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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>.
  
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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>.
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Solution by awesomechoco
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== See Also ==
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{{AHSME 4p box|year=1955|num-b=3|after=Last Question}}
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{{MAA Notice}}

Revision as of 22:10, 9 July 2018

Problem

The equality $\frac{1}{x-1}=\frac{2}{x-2}$ is satisfied by:

$\textbf{(A)}\ \text{no real values of }x\qquad\textbf{(B)}\ \text{either }x=1\text{ or }x=2\qquad\textbf{(C)}\ \text{only }x=1\\ \textbf{(D)}\ \text{only }x=2\qquad\textbf{(E)}\ \text{only }x=0$

Solution

From the equality, $\frac{1}{x-1}=\frac{2}{x-2}$, we get ${(x-1)}\times2={(x-2)}\times1$.

Solving this, we get, ${2x-2}={x-2}$.

Thus, the answer is $\fbox{{\bf(E)} \text{only} x = 0}$.

Solution by awesomechoco

See Also

Template:AHSME 4p box These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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