Question 1045842
<font face="Times New Roman" size="+2">


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ |2\ -\ x|\ >\ |1\ -\ 2x|]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \ \ 2\ -\ x\ >\ |1\ -\ 2x|]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \ \  2\ -\ x\ >\ 1\ -\ 2x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \ \  x\ >\ -1]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \ \  2\ -\ x\ >\ 2x\ -\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \ \ -3x\ >\ -3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \ \  x\ <\ 1]


Therefore:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  -1\ <\ x\ <\ 1]


You can also solve *[tex \Large x\ -\ 2\ <\ |1\ -\ 2x|] and get the same result.


*[illustration 2_AbsVal_Inequality_(2).jpg]


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 \ \ \ \ \ \ \ \ \ \  

</font>