[tex]y=\frac{1}{3}x-2[/tex]
Multiplying by 3 at both sides:
[tex]\begin{gathered} 3y=3(\frac{1}{3}x-2) \\ 3y=3\cdot\frac{1}{3}x-3\cdot2 \\ 3y=x-6 \end{gathered}[/tex]
Subtracting x at both sides:
[tex]\begin{gathered} 3y-x=x-6-x \\ -x+3y=-6 \end{gathered}[/tex]
Then, option D is correct
