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Normally for a problem like this, I would just test for n>0 or n<0 but apparently this is not the case.
Any question has always to be asked , especially before going to an exam such as the GMAT ... and u are welcome
Concerning the reason, the principle is to look for the crucial "point" or value s of n,x... when those values make flip the sign of the expression in the absolute value. This is done for every absolute values of an inequation.
Concretely, we decompose the whole inquation in module to study:
o |n+2|: # n+2 = 0 <=> n = -2
# n+2 > 0 <=> n > -2
# n+2 < 0 <=> n < -2
> The original inequation has to be studied on those 2 domains n < -2 and n > -2. Indeed, for those 2 domains, the simplified expression of |n+2| differs. That is |n+2| = n+2 when n > -2 and |n+2| = -(n+2) when n < -2.
o |n|: # n = 0
# n > 0
# n < 0
Finally, we have to study all combined domains, from |n+2| and from |n|, in the original inequation.