Difference between revisions of "2013 AMC 12A Problems/Problem 8"

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Cross multiply in either equation, giving us <math>xy=2</math>.
 
Cross multiply in either equation, giving us <math>xy=2</math>.
  
<math>\boxed{\textbf{(D) }{10}}</math>
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<math>\boxed{\textbf{(D) }{2}}</math>
  
 
==Solution 2==
 
==Solution 2==
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<math>xy = 2</math>
 
<math>xy = 2</math>
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== Video Solution ==
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https://youtu.be/ba6w1OhXqOQ?t=1129
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~ pi_is_3.14
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==Video Solution==
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https://youtu.be/CCjcMVtkVaQ
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~sugar_rush
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2013|ab=A|num-b=7|num-a=9}}
 
{{AMC12 box|year=2013|ab=A|num-b=7|num-a=9}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 04:15, 15 January 2021

Problem

Given that $x$ and $y$ are distinct nonzero real numbers such that $x+\tfrac{2}{x} = y + \tfrac{2}{y}$, what is $xy$?

$\textbf{(A)}\ \frac{1}{4}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 4\qquad$

Solution 1

$x+\tfrac{2}{x}= y+\tfrac{2}{y}$

Since $x\not=y$, we may assume that $x=\frac{2}{y}$ and/or, equivalently, $y=\frac{2}{x}$.

Cross multiply in either equation, giving us $xy=2$.

$\boxed{\textbf{(D) }{2}}$

Solution 2

$x+\tfrac{2}{x}= y+\tfrac{2}{y}$

$x-y+\frac{2}{x}-\frac{2}{y} = 0$

$(x-y)+2(\frac{y-x}{xy}) = 0$

$(x-y)(1-\frac{2}{xy})=0$

Since $x\not=y$

$1 = \frac{2}{xy}$

$xy = 2$

Video Solution

https://youtu.be/ba6w1OhXqOQ?t=1129

~ pi_is_3.14

Video Solution

https://youtu.be/CCjcMVtkVaQ ~sugar_rush

See also

2013 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
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 12 Problems and Solutions

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