Inverse Function Verification 
Are functions $f(x)=\frac{7x+4}{x+6}$ and $g(x)=\frac{6x-4}{7-x}$ inverses of each other?

I'm experiencing a very strange issue with this problem. When I find the inverse of $g$ I get $f(x)$. However, when I do the same thing for $f$ I do not get $g(x)$. 
\begin{align*}
f(x)&=\frac{7x+4}{x+6} \equiv y=\frac{7x+4}{x+6} \\ x&=\frac{7y+4}{y+6} \\ x(y+6)&=7y+4 \\ xy-7y&=-6x+4 \\ y(x-7)&=-6x+4\\ y&=\frac{-6x+4}{x-7} \\ \therefore f^{-1}(x)&=\frac{-6x+4}{x-7}
\end{align*}
It's very close, only the signs deviate. Yet when I try this method on $g(x)$ I get 
\begin{align*}
x&=\frac{6y-4}{7-y} \\ 7x-xy&=6y-4 \\ -y&=(\frac{-7x+4}{6+x}) \\ \therefore g^{-1}(x)&=\frac{7x+4}{x+6}
\end{align*}
Which says the functions are inverses of each other. What am I getting wrong with finding the inverse of function $f$?

It should be
\begin{align*}
-y&=-(\frac{7x+4}{6+x}) \\ \therefore g^{-1}(x)&=\frac{7x+4}{x+6}=\ f(x)
\end{align*}
And you already have $$f^{-1}(x)=\frac{-6x+4}{x-7}=\frac{6x-4}{7-x}=g(x)$$
