Let $\frac{2iz-2i}{i+3}=\frac{2+3i}{10}$ be an equation with the complex variable $z$. To solve it, I write the first member on the standard form ($z=z_1+iz_2$) and next I solve the linear system in the unknows $z_1$ and $z_2$ equationg real parts and imaginary parts, respectively.
However, if I solve it as a classical equation of first degree. May I write $$ z-1=\frac{i+3}{2i} \frac{2+3i}{10}? $$
Hint: By cross multiplication we get $$10(2iz-2i)=(2+3i)(ik+3)$$ or $$20iz-20i=2i-3+6+9i$$
so $$20iz=31i+3$$ so $$z=\frac{31i+3}{20i}$$ finally we get $$z=\frac{-3i+31}{20}$$