First, note that if #a<0# then there are no solutions, as the absolute value function always returns a nonnegative value. For the remainder of the problem, then, we will assume #a>=0#.
Using the definition of the absolute value function
#|x| = {(x if x>=0), (-x if x<0):}#
we consider two cases.
Case 1: #x-a >= 0#
#=> |x-a| = x-a#
#=> x-a <= a#
#=> x - a + a <= a+a#
#=> x <= 2a#
Note that the initial condition is equivalent to #x >= a#, however, so our solution set in this case is #x in [a, 2a]#.
Case 2: #x-a < 0#
#=> |x-a| = -(x-a)#
#=> -(x-a) <= a#
#=> -(x-a)+x-a <= a+x-a#
#=> 0 <= x#
Note that the initial condition in this case is equivalent to #x < a#, so together our solution set in this case becomes #x in [0, a)#.
Putting the two cases together, we get our total solution set for when #a>=0#:
#x in [0, a)uu[a, 2a] = [0, 2a]#.
and, for completion, we can include the results for when #a<0#:
#x in {(O/ if a < 0), ([0, 2a] if a>=0):}#
