let's recall that d = rt, distance = rate * time.
he went upstream to a distance "d", got tired and came back to his starting point, so he rowed back a distance "d" exactly.
we know the rates, we also know the trip took 2 hours, let's say on the way over he took "t" hours to get there, on the way back he lasted then "2 - t" hours.
[tex]\begin{array}{lcccl} &\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\ \cline{2-4}&\\ Upstream&d&2&t\\ Downstream&d&8&2-t \end{array}\qquad \qquad \begin{cases} d=2t\\\\ d=8(2-t) \end{cases}[/tex]
[tex]\stackrel{\textit{substituting on the 2nd equation}}{2t=8(2-t)\implies 2t=16-8t}\implies 10t=16\implies t=\cfrac{16}{10}\implies t=\cfrac{8}{5} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{we know that}}{d=2t}\implies d=2\left( \cfrac{8}{5} \right)\implies \stackrel{\textit{3 miles and 1056 feet}}{d=\cfrac{16}{5}\implies d=3\frac{1}{5}}[/tex]
