Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(→‎Solution)
Line 11: Line 11:
 
Solving this, we get, <math>{2x-2}={x-2}</math>.
 
Solving this, we get, <math>{2x-2}={x-2}</math>.
  
−
Thus, the answer is <math>{\bf(E)}</math>  <math>\text{only}</math> <math>{x}</math> = <math>{0}</math>.
+
Thus, the answer is <math>\fbox{{\bf(E)} \text{only} x = 0}</math>.

Revision as of 04:43, 8 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 $\fbox{{\bf(E)} \text{only} x = 0}$.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1955_AHSME_Problems/Problem_4&oldid=96025"