The system:
$$(10-i)z-(2-i)\bar{w}=26+8i$$ $$(3+2i)\bar{z}+(1-2i)w=8-4i$$
$w$ and $z$ are complex numbers.
So for this kind of problems I would usually put it in matrix form and solve from there. But here I cannot do that since some numbers are in conjugate form.
I tried getting from the conjugate number to the normal complex number by $z=\bar{z}+\frac{z}{\bar{z}}$ But it seems that this way I will get nowhere. How should I tackle this problem?
On
Here is the system of 4 equations, which Yves Daoust has mentioned in his comment. This system has no imaginary numbers, but has the same collection of solutions as your system with complex numbers. This system can be obtained by substituting $z=x+iy \space$ and $\space w=a+ib \space$ and then separating the real and imaginary parts.
$$ \begin{matrix} 10x & + & y & - & 2a & + & b & = & \space\space 26 \\ 3x & + & 2y & + & a & + &2 b & = & \space\space\space 8 \\ -x & + & 10y & + & a & + & 2b & = & \space\space\space 8 \\ 2x & - & 3y & - & 2a & + & b & = & -4 \\ \end{matrix} $$
Here is the matrix form of the above system of equations.
The solution can be obtained using Cramers's rule, Gaussian elimination, or any other method. The solutions are, $$x=3, y=1.5, a=1.4, \space\rm{and}\space \it{b} \rm{=-2.7}.$$
Therefore, $$z=3+1.5i \space\space\rm{and}\space \space \it{w}\rm{=1.4-2.7}\it{i}. $$
This method works for systems with and without complex conjugate.
Conjugate the entire second equation and you now have two equations in the unknowns $z$ and $\bar w$, that you can solve the usual way.