<PRE><B>Question 192238</B>
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Let &lt;i&gt;&lt;b&gt;x&lt;/b&gt;&lt;/i&gt; represent the number.


Then &lt;i&gt;&lt;b&gt;-x&lt;/b&gt;&lt;/i&gt; is the additive inverse.


The additive inverse divided by 12 is: *[tex \LARGE  \frac{-x}{12}]


The reciprocal of the number is *[tex \LARGE \frac{1}{x}]


Three times the reciprocal of the number is  *[tex \LARGE \frac{3}{x}]


One less than three times the reciprocal is *[tex \LARGE \frac{3}{x} - 1]


So, solve for &lt;i&gt;&lt;b&gt;x&lt;/b&gt;&lt;/i&gt;


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \frac{-x}{12}=\frac{3}{x} - 1]


&lt;i&gt;&lt;b&gt;Hint:&lt;/b&gt;&lt;/i&gt;  Multiply by &lt;i&gt;&lt;b&gt;-x&lt;/b&gt;&lt;/i&gt; first.  Then collect terms and solve the easily factorable quadratic.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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