<PRE><B>Question 754957</B>
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There is no unique solution for a single two-variable equation.  There are infinite ordered pairs in the solution set.  Pick any real number value you like for *[tex \LARGE x], substitute that value, and do the arithmetic required to determine the corresponding value for *[tex \LARGE y].  Then you know that the ordered pair *[tex \LARGE \left(x_i,y_i\right)] where *[tex \LARGE x_i] is the value you selected for *[tex \LARGE x] and *[tex \LARGE y_i] is the value calculated as a result of the selection, is an element of the solution set of the given equation (Given correctly performed arithmetic, of course)


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
&lt;font face=&quot;Math1&quot; size=&quot;+2&quot;&gt;Egw to Beta kai to Sigma&lt;/font&gt;
My calculator said it, I believe it, that settles it
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