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):}#