It should be 58. I’m sorry if I’m wrong!!
The picture is really blurry
Recall your d = rt, distance = rate * time.
going forth and back is 80 miles, thus upstream as well as downstream is just 80 miles.
now, bear in mind that say, the boat has a "still water" of "b", when going downstream, is not going "b" fast is going "b+7" because the current's is adding to it, likewise when is going upstream is going "b-7" because is going against the current and thus the current is eroding speed from it.
now, the whole trip took 3 hours and 20 minutes, bearing in mind that 20 minutes is 1/3 of an hour, so is 3 and 1/3 hours, or 10/3 of hours.
if say it took "t" hours going down, then going up it took the slack from 10/3 and "t", that is, it took "10/3 - t".
![\bf \begin{array}{lccclll} &\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\ &------&------&------\\ Downstream&80&b+7&t\\ Upstream&80&b-7&\frac{10}{3}-t \end{array} \\\\\\ \begin{cases} 80=t(b+7)\implies \frac{80}{b+7}=\boxed{t}\\\\ 80=(b-7)\left(\frac{10}{3}-t \right)\\ ----------\\ 80=(b-7)\left(\cfrac{10}{3}- \boxed{\frac{80}{b+7}} \right) \end{cases}](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%2680%26b%2B7%26t%5C%5C%0AUpstream%2680%26b-7%26%5Cfrac%7B10%7D%7B3%7D-t%0A%5Cend%7Barray%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Cbegin%7Bcases%7D%0A80%3Dt%28b%2B7%29%5Cimplies%20%5Cfrac%7B80%7D%7Bb%2B7%7D%3D%5Cboxed%7Bt%7D%5C%5C%5C%5C%0A80%3D%28b-7%29%5Cleft%28%5Cfrac%7B10%7D%7B3%7D-t%20%20%5Cright%29%5C%5C%0A----------%5C%5C%0A80%3D%28b-7%29%5Cleft%28%5Ccfrac%7B10%7D%7B3%7D-%20%5Cboxed%7B%5Cfrac%7B80%7D%7Bb%2B7%7D%7D%20%5Cright%29%0A%5Cend%7Bcases%7D)