$\displaystyle g(x)=\frac{2x-1}{x+3}$

My Attempt At The Inverse

Let $\displaystyle g(x)=z$

$\displaystyle (x+3)(z)=2x-1$

$\displaystyle xz+3z=2x-1$

$\displaystyle 1+3z=2x-xz$

$\displaystyle 1+3z=x(2-z)$

$\displaystyle \frac{1+3z}{2-z}=x$

$\displaystyle g^-1(x)=\frac{1+3x}{2-x}$

Where did I make a mistake?

Thanks!