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

Latest revision as of 22:14, 9 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 $\fbox{{\bf(E)} \text{only} x = 0}$ .

Solution by awesomechoco

1955 AHSME (Problems • Answer Key • Resources)
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.

@@ Line 6: / Line 6: @@
 ==Solution==
+From the equality, <math>\frac{1}{x-1}=\frac{2}{x-2}</math>, we get  <math>{(x-1)}\times2={(x-2)}\times1</math>.
-From this equality, we get  <math>{x-1}*2={x-2}</math>*1
+Solving this, we get, <math>{2x-2}={x-2}</math>.
+Thus, the answer is <math>\fbox{{\bf(E)} \text{only} x = 0}</math>.
+Solution by awesomechoco
+== See Also ==
+{{AHSME box|year=1955|num-b=3|num-a=5}}
+{{MAA Notice}}

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

Latest revision as of 22:14, 9 July 2018

Problem

Solution

See Also