How to do this modulus qestion

Solve for x
4|x| + |2x -2| = |3x -3|

ans: 1/5 and - 1/3

_______________________

    4|x| + |2x -2| = |3x -3|
    |4x| + 2|x -1| = 3|x -1|
                  |4x| = 3|x -1| - 2|x -1|
                  |4x| = |x -1|                  ← This is true only if 4x and x-1
                                                                are equal or opposites
So,
      4x = x-1   OR   4x = -(x-1)
      3x = -1            4x = 1-x
        x = -⅓              x = ⅕

                ANSWER
                x = -⅓, ⅕


Have a good one!!
____________________________

Hello,

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

► To answer any modulus equation you have to find the root of every single modulus expression:

|x|=0 → x=0
|2x-2| = 0 → x=1
|3x-3| = 0 → x=1

Thus there are three intervals to consider: ]-∞; 0], [0; 1] and [1; + ∞[.
We will study them one after the other.

► On interval ]-∞; 0]:
|x| = -x
|2x - 2| = 2 - 2x
|3x - 3| = 3 - 3x

Thus we get:
4|x| + |2x - 2| = |3x - 3|
-4x + 2 - 2x = 3 - 3x
-4x = 1 - x
3x = -1
x = -⅓

Since -⅓ is actually in interval ]-∞; 0], x=-⅓ is really a solution of the equation.

► On interval [0; 1]:
|x| = x
|2x - 2| = 2 - 2x
|3x - 3| = 3 - 3x

Thus we get:
4|x| + |2x - 2| = |3x - 3|
4x + 2 - 2x = 3 - 3x
5x = 1
x = 1/5

Since 1/5 is actually in interval [0; 1], x=1/5 is really a solution of the equation.

► On interval [1; +∞[:
|x| = x
|2x - 2| = 2x - 2
|3x - 3| = 3x - 3

Thus we get:
4|x| + |2x - 2| = |3x - 3|
4x + 2x - 2 = 3x - 3
3x = -1
x = -⅓

Since -⅓ is actually NOT in interval [0; +∞[, x=-⅓ would not be solution of the equation.
However, since the exact same value is solution elsewhere, we can conclude:

► The only real solutions of the modulus equation are