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Difference between revisions of "1976 AHSME Problems/Problem 1"

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<math>\textbf{(A) }-2\qquad \textbf{(B) }-1\qquad \textbf{(C) }1/2\qquad \textbf{(D) }2\qquad  \textbf{(E) }3</math>
 
<math>\textbf{(A) }-2\qquad \textbf{(B) }-1\qquad \textbf{(C) }1/2\qquad \textbf{(D) }2\qquad  \textbf{(E) }3</math>
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== Solution ==
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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>.

Revision as of 20:47, 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 $x=-2\Rightarrow \boxed{A}$.

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