Question 343447
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a(bx\ +\ 2)\ =\ cx\ -\ 12]


Distribute the *[tex \Large a]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ abx\ +\ 2a\ =\ cx\ -\ 12]


Add *[tex \Large -2a] and *[tex \Large -cx] to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ abx\ -\ cx\ =\ -2a\ -\ 12]


Factor out *[tex \Large x] in the LHS:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x(ab\ -\ c)\ =\ -2a\ -\ 12]


Multiply both sides by *[tex \Large \frac{1}{ab\ -\ c}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{-2a\ -\ 12}{ab\ -\ c}]


And that, as they say, is that.


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