imc

Note that when $y=0$, the $x$ values of the equations should be equal by the problem statement. We have that $$0=ax+5⟹x=-\frac{5}{a}$$ $$0=3x+b⟹x=-\frac{b}{3}$$ Which means that $$-\frac{5}{a}=-\frac{b}{3}⟹ab=15$$ The only possible pairs $(a,b)$ then are $(a,b)=(1,15),(3,5),(5,3),(15,1)$. These pairs give respective $x$-values of $-5,-\frac{5}{3},-1,-\frac{1}{3}$ which have a sum of $\boxed{-8}$.