There are three regions to consider, and these regions are determined by looking at where x+1 is negative or nonnegative, and where 2x-3 is negative or nonnegative:
(1) : x+1 < 0 and 2x-3 < 0
(2) : and 2x-3 < 0
(3) : and
Case (2) can be discarded as there are no solutions (the values are
simply too small in magnitude to add to 8)
Case (1)
Since both terms on the LHS are negative, we can remove the absolute value signs and write -8 on the RHS:
x + 1 + 2x - 3 = -8
3x - 2 = -8
x = -6/3 = -2
is one solution
Case (3)
Since both terms on the LHS are positive or zero, we can remove the absolute value signs and write:
x + 1 + 2x - 3 = 8
3x = 8+2 is another solution
