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Difference between revisions of "1951 AHSME Problems/Problem 20"

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== Problem ==
 
== Problem ==
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When simplified and expressed with negative exponents, the expression <math> (x \plus{} y)^{ \minus{} 1}(x^{ \minus{} 1} \plus{} y^{ \minus{} 1})</math> is equal to:
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When simplified and expressed with negative exponents, the expression <math> (x + y)^{ - 1}(x^{ - 1} + y^{ - 1})</math> is equal to:
  
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<math> \textbf{(A)}\ x^{ \minus{} 2} \plus{} 2x^{ \minus{} 1}y^{ \minus{} 1} \plus{} y^{ \minus{} 2} \qquad\textbf{(B)}\ x^{ \minus{} 2} \plus{} 2^{ \minus{} 1}x^{ \minus{} 1}y^{ \minus{} 1} \plus{} y^{ \minus{} 2} \qquad\textbf{(C)}\ x^{ \minus{} 1}y^{ \minus{} 1}</math>
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<math> \textbf{(A)}\ x^{ - 2} + 2x^{ - 1}y^{ - 1} + y^{ - 2} \qquad\textbf{(B)}\ x^{ - 2} + 2^{ - 1}x^{ - 1}y^{ - 1} + y^{ - 2} \qquad\textbf{(C)}\ x^{ - 1}y^{ - 1}</math>
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<math> \textbf{(D)}\ x^{ \minus{} 2} \plus{} y^{ \minus{} 2} \qquad\textbf{(E)}\ \frac {1}{x^{ \minus{} 1}y^{ \minus{} 1}}</math>
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<math> \textbf{(D)}\ x^{ - 2} + y^{ - 2} \qquad\textbf{(E)}\ \frac {1}{x^{ - 1}y^{ - 1}}</math>
  
 
== Solution ==  
 
== Solution ==  
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{{solution}}
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Note that <math>(x + y)^{-1}(x^{-1} + y^{-1}) = \dfrac{1}{x + y}\cdot\left(\dfrac{1}{x} + \dfrac{1}{y}\right) = \dfrac{1}{x + y}\cdot\dfrac{x + y}{xy} = \dfrac{1}{xy} = x^{-1}y^{-1}</math>. The answer is <math>\textbf{(C)}</math>.
  
 
== See Also ==
 
== See Also ==
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[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 21:58, 13 March 2015

Problem

When simplified and expressed with negative exponents, the expression $(x + y)^{ - 1}(x^{ - 1} + y^{ - 1})$ is equal to:

$\textbf{(A)}\ x^{ - 2} + 2x^{ - 1}y^{ - 1} + y^{ - 2} \qquad\textbf{(B)}\ x^{ - 2} + 2^{ - 1}x^{ - 1}y^{ - 1} + y^{ - 2} \qquad\textbf{(C)}\ x^{ - 1}y^{ - 1}$

$\textbf{(D)}\ x^{ - 2} + y^{ - 2} \qquad\textbf{(E)}\ \frac {1}{x^{ - 1}y^{ - 1}}$

Solution

Note that $(x + y)^{-1}(x^{-1} + y^{-1}) = \dfrac{1}{x + y}\cdot\left(\dfrac{1}{x} + \dfrac{1}{y}\right) = \dfrac{1}{x + y}\cdot\dfrac{x + y}{xy} = \dfrac{1}{xy} = x^{-1}y^{-1}$. The answer is $\textbf{(C)}$.

See Also

1951 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

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