How can I find x, y values for $\frac{(1+i)x-2i}{3+i}+\frac{(2-3i)y+i}{3-i}=i$

I believe the format I need in order to solve this problem should be such that the real parts and imaginary parts are separated, $\{\mathrm{R}\mathrm{e}\mathrm{a}\mathrm{l}\}+\{\mathrm{I}\mathrm{m}\mathrm{a}\mathrm{g}\mathrm{i}\mathrm{n}\mathrm{a}\mathrm{r}\mathrm{y}\}i=i$

Then I can equate the real and imaginary parts of the equation and solve for x and y. I tried to multiply by the conjugate so that I would get a real number at the denominator.

Rearranged:

$\frac{x+i(x-2)}{3+i}+\frac{2y+i(-3y+1)}{3-i}=i$

Multiplied by conjugate:

$\frac{x(3-i)+i(x-2)(3-i)}{10}+\frac{2y(3+i)+i(-3y+1)(3+i)}{10}=i$

And at this step I started to feel as if I made a mistake because I weren't sure how to proceed. Would someone let me know if my approach was correct and show me how to do this?

I believe the format I need in order to solve this problem should be such that the real parts and imaginary parts are separated, $\{\mathrm{R}\mathrm{e}\mathrm{a}\mathrm{l}\}+\{\mathrm{I}\mathrm{m}\mathrm{a}\mathrm{g}\mathrm{i}\mathrm{n}\mathrm{a}\mathrm{r}\mathrm{y}\}i=i$

Then I can equate the real and imaginary parts of the equation and solve for x and y. I tried to multiply by the conjugate so that I would get a real number at the denominator.

Rearranged:

$\frac{x+i(x-2)}{3+i}+\frac{2y+i(-3y+1)}{3-i}=i$

Multiplied by conjugate:

$\frac{x(3-i)+i(x-2)(3-i)}{10}+\frac{2y(3+i)+i(-3y+1)(3+i)}{10}=i$

And at this step I started to feel as if I made a mistake because I weren't sure how to proceed. Would someone let me know if my approach was correct and show me how to do this?