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

Revision as of 17:39, 20 April 2014

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) $x+\frac{1}{x}$

$\frac{a+b}{a-b}=$ \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.

@@ Line 1: / Line 1: @@
-\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>
+==Problem==
+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> (C) <math>1</math> (D) <math>x-\frac{1}{x}</math> (E) <math>x+\frac{1}{x}</math>
+==Solution==
+<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}$.
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