recall your d = rt, distance = rate * time
let's say the boat's rate is "b" in still water, and the current's rate is "c"
as the boat goes upstream, is not really going "b" fast, the water is not still anyway, is going much slower, is going " b - c ", because the current's rate is eroding speed because is going against the current.
as the boat goes downstream, is not going "b" fast either, is going faster, is going " b + c " fast, because is going downstream and thus with the current and the current is adding speed to it. notice, it takes longer going up, 5hrs, then coming down, 3hrs, same 150 on each way.
[tex]\bf \begin{array}{llll}
&distance&rate&time\\
&-----&-----&-----\\
upstream&150&b-r&5\\
downstream&150&b+r&3
\end{array}
\\\\\\
\begin{cases}
150=(b-r)5\\
\qquad \frac{150}{5}=b-r\\
\qquad 30=b-r\\
\qquad 30+r=\boxed{b}\\
150=(b+r)3\\
\qquad \frac{150}{3}=b+r\\
\qquad 50=b+r\\
\qquad 50=\boxed{30+r}+r
\end{cases}[/tex]
solve for "r"
what's the boat's rate? well, 30+r = b
