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

(Created page with "<math> x+\tfrac{2}{x}= y+\tfrac{2}{y} </math> Since <math>x\not=y</math>, we may assume that <math>x=\frac{2}{y}</math> and/or, equivalently, <math>y=\frac{2}{x}</math>. Cross ...")
 
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== Problem==
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Given that <math>x</math> and <math>y</math> are distinct nonzero real numbers such that <math>x+\tfrac{2}{x} = y + \tfrac{2}{y}</math>, what is <math>xy</math>?
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<math> \textbf{(A)}\ \frac{1}{4}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 4\qquad </math>
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==Solution==
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<math> x+\tfrac{2}{x}= y+\tfrac{2}{y} </math>
 
<math> x+\tfrac{2}{x}= y+\tfrac{2}{y} </math>
  
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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>.
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== See also ==
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{{AMC12 box|year=2013|ab=A|num-b=7|num-a=9}}

Revision as of 18:36, 22 February 2013

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

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

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 $xy=2$.

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