1955 AHSME Problems/Problem 4: Difference between revisions
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==Solution== | ==Solution== | ||
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>. | ||
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Thus, the answer is <math>\fbox{{\bf(E)} \text{only} x = 0}</math>. | Thus, the answer is <math>\fbox{{\bf(E)} \text{only} x = 0}</math>. | ||
Solution by awesomechoco | |||
== See Also == | |||
{{AHSME 4p box|year=1955|num-b=3|after=Last Question}} | |||
{{MAA Notice}} | |||
Revision as of 22:10, 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
Template:AHSME 4p box
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