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Difference between revisions of "1991 AHSME Problems/Problem 7"
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==Solution==
 
==Solution==
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<math>\frac{a+b}{a-b}= </math>\frac{\frac{a}{b} + 1}{\frac{a}{b} - 1} = \frac{x+1}{x-1}<math>, so the answer is </math>\boxed{B}$.
+
<math>\frac{a+b}{a-b}= \frac{\frac{a}{b} + 1}{\frac{a}{b} - 1} = \frac{x+1}{x-1}</math>, so the answer is <math>\boxed{B}</math>.
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 17:40, 20 April 2014

Problem

If $x=\frac{a}{b}$, $a\neq b$ and $b\neq 0$, then $\frac{a+b}{a-b}=$

(A) $\frac{x}{x+1}$ (B) $\frac{x+1}{x-1}$ (C) $1$ (D) $x-\frac{1}{x}$ (E) $x+\frac{1}{x}$

Solution

$\frac{a+b}{a-b}= \frac{\frac{a}{b} + 1}{\frac{a}{b} - 1} = \frac{x+1}{x-1}$, so the answer is $\boxed{B}$. The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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