Difference between revisions of "1979 AHSME Problems/Problem 2"

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==Solution==
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== Problem 2 ==
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For all non-zero real numbers <math>x</math> and <math>y</math> such that <math>x-y=xy, \frac{1}{x}-\frac{1}{y}</math> equals
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<math>\textbf{(A) }\frac{1}{xy}\qquad
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\textbf{(B) }\frac{1}{x-y}\qquad
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\textbf{(C) }0\qquad
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\textbf{(D) }-1\qquad
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\textbf{(E) }y-x    </math>
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==Solution 1==
  
 
Moving all variables to one side of the equation, we can use Simon's Favorite Factoring Trick to factor the equation into <cmath>(x+1)(y-1) = -1</cmath> Plugging in <math>-1</math> and <math>1</math> as the <math>x</math> and <math>y</math> sides respectively, we get <math>x = -2</math> and <math>y = 2</math>. Plugging this in to <math>\frac{1}{x}-\frac{1}{y}</math> gives us <math>\boxed{-1}</math> as our final answer.
 
Moving all variables to one side of the equation, we can use Simon's Favorite Factoring Trick to factor the equation into <cmath>(x+1)(y-1) = -1</cmath> Plugging in <math>-1</math> and <math>1</math> as the <math>x</math> and <math>y</math> sides respectively, we get <math>x = -2</math> and <math>y = 2</math>. Plugging this in to <math>\frac{1}{x}-\frac{1}{y}</math> gives us <math>\boxed{-1}</math> as our final answer.
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==Solution 2==
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Notice that we can do <math>\frac{x-y}{xy} = \frac{xy}{xy}</math>. We are left with <math>\frac{1}{y} - \frac{1}{x} = 1</math>. Multiply by <math>-1</math> to achieve <math>\frac{1}{x} - \frac{1}{y} = \boxed{-1}</math>.
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== See also ==
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{{AHSME box|year=1979|num-b=1|num-a=3}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 23:29, 6 February 2018

Problem 2

For all non-zero real numbers $x$ and $y$ such that $x-y=xy, \frac{1}{x}-\frac{1}{y}$ equals

$\textbf{(A) }\frac{1}{xy}\qquad \textbf{(B) }\frac{1}{x-y}\qquad \textbf{(C) }0\qquad \textbf{(D) }-1\qquad \textbf{(E) }y-x$

Solution 1

Moving all variables to one side of the equation, we can use Simon's Favorite Factoring Trick to factor the equation into \[(x+1)(y-1) = -1\] Plugging in $-1$ and $1$ as the $x$ and $y$ sides respectively, we get $x = -2$ and $y = 2$. Plugging this in to $\frac{1}{x}-\frac{1}{y}$ gives us $\boxed{-1}$ as our final answer.

Solution 2

Notice that we can do $\frac{x-y}{xy} = \frac{xy}{xy}$. We are left with $\frac{1}{y} - \frac{1}{x} = 1$. Multiply by $-1$ to achieve $\frac{1}{x} - \frac{1}{y} = \boxed{-1}$.

See also

1979 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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