I actually am not sure of the general solution to those sorts of equations. What I would do is consider the possible cases separately: first assume that x > 1.5. Then assume -3 < x <= 1.5. Lastly, assume x <= -3.
If x > 1.5, you can simply drop the absolute values because both terms are non-negative. If -3 < x <= 1.5, then the left absolute value term is positive, so you can drop that absolute value and then solve the equation as you normally would with an absolute value problem. If x <= -3, then the first term can be replaced by "-x - 3", and the second term can be replaced by "-x + 1.5" (you can reason through why yourself, or ask me to explain why).
In each case, make sure that the answers you get when you solve for x match your assumption (so for the first case, throw out any answers that are less than or equal to 1.5).
I hope that helps. Perhaps someone else can come up with a more general method for solving it.
Here's a similar example I contrived to illustrate the process without giving the answer away:
|x| + |x - 2| = 2
First assume x > 2.
- x + x - 2 = 2
- 2x = 4
- x = 1/2
- Since we assumed that x > 2, we can't use this solution.
Next assume x > 0 but x <= 2.
- x + |x - 2| = 2
- |x - 2| = 2 - x
- x - 2 = 2 - x
- OR x - 2 = x - 2
- In the first case, 2x = 4, so x = 1/2 (valid solution)
- The second case yields 0 = 0, which gives us infinitely many solutions (so any value of x that is > 0 but <= 2 is a solution)
Lastly, assume that x <= 0
- -x + -x + 2 = 2
- -2x = 0
- x = 0 (valid solution)
If we combine all these answers, we learn that the value of x is anything from 0 to 2 (i.e. the closed interval [0, 2]). I guess this was actually a bad example because the solution to your problem is simpler than this one (it only has two distinct solutions, not infinitely many.) But I hope this example at least shows you how to proceed.