Difference between revisions of "2014 AMC 10B Problems/Problem 9"
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==Solution 2== | ==Solution 2== | ||
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+ | 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>. | 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 | + | ==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>. | 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 | ~idk12345678 | ||
− | ==Solution | + | ==Solution 5== |
Set \( w = 2 \) and \( z = 1 \). | Set \( w = 2 \) and \( z = 1 \). | ||
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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> | ||
− | Therefore, the answers to the equations are the negatives of each other. Thus | + | Therefore, the answers to the equations are the negatives of each other. Thus the answer is (A) |
+ | |||
~WalkEmDownTrey | ~WalkEmDownTrey | ||
+ | |||
+ | |||
==Video Solution (CREATIVE THINKING)== | ==Video Solution (CREATIVE THINKING)== |
Latest revision as of 00:25, 3 November 2024
Contents
Problem
For real numbers and , What is ?
Solution
Multiply the numerator and denominator of the LHS (left hand side) by to get . Then since and , , or choice .
Solution 2
Basic algebra at the end of the day, so simplify the numerator and the denominator. The numerator simplifies out to and the denominator simplifies out to .
This results in .
Division results in the elimination of , so we get .
is just so the equation above is .
Solving this results in .
~AkCANdo
Solution 3
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 4
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
Solution 5
Set \( w = 2 \) and \( z = 1 \).
Substitute the new values into the first equation
,
,
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): and
Therefore, the answers to the equations are the negatives of each other. Thus the answer is (A)
~WalkEmDownTrey
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.