Solve for

We have, $(1-\mathrm{i}) \mathrm{x}+(1+\mathrm{i}) \mathrm{y}=1-3 \mathrm{i}$

$\Rightarrow x-i x+y+i y=1-3 i$

$\Rightarrow(x+y)+i(-x+y)=1-3 i$

On equating the real and imaginary coefficients we get,

$\Rightarrow x+y=1$ (i) and $-x+y=-3$ (ii)

From (i) we get

$x=1-y$

Substituting the value of x in (ii), we get

$-(1-y)+y=-3$

$\Rightarrow-1+y+y=-3$

$\Rightarrow 2 y=-3+1$

$\Rightarrow y=-1$

$\Rightarrow x=1-y=1-(-1)=2$

Hence, $x=2$ and $y=-1$