<PRE><B>Question 608423</B>
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Just add the two equations just the way they are.  The *[tex \LARGE y] variable will go away.  Then collect like terms on one side of the equals sign leaving yourself with a factorable quadratic.  Once you have the two roots of the quadratic, substitute back into either of the given equations to determine the *[tex \LARGE y] value that corresponds to that *[tex \LARGE x] value.  The book&#39;s answers are correct.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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