Difference between revisions of "1976 AHSME Problems/Problem 1"
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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=- | + | 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=-1\Rightarrow \textbf{(B)}</math>.~MathJams |
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| + | {{AHSME box|year=1976|before=[[1975 AHSME]]|after=[[1977 AHSME]]}} | ||
Latest revision as of 16:34, 29 November 2020
Problem 1
If one minus the reciprocal of 
equals the reciprocal of , then 
equals
Solution
The reciprocal of is 
, so our equation is ![\[1-\frac{1}{1-x}=\frac{1}{1-x},\]](http://latex.artofproblemsolving.com/f/f/3/ff381ea1ef9861150899951ec43cba92a5a1eb86.png)
which is equivalent to 
. So, 
and 
.~MathJams
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