# How can I find x, y values for ( 1 + i ) x &#x221

How can I find x, y values for $\frac{\left(1+i\right)x-2i}{3+i}+\frac{\left(2-3i\right)y+i}{3-i}=i$
I believe the format I need in order to solve this problem should be such that the real parts and imaginary parts are separated, $\left\{\mathrm{R}\mathrm{e}\mathrm{a}\mathrm{l}\right\}+\left\{\mathrm{I}\mathrm{m}\mathrm{a}\mathrm{g}\mathrm{i}\mathrm{n}\mathrm{a}\mathrm{r}\mathrm{y}\right\}i=i$
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:
$\frac{x+i\left(x-2\right)}{3+i}+\frac{2y+i\left(-3y+1\right)}{3-i}=i$
Multiplied by conjugate:
$\frac{x\left(3-i\right)+i\left(x-2\right)\left(3-i\right)}{10}+\frac{2y\left(3+i\right)+i\left(-3y+1\right)\left(3+i\right)}{10}=i$
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?
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toriannucz
$\frac{x\left(3-i\right)+i\left(x-2\right)\left(3-i\right)}{10}+\frac{2y\left(3+i\right)+i\left(-3y+1\right)\left(3+i\right)}{10}=i$
$\frac{3x-ix+3ix+x-6i-2+6y+2iy-9iy+3y+3i-1}{10}=i$
$\frac{4x+9y-3}{10}+\frac{i\left(2x-7y-3\right)}{10}=0+i$

solve simultaneously to obtain values for x and y

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