Question 447716
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Like Dr. Livingstone said, "Safari, so goodie."  Almost, that is.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\left(y\ -\ \frac{1}{2}\right)\ \neq\ 2y\ +\ 1]


Rather:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\left(y\ -\ \frac{1}{2}\right)\ =\ 2y\ -\ 1]



Now put it into standard form.  Add *[tex \Large -2y\ +\ 1] to both sides.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y^2\ -\ 2y\ +\ 1\ =\ 0]


Then just solve the factorable quadratic.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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