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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>.
  
 
Solving this, we get, <math>{2x-2}={x-2}</math>.
 
Solving this, we get, <math>{2x-2}={x-2}</math>.
  
 
Thus, \ \text{only }x=0$ can satisfy the equation, {(E)}.
 
Thus, \ \text{only }x=0$ can satisfy the equation, {(E)}.

Revision as of 22:59, 6 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, \ \text{only }x=0$ can satisfy the equation, {(E)}.

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