Question 252038
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Since


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ -\ b\ =\ -(b\ -\ a)] for all real *[tex \LARGE  a] and *[tex \LARGE b],


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ |a\ -\ b|\ =\ |b\ -\ a|] for all real *[tex \LARGE  a] and *[tex \LARGE b]


Hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ |x\ -\ 3|\  +\  |3\ -\ x|\ =\ 12\ \Rightarrow\ 2|x\ -\ 3|\ =\ 12\ \Rightarrow\ |x\ -\ 3|\ =\ 6]


You should be able to take it from there.



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