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 "1976 AHSME Problems/Problem 1"

(→‎Problem 1)
(→‎Solution)
Line 7: Line 7:
 
== Solution ==
 
== Solution ==
  
−
The reciprocal of <math>(1-x)</math> is <math>\frac{1}{1-x}</math>, so our equation is <cmath>1-\frac{1}{1-x}=\frac{1}{1-x},</cmath> which is equivalent to <math>\frac{1}{1-x}=\frac{1}{2}</math>. So, <math>1-x=2</math> and <math>x=-2\Rightarrow \boxed{A}</math>.
+
The reciprocal of <math>(1-x)</math> is <math>\frac{1}{1-x}</math>, so our equation is <cmath>1-\frac{1}{1-x}=\frac{1}{1-x},</cmath> which is equivalent to <math>\frac{1}{1-x}=\frac{1}{2}</math>. So, <math>1-x=2</math> and <math>x=-2\Rightarrow \textbf{(A)}</math>.

Revision as of 21:48, 11 July 2020

Problem 1

If one minus the reciprocal of $(1-x)$ equals the reciprocal of $(1-x)$, then $x$ equals

$\textbf{(A) }-2\qquad \textbf{(B) }-1\qquad \textbf{(C) }1/2\qquad \textbf{(D) }2\qquad \textbf{(E) }3$

Solution

The reciprocal of $(1-x)$ is $\frac{1}{1-x}$, so our equation is \[1-\frac{1}{1-x}=\frac{1}{1-x},\] which is equivalent to $\frac{1}{1-x}=\frac{1}{2}$. So, $1-x=2$ and $x=-2\Rightarrow \textbf{(A)}$.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1976_AHSME_Problems/Problem_1&oldid=128114"