Question

|(2x-1)/3|>6

Answer 1

(2x-1)/3>6
2x-1>18
2x>19
x>19/2
and
(2x-1)/3<-6
2x-1<-18
2x<-17
x<-17/2

Answer 2

one case is (2x-1)/3>6, and the other is -(2x-1)/3>6
So, for the first:
2x-1>18
x>19/2. The other is
2x-1<-18
x<-17/2. Either will work, so your interval is \[\left[ -\infty,-17/2 \right]\cup \left[ 19/2,\infty \right]\]