Solving $(1 + i) z - 2 \overline{z} = -11 + 25i$ How do we find the complex number $z$ that satisfies
$$(1 + i) z - 2 \overline{z} = -11 + 25i.$$
How should I start? Am I supposed to express $z$ as $x+yi$?

Since both, left hand side and right hand side of
\begin{align*}
(1 + i) z - 2 \overline{z} = -11 + 25i\tag{1}
\end{align*}
represent a complex number, we can equate real and  imaginary part  of them.  We  set $z=x+iy$ with real part $x$ and imaginary part $y$.

We obtain
  \begin{align*}
\Re\left((1+i)(x+iy)-2(x-iy)\right)&=(x-y)-2x\\
&=-x-y\tag{2}\\
\Im\left((1+i)(x+iy)-2(x-iy)\right)&=(x+y)+2y\\
&=x+3y\tag{3}
\end{align*}
Equating real part (2) and imaginary part (3) with the RHS of (1) we get
  \begin{align*}
-x-y&=-11\\
x+3y&=25
\end{align*}
  and finally obtain 
  \begin{align*}
z&=x+iy\color{blue}{=4+7i}
\end{align*}

