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Difference between revisions of "1984 AHSME Problems/Problem 2"

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==Solution==
 
==Solution==
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Multiply the expression by <math> \frac{xy}{xy} </math> to get rid of the fractional numerator and denominator: <math> \frac{x^2y-x}{xy^2-y} </math>. This can be factored as <math> \frac{x(xy-1)}{y(xy-1)} </math>. The <math> xy-1 </math> terms cancel out, leaving <math> \frac{x}{y}, \boxed{\text{B}} </math>.
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Multiply the expression by <math> \frac{xy}{xy} </math> to get rid of the fractional [[numerator]] and [[denominator]]: <math> \frac{x^2y-x}{xy^2-y} </math>. This can be [[Factoring|factored]] as <math> \frac{x(xy-1)}{y(xy-1)} </math>. The <math> xy-1 </math> terms cancel out, leaving <math> \frac{x}{y}, \boxed{\text{B}} </math>.
  
 
==See Also==
 
==See Also==
 
{{AHSME box|year=1984|num-b=1|num-a=3}}
 
{{AHSME box|year=1984|num-b=1|num-a=3}}
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{{MAA Notice}}

Latest revision as of 11:48, 5 July 2013

Problem

If $x, y$, and $y-\frac{1}{x}$ are not $0$, then

$\frac{x-\frac{1}{y}}{y-\frac{1}{x}}$ equals

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

Solution

Multiply the expression by $\frac{xy}{xy}$ to get rid of the fractional numerator and denominator: $\frac{x^2y-x}{xy^2-y}$. This can be factored as $\frac{x(xy-1)}{y(xy-1)}$. The $xy-1$ terms cancel out, leaving $\frac{x}{y}, \boxed{\text{B}}$.

See Also

1984 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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