$\begin{aligned} & \frac{(1+i) x-2 i}{3+i}+\frac{(2-3 i) y}{3-i}=i \\ & \Rightarrow((1+i) x-2 i)(3-i)+((2-3 i) y)(3+i)=10 i \\ & \Rightarrow(x+i x-2 i)(3-i)+(2 y-3 i y)(3+i)=10 i \\ & \Rightarrow 3 x+3 i x-6 i-i x+x-2 \\ & \quad+6 y-9 i y+2 i y+3 y=10 i \\ & \Rightarrow(3 x+x-2+6 y+3 y)+ \\ & \Rightarrow 4 x+9 y-2=0,2 x-7 y-16=0 \\ & \Rightarrow y=\frac{-30}{23}, x=\frac{79}{23} . \\ & \therefore x+y=\frac{79}{23}-\frac{30}{23}=\frac{49}{23}\end{aligned}$