Question 885016
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You aren't very clear about what you want.  Contrary to my normal modus operandi, I'm going to make the assumption that you want to solve for *[tex \Large x] in terms of *[tex \Large a] and *[tex \Large b].


Distribute the *[tex \Large a] in the LHS.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ ax\ +\ 2a\ =\ bx\ +\ a]


Add *[tex \Large -bx] to both sides


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ ax\ -\ bx\ +\ 2a\ =\ a]


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ ax\ -\ bx\ =\ -a]


Factor *[tex \Large x] out of the LHS expression


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (a\ -\ b)x\ =\ -a]


Multiply both sides by *[tex \Large \frac{1}{a\ -\ b}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ -\frac{a}{a\ -\ b}]


Going into the ninth grade is the time to stop taking the "cookbook" approach to mathematics.  Learn the PRECISE definitions of things and use them to say exactly what you mean.


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