Hi Candice! I will write up the workings by 4 am if no one has done them yet.

Ah its fine now! Someone answered already but thanks :)

I type here since I do not have access to writing materials for now.

7 - |3x - 1| = |9 - 27x| + 2 |6x - 2|

The nice thing about modulus expressions is that for a two term expression in the brackets not involving division (addition, subtraction or multiplication), we can swap over the terms without penalty, since the expression remains the same.

7 - |3x - 1| = |27x - 9| + 2 |6x - 2|

Positive constant multipliers of factorisation can also be taken out.

7 - |3x - 1| = 9 * |3x - 1| + 2 * 2 * |3x - 1|

To simplify matters, let u = |3x - 1|.

7 - u = 9u + 4u
7 = 14u
u = 1/2

And there you are. Now we convert back the equation back into x.

|3x - 1| = 1/2
3x - 1 = 1/2 or -1/2
3x = 3/2 or -3/2
x = 1/2 or -1/2

Both solutions are valid since x remains in the modulus brackets and there is no “wild” x running around outside the modulus brackets.

Oh i see!! Thanks so much :)