The given expression is,
[tex](\frac{1}{3}x)+y=4[/tex]On solving,
[tex]\begin{gathered} \frac{x}{3}+y=4 \\ x=12-3y \end{gathered}[/tex]Substituting the value of x in the first equation,
[tex]\begin{gathered} \frac{12-3y}{3}+y=4 \\ 12-3y+3y=12 \\ y=0 \end{gathered}[/tex]Thus, y =0
