If the point $\left(x,y\right)$ satisfies the equation $\frac{x+i\left(x-2\right)}{3+i}-i=\frac{2y+i\left(1-3y\right)}{i-3}$, then $x+y=$

Given,

The point $\left(x,y\right)$ satisfies the equation $\frac{x+i\left(x-2\right)}{3+i}-i=\frac{2y+i\left(1-3y\right)}{i-3}$

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

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

$\Rightarrow \left(x+i\left(x-2\right)\right)\left(3-i\right)-10i=-\left(2y+i\left(1-3y\right)\right)\left(i+3\right)$

$\Rightarrow \left(3x-ix+i\left(x-2\right)\left(3-i\right)\right)-10i=-\left(2iy+6y+i\left(1-3y\right)\left(i+3\right)\right)$

$\Rightarrow \left(3x-ix+i\left(3x-ix-6+2i\right)\right)-10i=-\left(2iy+6y+i\left(i+3-3iy-9y\right)\right)$

$\Rightarrow \left(3x-ix+3ix+x-6i-2\right)-10i=-\left(2iy+6y-1+3i+3y-9iy\right)$

$\Rightarrow 4x+2ix-13i+9y-3-7iy=0$

$\Rightarrow 4x+9y-3+i\left(2x-13-7y\right)=0$

Now comparing real part and imagenary part we get,

$\Rightarrow 4x+9y-3=0 ........\left(1\right)$ and $\Rightarrow \left(2x-13-7y\right)=0 ..........\left(2\right)$

Now solving equation $\left(1\right) \& \left(2\right)$ we get, $x=3 \& y=-1$

Then $x+y=2$