Working with $z$, $\overline{z}$ instead of $\operatorname{Re}(z)$, $\operatorname{Im}(z)$

Working with $z$, $\overline{z}$ instead of $\operatorname{Re}(z)$,
$\operatorname{Im}(z)$

The problem is the following:
Determine conditions for $a,b,c\in\mathbb{C}$ such that
$az+b\overline{z}+c=0$ has unique solution in $\mathbb{C}$.
Teacher answer:
If $az+b\overline{z}+c=0$, then
$\overline{b}z+\overline{a}\overline{z}+\overline{c}=0$. So
$\left(\begin{array}\\ a & b \\ \overline{b}&\overline{a}
\end{array}\right)\left(\begin{array}\\ z \\ \overline{z}
\end{array}\right)=\left(\begin{array}\\ -c \\ -\overline{c}
\end{array}\right)$. Like in linear algebra, we have only one solution iff
$|a|^2-|b|^2\neq 0$.
My answer is too long (equaling real and imaginary parts and taking a 2x2
system of linear equations), and teacher answer didn't convince me because
teacher is solving a system where one component $z$ depends of another
component which is $\overline{z}$ (if we obtain $z$ we automatically
obtain $\overline{z}$ and viceversa).