-(x+2)+x+5<4 works for any -5 < [or =] x < [or =] -2.
for the other neg/pos case..
lx+2l + lx+5l<4
x + 2 -(x + 5)
x + 2 -x -5
-3 < 4
There are no solutions for this one.
(As [x>-2 and x<-5] makes no sense/has no values)
hopefully got that right
I don't get the answer to Q1. Wouldn't it be |x| > |a|, which then would have 4 solutions (each depending on whether x and/or a are negative)? I haven't done the graphing stuff yet.