Q13/094) Solve |x+1|+|x-2| < 5

As -1 < 2 and in a line the distance between -1 and 2 is 3 so all the points from -1 to 2 satisfy the relation as the sum is the distance from -1 to 2

Now if it to the left of -1 the distance from -1 shall be added twice once for |x+1| and another time for |x-2| the sum < 2 and hence x > - 2

For the point to the right of 2 similarly x < 3

So -2 < x < 3