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.

![\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(1)/(5)}](https://img.qammunity.org/2023/formulas/mathematics/high-school/niwfmr34d89try93y37a1h0bgfrsab52rj.png)