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 1

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

Solution 2 (answer choices)

Looking at the answer choices, we can see B, C, and D are incorrect because $x = 1$ and $x = 2$ result in division by $0$ .

Plugging in $0$ gives:

$\frac{1}{0 - 1} = \frac{1}{-1} = -1$

$\frac{2}{0 - 2} = \frac{2}{-2} = -1$

These two are equal, so the answer is $\fbox{{\bf(E)} \text{only} x = 0}$ .

~anabel.disher

1955 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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1955 AHSME Problems/Problem 4

Contents

Problem

Solution 1

Solution 2 (answer choices)

See Also