1991 AHSME Problems/Problem 7

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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 $\boxed{B}$.

See also

1991 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
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