Difference between revisions of "1991 AHSME Problems/Problem 7"
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| − | + | ==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> | (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}} | ||
Latest revision as of 03:10, 28 September 2014
Problem
If 
, 
and 
, then 
(A) 
(B) 
(C) 
(D) 
(E) 
Solution
, so the answer is 
.
See also
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
