Is lxl< 1?

Can you explain further how you solved lx + 1l = 2lx - 1l for x? I am confused w/ the absolute value and inequality signs.

Here's my approach:
I'll begin with an aside: if |x| = 5 then x=5 or -5
In general if k = |something| then k = something or k = -(something)
If lx + 1l = 2lx - 1l then either
a) x+1 = 2(x-1) --> x=3
or
b) -(x+1) = 2(x-1) --> x= 1/3

Notice that, since we have two sets of absolutes in the original question, we could have created additional equations: -(x+1) = (2)[-(x-1)] and x+1 = (2)[-(x-1)], but these are simply versions of equations (a) and (b)

I hope that helps.