Let v_b be the speed of the boat in still water; on the other hand, the relation between speed, distance, and time is
[tex]v=\frac{d}{t}[/tex]Transform minutes into hours as shown below
[tex]\begin{gathered} \frac{15}{60}=0.25 \\ \frac{5}{60}=\frac{1}{12} \end{gathered}[/tex]Therefore, in our case, the two equations are
[tex]\begin{gathered} d=0.25(v_b-9)\rightarrow\text{ upstream} \\ d=\frac{1}{12}(v_b+9)\rightarrow\text{ downstream} \end{gathered}[/tex]Thus,
[tex]\begin{gathered} d=d \\ \Rightarrow0.25(v_b-9)=\frac{1}{12}(v_b+9) \\ \Rightarrow0.25v_b-2.25=\frac{v_b}{12}+\frac{3}{4} \\ \Rightarrow(\frac{1}{4}-\frac{1}{12})v_b=2.25+\frac{3}{4} \\ \Rightarrow\frac{1}{6}v_b=3 \\ \Rightarrow v_b=18 \end{gathered}[/tex]Therefore, the answer is 18km/h