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

Revision as of 10:07, 11 July 2013

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

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

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

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 $xy=2$ .

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

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

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

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

Since $x\not=y$

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

$xy = 2$

2013 AMC 12A (Problems • Answer Key • Resources)
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

@@ Line 5: / Line 5: @@
 <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>
-==Solution==
+==Solution 1==
 <math> x+\tfrac{2}{x}= y+\tfrac{2}{y} </math>
@@ Line 12: / Line 12: @@
 Cross multiply in either equation, giving us <math>xy=2</math>.
+==Solution 2==
+<math>x+\tfrac{2}{x}= y+\tfrac{2}{y}</math>
+<math>x-y+\frac{2}{x}-\frac{2}{y} = 0</math>
+<math>(x-y)+2(\frac{y-x}{xy}) = 0</math>
+<math>(x-y)(1-\frac{2}{xy})=0</math>
+Since <math>x\not=y</math>
+<math>1 = \frac{2}{xy}</math>
+<math>xy = 2</math>
 == See also ==
 {{AMC12 box|year=2013|ab=A|num-b=7|num-a=9}}
 {{MAA Notice}}