It is an absolute value inequality.

The method of solving such inequalities is to divide the number line in intervals where each term of the equation is a linear function,

and then solve the resulting inequalities for linear functions for each separate interval.


In this case the critical values, where functions change their behavior, are x= 0 and x= 2.  They divide the number line

in three intervals:  1) x < 0;  2) 0 < x < 2,  and  3) x > 2.





Plot y =  (red) and y = x + abs(x) (green)



In the FIRST interval,  x < 0,  we have (x-2) < 0;  therefore  |x-2| = -(x-2).

Also, in this interval x < 0, therefore  |x| = -x  and then  x + |x| = x + (-x) = 0.

Thus in the first interval the inequality takes the form 

     -(x-2) > 0 ,  or, equivalently,

      x - 2 < 0,   or

      x < 2,   which is always TRUE in the interval x < 0.

Thus the first interval,  x < 0, is the part of the solution to the original inequality.  


    At this point I completed with the first interval and now start working with the second interval.



In the SECOND interval,  0 < x < 2,  we have (x-2) < 0;  therefore  |x-2| = -(x-2).

Also, in this interval x > 0, therefore  |x| = x,  and then  x + |x| = x + x = 2x.

Thus in the second interval the inequality takes the form 

     -(x-2) > 2x ,  or, equivalently,

      2x + (x - 2) < 0,   or

      3x < 2,   which is  TRUE  if  x < .

Thus for the second interval,  0 < x < 2,  ONLY PART OF IT  x <  is the solution to the original inequality.  



In the THIRD interval,  x > 2,  we have (x-2) > 0;  therefore  |x-2| = (x-2).

Also, in this interval x > 0, therefore  |x| = x,  and then  x + |x| = x + x = 2x.

Thus in the third interval the inequality takes the form 

     (x-2) > 2x ,  or, equivalently,

      2x - (x - 2) < 0,   or

      x < 2,   which is always FALSE  in the interval x > 2.

Thus for the third interval, x > 2.  NO PART OF IT  is the solution to the original inequality.  



After this analysis you can conclude that the solution set to the original inequality is  x < ,  or, in interval form,  (,).


Answer.  The solution set to the original inequality is  x < ,  or, in interval form,  (,).

     I think that these sources are the only place in the american school math literature, where you can learn the subject adequately.