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.
[tex]\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}[/tex]
what's the speed of the boat? well, 40 - c = b.