I am looking on the answers, and there is only one case, when a or b or c or d pass: 3|x-3| + 2 = 14. So I assume, that before two is plus. Then:
3|x-3|+2=14 |minus 2
3|x-3|=12 |divide 3
|x-3|=4
From absolute value definition you've got two ways:
x-3=4 or x-3=-4
x=7 or x=-1
And answer d) passes
