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







 

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