1955 AHSME Problems/Problem 4: Difference between revisions

Revision as of 07:06, 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

Revision as of 07:06, 7 July 2018 view source Awesomechoco (talk \| contribs) editor 60 edits →Solution ← Older edit		Revision as of 07:06, 7 July 2018 view source Awesomechoco (talk \| contribs) editor 60 edits →Solution Newer edit →
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	Solving this, we get, <math>{2x-2}={x-2}</math>.		Solving this, we get, <math>{2x-2}={x-2}</math>.

	Thus, the answer is <math>{\bf(E)}</math> \text{only} <math>{x}</math> = 0		Thus, the answer is <math>{\bf(E)}</math> <math>\text{only}</math> <math>{x}</math> = 0

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1955 AHSME Problems/Problem 4: Difference between revisions

Revision as of 07:06, 7 July 2018

Problem

Solution