solve for |x+1| + |x-2|< 2 
absolute value of x+1 and absolute value 
of x-2 is less than or equal to 2

Use the principle that

if A > 0 then |A| = A 
if A < 0 then |A| = -A
if A = 0 then |A| = 0

There are four cases to consider:

Case 1:  x+1 > 0 AND x-2 > 0
           x > -1 AND x > 2

This means x > 2

|x+1| + |x-2| < 2
    x+1 + x-2 < 2
       2x - 1 < 2
           2x < 3
            x < 3/2

This contradicts x > 2, so Case 1 is impossible.

Case 2:  x+1 > 0 AND x-2 < 0
           x > -1 AND x < 2

This means -1 < x < 2

 |x+1| + |x-2| < 2
  x+1 + -(x-2) < 2
 x + 1 - x + 2 < 2
             3 < 2
  
This is never true, so Case 2 is impossible.

Case 3:  x+1 < 0 AND x-2 > 0
           x < -1 AND x > 2

This is impossible so Case 3 is impossible.

Case 4:  x+1 < 0 AND x-2 < 0
           x < -1 AND x < 2

This means x < -1

   |x+1| + |x-2| < 2
 -(x+1) + -(x-2) < 2
     -x-1 - x+2  < 2
           -2x+1 < 2
             -2x < 1
               x > -1/2
     
But this contradicts x < -1

So Case 4 is impossible also.

There is no solution.

Edwin