|x+1| + |x-1| = 2x

We consider 4 cases:

Case 1. x+1≧0 and x-1≧0
     x≧-1 and x≧1 which amounts to x≧1

Then we have

x+1 + x-1 = 2x
       2x = 2x
        
So case 1 is an identity.

So all values of x where x≧1 are solutions.  

Case 2. x+1≧0 and x-1≦0
     x≧-1 and x≦1 which amounts to -1≦x≦1

Then we have

x+1 - (x-1) = 2x
x+1 - x + 1 = 2x
          2 = 2x
          1 = x, which is a solution since -1≦1≦1,

However 1 is already included as a solution in case 1,
so 1 is not a "new" solution.

Case 3. x+1≦0 and x-1≧0
     x≦-1 and x≧1 which is impossible,

so case 3 is out.

Case 4. x+1≦0 and x-1≦0
     x≦-1 and x≦1 which amounts to x≦-1

Then we have

-(x+1) - (x-1) = 2x
-x-1 - x + 1 = 2x
         -2x = 2x
           0 = 4x
           0 = x
But that is incompatible with x≦-1,
So case 4 is out.

Therefore the solution set is from case 1 (which includes 
the solution from case 2).

Answer: {x|x≧1} or [1,∞)

Edwin