By finding where each of the absolute value terms equals 0, we can separate the equation into three regions (while you would think +- for each of the two terms, you might think that there would be four regions, but one of these four will not exist).
1/2 x - 2 = 0; 1/2 x = 2; x = 4
x - 1/2 = 0; x = 1/2
Thus, we have three regions (-∞, 1/2), (1/2, 4), (4, ∞)
We also note that the original coefficient on the x terms is positive in both cases.
Lastly, we can note that, while we have +- both terms, we will only have equality of the terms with two expressions, so with at least one of the three intervals, we will have no exact equalities, so it will either be the entire interval or none of it.
Thus, first solve.
On (4, ∞), x - 1/2 = 1/2 x - 2; 1/2 x = -3/2; x = -3. -3 is not in the interval, which means we either have the entire interval or none of it. Consider x = 6. |6 - 1/2| = 5 1/2 |1/2 (6) - 2| = 1. 5 1/2 > 1, so this is the entire interval (and we will also include 4, as the function is continuous, and we have shown the expressions are equal at -3; we can verify by plugging in 4 and seeing that |4 - 1/2| = 3 1/2 > 0
Note that the equation we just solved is also the solution for x < 1/2, as both expressions are the negative.
-3 is in this interval. Thus, we will have a subset of the interval.
Plugging in x = 0, we have |1/2| > |2|, which is not true, so the interval will be (-∞, -3) (We don't include -3, as the sign is >, not >=; We could also solve this using
-(x - 1/2) > - (1/2 x - 2); 1/2 - x > 2 - 1/2x; -3/2 > 1/2 x; -3 > x
Last, we consider 1/2, 4. Given that we are increasing each expression from 0, we know we will have a place of equality. Now, to find it.
If we had -(1/2 x - 2) = x - 1/2; 2 - 1/2 x = x - 1/2; 5/2 = 3/2 x; x = 5/3.
Now, as this is an inequality, and we already know that 1/2 is not in the interval and 4 is, our solution is (5/3, 4)
We can actually pay attention to the equality sign and verify this.
x - 1/2 > 2 - 1/2 x
3/2 x > 5/2
x > 5/3
Thus, our solution is (-∞, -3) U (5/3, 4) U [4, ∞) = (-∞, -3) U (5/3, ∞)
