$z=2+i, w=-i$

$\begin{aligned}
-z + 2 i w &=-i \\2 i w &=z-i
\\ i w &=\frac{1}{2}(z-i) — (1)
\\ \text { sub (1) in the second given equation }
\\(2-i) z+\frac{1}{2}(z-i) &=6 \\\left(2-i+\frac{1}{2}\right) z &=6+\frac{1}{2} i \\\left(\frac{5}{2}-i\right) z &=6+\frac{1}{2} i \\(5-2 i) z &=12+i
\end{aligned}$

$$\\ z =\frac{12+i}{5-2 i} \times \frac{5+2 i}{5+2 i}=2+i$$
$\begin{aligned}
\\ \Rightarrow 2 i-1 &+2 w=-1
\\ w &=-i
\end{aligned}$