Recall your d = rt, distance = rate * time
bearing in mind that the trip over, upstream, is the same distance as back, 280 miles.
if say, the boat has a speed rate of say "b", and the current has a speed rate of "c", so, when the boat was going upstream, it really wasn't going "b" fast, it was going "b-c" fast, because the stream is subtracting speed from it, because is going against the stream.
And when the boat was going downstream, is not going "b" fast either, is going "b+c" because, since it's going with the current, the current's rate is adding speed to it.
![\bf \begin{array}{lccclll} &\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\ &------&------&------\\ Downstream&280&b+c&7\\ Upstream&280&b-c&14 \end{array} \\\\\\ \begin{cases} 280=7(b+c)\implies \frac{280}{7}=b+c\\ 40=b+c\implies 40-c=\boxed{b}\\ -------------\\ 280=14(b-c)\implies \frac{280}{14}=b-c\\ 20=b-c\\ ----------\\ 20=\left( \boxed{40-c} \right)-c \end{cases} \\\\\\ 20=40-2c\implies 2c=40-20\implies c=\cfrac{20}{2}](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7Blccclll%7D%0A%26%5Cstackrel%7Bmiles%7D%7Bdistance%7D%26%5Cstackrel%7Bmph%7D%7Brate%7D%26%5Cstackrel%7Bhours%7D%7Btime%7D%5C%5C%0A%26------%26------%26------%5C%5C%0ADownstream%26280%26b%2Bc%267%5C%5C%0AUpstream%26280%26b-c%2614%0A%5Cend%7Barray%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Cbegin%7Bcases%7D%0A280%3D7%28b%2Bc%29%5Cimplies%20%5Cfrac%7B280%7D%7B7%7D%3Db%2Bc%5C%5C%0A40%3Db%2Bc%5Cimplies%2040-c%3D%5Cboxed%7Bb%7D%5C%5C%0A-------------%5C%5C%0A280%3D14%28b-c%29%5Cimplies%20%5Cfrac%7B280%7D%7B14%7D%3Db-c%5C%5C%0A20%3Db-c%5C%5C%0A----------%5C%5C%0A20%3D%5Cleft%28%20%5Cboxed%7B40-c%7D%20%5Cright%29-c%0A%5Cend%7Bcases%7D%0A%5C%5C%5C%5C%5C%5C%0A20%3D40-2c%5Cimplies%202c%3D40-20%5Cimplies%20c%3D%5Ccfrac%7B20%7D%7B2%7D)
what's the speed of the boat? well, 40 - c = b.