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$\displaystyle (1 - i)w+z = 2i \quad ............(1)$
$\displaystyle w + (1 - i)z = 1 \quad ............(2)$
multiply (2) by (1 - i) and subtract from (1), we get
$\displaystyle \{ (1 - i)w+z\} -\{ (1 - i)w + (1 - i)^2z \}=2i-(1-i)$
$\displaystyle \Rightarrow \quad z-(1 - i)^2z=(3i-1)\quad \Rightarrow \quad z-(-2 i)z=(-1+3i) $
$\displaystyle \Rightarrow \quad(1+2i)z=-1+3i\quad \Rightarrow \quad z=\frac{-1+3i}{1+2i}$
rationalize $\displaystyle z=\frac{-1+3i}{1+2i}$ we get$\displaystyle \boxed{z=1+i}$
substitute value of z in (2),we get $\displaystyle \boxed{w=-1}$