Question 592763
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You went to way more trouble than was necessary, and then made algebra errors in the process.  See annotation in your problem statement.  If you take it from there, it works out correctly.  However, it would have been much easier had you first factored the LHS numerator:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{3x\ -\ 9}{x\ -\ 3}\ =\ \frac{-3x}{2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{3(x\ -\ 3)}{x\ -\ 3}\ =\ \frac{-3x}{2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3\ =\ \frac{-3x}{2}]


Solve from there.


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