Since absolute value has two separate formulas depending on whether the number inside the absolute value is positive or negative. That means you need to look at x+1< 0 (x< -1), x+1> 0 (x> -1), x-1< 0 (x< 1), and x-1> 0 (x> 1).

That means we need to look at three intervals: x< -1, -1< x< 1, and x> 1.

If x< -1, both x+1< 0 and x-1< 0 so the equation becomes -(x+1)-(-(x-1))= -2= 2k+1 which has no solution for x if x is not -3/2 but an infinite number of solutions if k= 3/2 (since then the equation reduces to -2= -2 which is true for all x).

If -1< x< 1, x+1> 0 but x-1< 0 so the equation becomes (x+1)-(-(x-1))= 2x= 2k+1 which has a single solution for x no matter what k is.

If x> 1, both x+1>0 and x-1> 0 so the equation becomes (x+1)- (x-1)= 2= 2k+1 which has no solution for x is k is not 1/2 but has an infinite number of solutions if k= 1/2 (because then the equation reduces to 2= 2 which is true for all x).