smc

The man has three possible combinations of transportation: $$\text{Morning train, Afternoon bus (m.t., a.b.)}$$ $$\text{Morning bus, Afternoon train (m.b., a.t.)}$$ $$\text{Morning bus, Afternoon bus (m.b, a.b.)}$$. Let $y$ be the number of times the man takes the $\text{a.t.}$. Then, $9-y$ is the number of times he takes the $\text{m.t.}$. Keep in mind that $\text{m.b.}=y$ and $\text{a.b.}=9-y$. Let $z$ be the number of times the man takes the $\text{m.b.}$ and $\text{a.b.}$. Now, we get the two equations $$y+z=8$$ and $$9-y+z=15.$$ Solving the system of equations, we get $y=1$ and $z=7$. So during the $x$ working days, the man took the $\text{(m.t., a.b.)}$ on $9-1=8$ days, the $\text{(m.b., a.t.)}$ on $1$ day, and the $\text{(m.b., a.b.)}$ on $7$ days. Therefore, $x=8+1+7=\boxed{16}$.