1991 AHSME Problems/Problem 7: Difference between revisions

Latest revision as of 02:10, 28 September 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 $\boxed{B}$ .

1991 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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-\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}= \frac{\frac{a}{b} + 1}{\frac{a}{b} - 1} = \frac{x+1}{x-1}</math>, so the answer is <math>\boxed{B}</math>.
+== See also ==
+{{AHSME box|year=1991|num-b=6|num-a=8}}
+[[Category: Introductory Algebra Problems]]
 {{MAA Notice}}