more about absolute values

Since for |a| = b, a = b if a greater than or equal to 0 or -a = b if a is smaller than 0
Therefore,
|3-2x| = 6x-5

Case 1:
Assume 3-2x is greater than or equal to 0
|3-2x| = 3-2x = 6x-5
Therefore, 8x = 8 and x = 1
Then you put back the answer to our assumption 3-2x greater or equal to 0
x = 1, 3-2x = 3-2(1) = 1 which is still greater than or equal to 0
Therefore, no contradiction between our assumption (3-2x greater or equal to 0) and the result (x=1)

Case 2:
Assume 3-2x is less than 0
|3-2x| = 2x-3 = 6x-5
4x = 2
x = 0.5
Once again, you put back the answer to the assumption 3-2x smaller than 0
when x = 0.5, 3-2x = 3 - 2(0.5) = 2 which is greater than 0
Therefore, there is contradiction between the assumption (3-2x smaller than 0) and the result (x=0.5) and this result must be rejected.

Therefore, the only answer is x = 1

|x-2| + |2x+1| = 4
There are 2 numbers within the absolute value: x-2 and 2x+1
As you mentioned, the critical points are the numbers changing sign (from -ve to +ve or vice versa)
We test the critical points as x-2 = 0 (ie. x = 2) and 2x+1 = 0 (ie. x = -0.5)
Since -0.5 is less than 2
Therefore, we have to consider 3 cases:
Case 1:
x is less than -0.5
Then 2x + 1 is less than 0 and so |2x+1| = -2x - 1
Since x - 2 is always less than 0 if x is less than -0.5, so |x-2| = -x+2
Therefore,
|x-2| + |2x+1| = 4
-(x-2) - (2x+1) = 4
-x+2-2x-1 = 4
3x = -3
x = -1 (which is also less than -0.5)

Case 2:
x is greater or equal to -0.5 and less than or equal to 2
Therefore, x-2 is less than 0 and so |x-2| = -x+2
However, 2x+1 is greater than 0 in this case, so |2x+1| = 2x+1
|x-2| + |2x+1| = 4
-(x-2) + (2x+1) = 4
x = 1 (which is also between 2 and -0.5)

Case 3:
x is greater than 2
Therefore,
|x-2| + |2x+1| = 4
(x-2) + (2x+1) = 4
3x = 5
x = 5/3 which is less than 2 (Contradiction)

Therefore, there are only 2 correct answers:
x = 1 or x = -1

lxl = x if x is positive lxl = -x if x is negative.
because we need to change sign, so we need to check when the polynomial inside the absolute sign is positive or negative.

Inside case, we use "and". Between cases, we use "or"

|3-2x| = 6x-5

This is an easy question, we know that 3-2x changes sign when 3-2x=0, that is the critical point is x=3/2

Case 1: x is less than 3/2
3-2x = 6x-5
x = 1
so x is less than 3/2 and x = 1
thus x = 1

Case 2: x is greater than or equal to 3/2
3-2x<0 thus,
|3-2x| = 2x-3 = 6x-5
4x = 2
x = 0.5
x is greater than or equal to 3/2 and x = 0.5
so no real answer

Conclusion x=1 or x = no real answer

Therefore, the only answer is x = 1

|x-2| + |2x+1| = 4

Similarly, critical points as x-2 = 0 (ie. x = 2) and 2x+1 = 0 (ie. x = -0.5)

Case 1: x is less than -0.5
Then 2x + 1 is less than 0 and so |2x+1| = -2x - 1
Since x - 2 is always less than 0 if x is less than -0.5, so |x-2| = -x+2
Therefore,
|x-2| + |2x+1| = 4
-(x-2) - (2x+1) = 4
-x+2-2x-1 = 4
3x = -3
x = -1

thus x= -1 and x is less than - 0.5, so x = -1

Case 2: x is greater or equal to -0.5 and less than 2
Therefore, x-2 is less than 0 and so |x-2| = -x+2
However, 2x+1 is greater than 0 in this case, so |2x+1| = 2x+1
|x-2| + |2x+1| = 4
-(x-2) + (2x+1) = 4
x = 1

x is greater or equal to -0.5 and less than 2 and x = 1, thus x = 1

Case 3: x is greater than or equal to 2
All are positive, therefore
|x-2| + |2x+1| = 4
(x-2) + (2x+1) = 4
3x = 5
x = 5/3

x is greater than or equal to 2 and x= 5/3, thus no solution

Conclusion:
x = -1 or x =1 or x has ho solution
thus
x = 1 or x = -1

Critical point is the point when the polynomial inside absolute sign is 0