{w+z}{w-z} = 2014</math>.
 +
 
 +
Solving this results in <math>\frac{w+z}{w-z} = \boxed{\textbf{(A)}\ -2014}</math>.
 +
 
 +
~AkCANdo
 +
 
 +
==Solution 3==
 +
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>.
 +
 
 +
==Solution 4==
 +
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>.
 +
 
 +
~idk12345678
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 +
==Solution 5==
 +
 
 +
Set \( w = 2 \) and \( z = 1 \).
 +
 
 +
Substitute the new values into the first equation
 +
 
 +
<math>1/2 + 1 = 3/2</math>,
 +
 
 +
<math>1/2 - 1 = -1/2</math>,
 +
 
 +
<math>(3/2) / (-1/2) = -3</math>
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 +
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): <math>x</math> and <math>-x</math>
 +
 
 +
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==
 
==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 00:25, 3 November 2024

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

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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