|x-1|+|x-2|+|x-3|=4 

We have to consider 8 feasible cases, some of which
will be contradictory, and some which although not
contradictor, will produce a solution which contradicts
them, i.e., does not satisfy the conditions of the
case.

1. x-1³0 and x-2³0 and x-3³0        
     x³1 and   x³2 and   x³3
 This simplifies to x³3 for if x is greater than or 
 equal to 3 it is not necessary to say it is greater than or
 equal to 2 or 1.
 x³3 "takes care of" the first two 
 The equation in this case becomes
  (x-1)+(x-2)+(x-3)=4  which gives answer x=, and since
   this satisfies x³3 then  is a solution. 
     
2. x-1³0 and x-2³0 and x-3£0        
     x³1 and   x³2 and   x£3 
  This simplifies to  2 £ x £ 3
 The equation in this case becomes
  (x-1)+(x-2)-(x-3)=4  which gives answer x=4, but this 
 does not satisfies 2 £ x £ 3 so 4 is not a solution. 

3. x-1³0 and x-2£0 and x-3³0      
     x³1 and   x£2 and   x³3
  This is contradictory so we
  ignore this case  

4. x-1³0 and x-2£0 and x-3£0      
     x³1 and   x£2 and   x£3
   This simplifies to 1 £ x £ 2 
 The equation in this case becomes
  (x-1)-(x-2)-(x-3)=4  which gives answer x=0, but this 
 does not satisfies 1 £ x £ 2 so 0 is not a solution. 


5. x-1£0 and x-2³0 and x-3³0    
     x£1 and   x³2 and   x³3 
  This is contradictory so we
  ignore this case.

6. x-1£0 and x-2³0 and x-3£0    
     x£1 and   x³2 and   x£3
  This is contradictory so we
  ignore this case 

7. x-1£0 and x-2£0 and x-3³0     
     x£1 and   x£2 and   x³3
  This is contradictory so we
  ignore this case 

8. x-1£0 and x-2£0 and x-3£0     
     x£1 and   x£2 and   x£3
   This simplifies to x£1.  
 The equation in this case becomes
 -(x-1)-(x-2)-(x-3)=4  which gives answer x=, and since
   this satisfies x£1 then  is a solution. 

So there are exactly two solutions:

x =  and x = 

Edwn