Difference between revisions of "1955 AHSME Problems/Problem 4"
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==Solution== | ==Solution== | ||
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From the equality, <math>\frac{1}{x-1}=\frac{2}{x-2}</math>, we get <math>{(x-1)}\times2={(x-2)}\times1</math>. | From the equality, <math>\frac{1}{x-1}=\frac{2}{x-2}</math>, we get <math>{(x-1)}\times2={(x-2)}\times1</math>. | ||
Solving this, we get, <math>{2x-2}={x-2}</math>. | Solving this, we get, <math>{2x-2}={x-2}</math>. | ||
| − | Thus, the answer is <math>{\bf(E)} | + | Thus, the answer is <math>\fbox{{\bf(E)} \text{only} x = 0}</math>. |
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| + | Solution by awesomechoco | ||
| + | |||
| + | == See Also == | ||
| + | {{AHSME box|year=1955|num-b=3|num-a=5}} | ||
| + | |||
| + | {{MAA Notice}} | ||
Latest revision as of 22:14, 9 July 2018
Problem
The equality 
is satisfied by:
Solution
From the equality, , we get 
.
Solving this, we get, 
.
Thus, the answer is 
.
Solution by awesomechoco
See Also
| 1955 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 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. 
