Question 124729
<pre><font size = 3><b>
 |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-1<font face = "symbol">³</font>0 and x-2<font face = "symbol">³</font>0 and x-3<font face = "symbol">³</font>0        
     x<font face = "symbol">³</font>1 and   x<font face = "symbol">³</font>2 and   x<font face = "symbol">³</font>3
 This simplifies to x<font face = "symbol">³</font>3 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.
 x<font face = "symbol">³</font>3 "takes care of" the first two 
 The equation in this case becomes
  (x-1)+(x-2)+(x-3)=4  which gives answer x={{{10/3}}}, and since
   this satisfies x<font face = "symbol">³</font>3 then {{{10/3}}} is a solution. 
     
2. x-1<font face = "symbol">³</font>0 and x-2<font face = "symbol">³</font>0 and x-3<font face = "symbol">£</font>0        
     x<font face = "symbol">³</font>1 and   x<font face = "symbol">³</font>2 and   x<font face = "symbol">£</font>3 
  This simplifies to  2 <font face = "symbol">£</font> x <font face = "symbol">£</font> 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 <font face = "symbol">£</font> x <font face = "symbol">£</font> 3 so 4 is not a solution. 

3. x-1<font face = "symbol">³</font>0 and x-2<font face = "symbol">£</font>0 and x-3<font face = "symbol">³</font>0      
     x<font face = "symbol">³</font>1 and   x<font face = "symbol">£</font>2 and   x<font face = "symbol">³</font>3
  This is contradictory so we
  ignore this case  

4. x-1<font face = "symbol">³</font>0 and x-2<font face = "symbol">£</font>0 and x-3<font face = "symbol">£</font>0      
     x<font face = "symbol">³</font>1 and   x<font face = "symbol">£</font>2 and   x<font face = "symbol">£</font>3
   This simplifies to 1 <font face = "symbol">£</font> x <font face = "symbol">£</font> 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 <font face = "symbol">£</font> x <font face = "symbol">£</font> 2 so 0 is not a solution. 


5. x-1<font face = "symbol">£</font>0 and x-2<font face = "symbol">³</font>0 and x-3<font face = "symbol">³</font>0    
     x<font face = "symbol">£</font>1 and   x<font face = "symbol">³</font>2 and   x<font face = "symbol">³</font>3 
  This is contradictory so we
  ignore this case.

6. x-1<font face = "symbol">£</font>0 and x-2<font face = "symbol">³</font>0 and x-3<font face = "symbol">£</font>0    
     x<font face = "symbol">£</font>1 and   x<font face = "symbol">³</font>2 and   x<font face = "symbol">£</font>3
  This is contradictory so we
  ignore this case 

7. x-1<font face = "symbol">£</font>0 and x-2<font face = "symbol">£</font>0 and x-3<font face = "symbol">³</font>0     
     x<font face = "symbol">£</font>1 and   x<font face = "symbol">£</font>2 and   x<font face = "symbol">³</font>3
  This is contradictory so we
  ignore this case 

8. x-1<font face = "symbol">£</font>0 and x-2<font face = "symbol">£</font>0 and x-3<font face = "symbol">£</font>0     
     x<font face = "symbol">£</font>1 and   x<font face = "symbol">£</font>2 and   x<font face = "symbol">£</font>3
   This simplifies to x<font face = "symbol">£</font>1.  
 The equation in this case becomes
 -(x-1)-(x-2)-(x-3)=4  which gives answer x={{{2/3}}}, and since
   this satisfies x<font face = "symbol">£</font>1 then {{{2/3}}} is a solution. 

So there are exactly two solutions:

x = {{{2/3}}} and x = {{{10/3}}}

Edwn</pre>