recall your d = rt, that is, distance = rate * time
now, bear in mind, that hmm, if say, the boat's speed in still water is say "r", when the boat is going upstream, is not really going at "r" speed, but the stream is subtracting from it, so is really going at " r - 4 " mph fast, since the speed rate of the stream is 4
now, when the boat is going downstream, is going "with" the stream, so is really going faster than "r", is actually going at " r + 4 " mph fast
now, the time it took upstream, is the same time it took downstream... say it took "t" time
thus [tex]\bf \begin{array}{lccclll}
&distance&rate&time\\
&-----&-----&-----\\
upstream&11&r-4&t\\
downstream&19&r+4&t
\end{array}
\\\\\\
\begin{cases}
11=t(r-4)\implies \frac{11}{r-4}=\boxed{t}\\\\
19=t(r+4)\\
----------\\
19=\boxed{\frac{11}{r-4}}(r+4)
\end{cases}[/tex]
solve for "r"
