Question 12890
  |x-1|+|x-2|+|x-3|=4
 Note 1,2 and 3 divide the real line into four parts (intervals )
 nameley, (-oo,1),[1,2),[2,3) & [3,+oo)
 case (i) when {{{ x>= 3}}} , we get  x-1+ x-2+ x-3=4 
          or 3x = 10, so x = 10/3 ({{>= 3}}} OK.
 case (ii) when {{{ 2 <= x < 3}}} , we get  x-1+ x-2 -(x-3)=4 
          or x = 4. Since 4 is not in [2,3), invalid answer.
 case (iii) when {{{ 1 <= x < 2}}} , we get  x-1 -(x-2) -(x-3)=4 
       or -x + 4 =  4 so x = 0. Since 0 is not in [1,2), invalid answer.
 case (iv) when {{{ x < 1}}} , we get  -(x-1)-(x-2) -(x-3) =4 
   or -3x = -2, so x = 2/3. Since 2/3 is not in (-oo,1), invalid answer.

 Hence, there is only one solution x = 10/3.

 Try to read carefully about the details. 
 Sorry, I will not give further explanations for this simple questions. 
 The reason I solved it for you is to aviod that you may get long solutions.

 Kenny