Question 772282
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You might have had better luck if you had checked your answers.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ |3(y\ -\ 1)\ -\ 12|\ =\ 0]


Let's try 5:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ |3(5\ -\ 1)\ -\ 12|\ =\ ?]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ |3(4)\ -\ 12|\ =\ ?]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ |12\ -\ 12|\ =\ ?]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ |0|\ =\ 0]


Let's try -5


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ |3(-5\ -\ 1)\ -\ 12|\ =\ ?]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ |3(-6)\ -\ 12|\ =\ ?]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ |-18\ -\ 12|\ =\ ?]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ |-30|\ =\ 30\ \not =\ 0]


Let's re-attempt solving from the beginning:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ |3(y\ -\ 1)\ -\ 12|\ =\ 0]


Remove the absolute value bars and put *[tex \LARGE \pm] in the RHS:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3(y\ -\ 1)\ -\ 12\ =\ \pm0]


But *[tex \LARGE \pm0] is just *[tex \LARGE 0], so


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3(y\ -\ 1)\ -\ 12\ =\ 0]


Which is just a simple linear equation in one variable with a single element in the solution set, namely 5 as we have seen earlier.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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