jmc

Since the two lines intersect at $(6,10)$, each must pass through that point. So, we can substitute this point for $x$ and $y$ in the two given equations, then solve for $m$ and $b$. In the first equation we find: $$\begin{array}{rl}10& =m(6)+4\\ \Rightarrow 6& =6m\\ \Rightarrow 1& =m\end{array}$$ Using the same method in the second equation gives: $$\begin{array}{rl}10& =3(6)+b\\ \Rightarrow 10& =18+b\\ \Rightarrow -8& =b\end{array}$$ The value of $b+m$ is equal to $1+(-8)=\boxed{-7}$.