Solve:|x-1|+|x-2|<3
Note that |a+b| is always +ve, while (a+b) can be either +ve or -ve.
These means that we have 4 cases
(x-1) +ve, (x-2) +ve,
(x-1) +ve, (x-2) -ve,
(x-1) -ve, (x-2) +ve,
(x-1) -ve, (x-2) -ve.
So, now we can write,
(x-1) + (x-2) < 3
(x-1) - (x-2) < 3
-(x-1) + (x-2) < 3
-(x-1) - (x-2) < 3
Solving these equations,
2x - 3 < 3
1 < 3
-1 < 3
-2x + 3 < 3
The middle two inequalities are simply true statements, and can be ignored, leaving us with
2x - 3 < 3
2x - 3 > -3
Or,
2x < 6
2x > 0
which simplifies to,
x < 3 and x > 0
Answer: 0 < x < 3