Difference between revisions of "1991 AHSME Problems/Problem 7"

(Created page with "\If <math>x=\frac{a}{b}</math>, <math>a\neq b</math> and <math>b\neq 0</math>, then <math>\frac{a+b}{a-b}=</math> (A) <math>\frac{x}{x+1}</math> (B) <math>\frac{x+1}{x-1}</math>...")
 
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(A) <math>\frac{x}{x+1}</math> (B) <math>\frac{x+1}{x-1}</math> (C) <math>1</math> (D) <math>x-\frac{1}{x}</math> (E) <math>x+\frac{1}{x}</math>
 
(A) <math>\frac{x}{x+1}</math> (B) <math>\frac{x+1}{x-1}</math> (C) <math>1</math> (D) <math>x-\frac{1}{x}</math> (E) <math>x+\frac{1}{x}</math>
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Revision as of 13:52, 5 July 2013

\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) $x+\frac{1}{x}$ The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png