<PRE><B>Question 639601</B>
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Since *[tex \LARGE y\ =\ y], you can write *[tex \LARGE x^2\ =\ 6x\ -\ 5] which is to say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 6x\ +\ 5\ =\ 0]


Factor and solve the quadratic for its two roots, one of which is, indeed, 5.  Then substitute back into *[tex \LARGE y\ =\ x^2] to find the two values of *[tex \LARGE y], one of which is, indeed, 25.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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