Difference between revisions of "1955 AHSME Problems/Problem 4"
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From the equality, <math>\frac{1}{x-1}=\frac{2}{x-2}</math>, we get <math>{(x-1)}*2={(x-2)}*1</math>. | From the equality, <math>\frac{1}{x-1}=\frac{2}{x-2}</math>, we get <math>{(x-1)}*2={(x-2)}*1</math>. | ||
| − | Solving this, we get, <math>{2x-2}={x-2}</math> | + | Solving this, we get, <math>{2x-2}={x-2}</math>. |
| + | |||
| + | Thus, ${x} can only equal 0 | ||
Revision as of 22:53, 6 July 2018
Problem
The equality 
is satisfied by:
Solution
From the equality, , we get 
.
Solving this, we get, 
.
Thus, ${x} can only equal 0
