recall your d = rt, distance = rate * time.
p = speed of the plane
w = speed of the wind
let's keep in mind that, when the plane is going with the wind, is not really going "p" fast is actually going "p + w" since the wind is adding speed to it, likewise, when the plane is going against the wind, the plane is going "p - w" fast, since the wind is subtracting speed from it.
[tex]\bf \begin{array}{lcccl} &\stackrel{km s}{distance}&\stackrel{kph}{rate}&\stackrel{hours}{time}\\ \cline{2-4}&\\ \textit{against the wind}&3605&p-w&5\\ \textit{with the wind}&4605&p+w&5 \end{array}~\hfill \begin{cases} 3605=(p-w)5\\ \frac{3605}{5}=p-w\\ 721=p-w\\ 721+w=\boxed{p}\\ \cline{1-1} 4605=(p+w)5 \end{cases} \\\\\\ 4605=(p+w)5\implies \cfrac{4605}{5}=p+w\implies \stackrel{\textit{substituting in the 2nd equation}}{921=\left( \boxed{721+w} \right)+w}[/tex]
[tex]\bf 921=721+2w\implies 200=2w\implies \cfrac{200}{2}=w\implies \blacktriangleright 100=w \blacktriangleleft \\\\\\ \stackrel{\textit{since we know that}}{p=721+w\implies }p=721+100\implies \blacktriangleright p=821 \blacktriangleleft[/tex]
