1955 AHSME Problems/Problem 4

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

1955 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
All AHSME Problems and Solutions


The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS