How to do this modulus qestion

modules x = x , x>= 0
= -x , x<0
2x-2 = 2x -2 , x>=1
= 2 -2x , x<1
3x - 3 = 3x - 3 , x>=3
= 3 - 3x , x<3

1. when x<0 ............. -4x + 2 - 2x = 3 -3x
x = -1/3
x<0, x = -1/3 gives x = -1/3

2. when 0<= x < 1 ........................ 4x -2x +2 = -3x +3
x = 1/5
0<= x <1 , x = 1/5 gives x = 1/5

3. when x>= 1.............. 4x +2x - 2 = 3x -3
x = -1/3
x>= 1 , x = -1/3 no solutions because x>=1

x = -1/3 and 1/5

4|x| + |2x -2| = |3x -3|
|4x| + 2|x -1| = 3|x -1|
|4x| = 3|x -1| - 2|x -1|
|4x| = |x -1|   
Square both sides, 16x^2 = x^2-2x+1, or 15x^2+2x-1 = 0.
(5x-1)(3x+1)=0.
x=1/5 or -1/3.
Check: x=1/5, 4|x| + |2x -2| = 4/5+|2/5 -2| = 4/5 + 8/5 = 12/5, and |3x -3| = 12/5
x=-1/3, LHS = 4/3 + 8/3 = 12/3 = 4, and RHS = 4

Case 1 (x<0)
−4x + 2−2x = 3−3x
3x = −1
x = −1/3

Case 2 (0≤x<1)
4x + 2−2x = 3−3x
5x = 1
x = 1/5

Case 3 (x≥1)
4x + 2x−2 = 3x−3
3x = −1
x = −1/3 (rejected because x≥1)

Hence x = 1/5 and −1/3.

u substitued both values for x!
for x =1/5

u get LHS=17/5

RHS=12/5

But x=-1/3

LHS=12/3=4


so ans is x=-1/3

Rhs=12/3=4

4|x| + |2x -2| = |3x -3|
4|x| + 2|x-1| = 3|x-1|

Let x<0
|x|=-x
|x-1|=-(x-1)=1-x
4|x| + 2|x-1| = 3|x-1| => -4x + 2(1-x) = 3(1-x) => x=-1/3

Let 0<=x<1
|x|=x
|x-1|=-(x-1)=1-x
4|x| + 2|x-1| = 3|x-1| => 4x + 2(1-x) = 3(1-x) => x=1/5


Let 1<=x
|x|=x
|x-1|=x-1
4|x| + 2|x-1| = 3|x-1| => 4x + 2(x-1) = 3(x-1) => x=-1/3 but 1<=x hence, no solution


x₁= 1/5
x₂=-1/3

PS:
4|x| + 2|x-1| = 3|x-1|
4|x| = |x-1|