1955 AHSME Problems/Problem 4: Difference between revisions

Revision as of 22:49, 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 this equality, we get ${(x-1)}*2=$ {x-2}*1

Revision as of 22:48, 6 July 2018 view source Awesomechoco (talk \| contribs) editor 60 edits →Solution ← Older edit		Revision as of 22:49, 6 July 2018 view source Awesomechoco (talk \| contribs) editor 60 edits →Solution Newer edit →
Line 7:		Line 7:
	==Solution==		==Solution==

	From this equality, we get <math>{x-1}2={x-2}~~</math>~~1		From this equality, we get <math>{(x-1)}2=</math>{x-2}1

Page

Toolbox

Search

1955 AHSME Problems/Problem 4: Difference between revisions

Revision as of 22:49, 6 July 2018

Problem

Solution