jmc

First we begin by solving the system of equations $$\begin{array}{rl}5a& =-4b+5,\\ 3a& =-2b+3.\end{array}$$Subtracting two times the second equation from the first equation, we get $5a-2(3a)=-4b+5-2(-2b+3)$, which simplifies to $-a=-1$. So $a=1$ and plugging this into the first equation we obtain $5=-4b+5$. Solving for $b$, we find that $b=0$. Hence $6b=6\cdot 0=\boxed{0}$.