Recall the definition of absolute value:
• |x| = x if x ≥ 0
• |x| = -x if x < 0
So you need to consider 4 different cases (2 absolute value expressions with 2 possible cases each).
(i) Suppose 2x + 3 < 0 and x - 2 < 0. The first inequality says x < -3/2 and the second says x < 2, so ultimately x < -3/2. Then
|2x + 3| + |x - 2| = 6x
-(2x + 3) - (x - 2) = 6x
-2x - 3 - x + 2 = 6x
-3x - 1 = 6x
9x = -1
x = -1/9
But -1/9 is not smaller than -3/2, so this case provides no valid solution.
(ii) Suppose 2x + 3 ≥ 0 and x - 2 < 0. Then x ≥ -3/2 and x < 2, or -3/2 ≤ x < 2. Under this condition,
|2x + 3| + |x - 2| = 6x
(2x + 3) - (x - 2) = 6x
2x + 3 - x + 2 = 6x
x + 5 = 6x
5x = 5
x = 1
This solution is valid because it does fall in the interval -3/2 ≤ x < 2.
(iii) Suppose 2x + 3 < 0 and x - 2 ≥ 0. Then x < -3/2 or x ≥ 2. So
|2x + 3| + |x - 2| = 6x
-(2x + 3) + (x - 2) = 6x
-2x - 3 + x - 2 = 6x
-x - 5 = 6x
7x = -5
x = -5/7
This isn't a valid solution, because neither -5/7 < -3/2 nor -5/7 ≥ 2 are true.
(iv) Suppose 2x + 3 ≥ 0 and x - 2 ≥ 0. Then x ≥ -3/2 and x ≥ 2, or simply x ≥ 2.
|2x + 3| + |x - 2| = 6x
(2x + 3) + (x - 2) = 6x
2x + 3 + x - 2 = 6x
3x + 1 = 6x
3x = 1
x = 1/3
This is yet another invalid solution since 1/3 is smaller than 2.
So there is one solution at x = 1.