jmc

While we can solve for $a$ and $b$ individually, it is simpler to note that $f(1)=a+b$. Thus, substituting $1$ into the given equation, we obtain $$g(f(1))=3\cdot 1+4=7.$$ Thus, $$g(f(1))=2\cdot f(1)-5=7⟹f(1)=\boxed{6}.$$