Question 250306
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2b\ +\ 4(b\ -\ 1)\ =\ 5(b\ -\ 1)]


Multiply 4 times b and 4 times -1:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2b\ +\ 4b\ -\ 4\ =\ 5(b\ -\ 1)]


Multiply 5 times b and 5 times -1:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2b\ +\ 4b\ -\ 4\ =\ 5b\ -\ 5]


Add -5b to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2b\ +\ 4b\ -\ 5b\ -\ 4\ =\ 5b\ -\ 5b\ -\ 5]


And collect like terms:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ b\ -\ 4\ =\ -5]


Add 4 to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ b\ -\ 4\ +\ 4\ =\ -5\ +\ 4]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ b\ =\ -1]


Check:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2(-1)\ +\ 4(-1\ -\ 1)\ =^?\ 5(-1\ -\ 1)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -2\ +\ 4(-2)\ =^?\ 5(-2)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -2\ +\ (-8)\ =^?\ -10]


Yep, every time.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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