jmc

Putting the $x$ and $y$ terms of the second equation on the left gives $6x+15y=16+a$. Multiplying the first equation by 3 gives $6x+15y=-24$. So, our system now is $$\begin{array}{rl}6x+15y& =-24,\\ 6x+15y& =16+a.\end{array}$$This system has infinitely many solutions only if the two right sides are the same, which would make the two equations the same. So, we must have $-24=16+a$, so $a=\boxed{-40}$.