1991 AHSME Problems/Problem 7: Difference between revisions

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 $\boxed{B}$ . These problems are copyrighted © by the Mathematical Association of America.

@@ Line 5: / Line 5: @@
 ==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}$.
+<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>.
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1991 AHSME Problems/Problem 7: Difference between revisions

Revision as of 17:40, 20 April 2014

Problem

Solution