Difference between revisions of "1955 AHSME Problems/Problem 4"
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<math>\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</math> | <math>\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</math> | ||
| − | ==Solution== | + | ==Solution 1== |
| + | From the equality, <math>\frac{1}{x-1}=\frac{2}{x-2}</math>, we get <math>{(x-1)}\times2={(x-2)}\times1</math>. | ||
| − | + | 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 | ||
| + | ==Solution 2 (answer choices)== | ||
| + | Looking at the answer choices, we can see B, C, and D are incorrect because <math>x = 1</math> and <math>x = 2</math> result in division by <math>0</math>. | ||
| + | |||
| + | Plugging in <math>0</math> gives: | ||
| + | |||
| + | <math>\frac{1}{0 - 1} = \frac{1}{-1} = -1</math> | ||
| + | |||
| + | <math>\frac{2}{0 - 2} = \frac{2}{-2} = -1</math> | ||
| + | |||
| + | These two are equal, so the answer is <math>\fbox{{\bf(E)} \text{only} x = 0}</math>. | ||
| − | + | ~anabel.disher | |
| + | == See Also == | ||
| + | {{AHSME box|year=1955|num-b=3|num-a=5}} | ||
| − | + | {{MAA Notice}} | |
Latest revision as of 16:18, 18 June 2025
Problem
The equality 
is satisfied by:
Solution 1
From the equality, , we get 
.
Solving this, we get, 
.
Thus, the answer is 
.
Solution by awesomechoco
Solution 2 (answer choices)
Looking at the answer choices, we can see B, C, and D are incorrect because 
and 
result in division by 
.
Plugging in gives:
These two are equal, so the answer is .
~anabel.disher
See Also
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
