Verified answer

[tex]\displaystyle\bf\\(1-i)z-3iz=2-i\\\\z\cdot(1-i-3i)=2-i\\\\z\cdot(1-4i)=2-i\\\\\\z=\frac{2-i}{1-4i} =\frac{(2-i)\cdot(1+4i)}{(1-4i)\cdot(1+4i)}=\frac{2+8i-i-4i^{2} }{1-16i^{2} } =\\\\\\=\frac{2+7i-4\cdot(-1)}{1-16\cdot(-1)} =\frac{2+7i+4}{1+16} =\frac{6+7i}{17} =\frac{6}{17} +\frac{7}{17} i[/tex]