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Difference between revisions of "1955 AHSME Problems/Problem 4"

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

Revision as of 07:04, 7 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 {\bf(E)} \text{only} ${x}$ = 0.

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