Question 1047632
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3(2x\ -\ 1)\ +\ 5\ =\ 6x\ +\ 2]


Distribute the 3 in the LHS


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6x\ -\ 3\ +\ 5\ =\ 6x\ +\ 2]


Collect like terms in the LHS


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6x\ +\ 2\ =\ 6x\ +\ 2]


Add *[tex \Large 6x] to both sides


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\ =\ 2]


Which is a true statement.  Therefore, the original equation is true for all real numbers *[tex \Large x].  The solution set is *[tex \Large \left{x\ :\ x\ \in\ \mathbb{R}\right}].


In fact, you could actually say that the equation is true for all complex numbers, and the solution set is *[tex \Large \left{x\ :\ x\ \in\ \mathbb{C}\right}]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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