recall your d = rt, distance = rate * time.
as the plane goes against the wind, the plane is not really flying at "x" mph, but is really going slower, at " x - y ", because the wind is subtracting speed from it.
likewise, when the plane is going with the wind, is not really going at "x" mph, but at " x + y ", is going faster due to the wind, thus
[tex]\bf \begin{array}{lccclll}
&\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\
&------&------&------\\
\textit{against the wind}&1720&x-y&5\\
\textit{with the wind}&1720&x+y&4
\end{array}
\\\\\\
\begin{cases}
1720=(x-y)5\implies \frac{1720}{5}=x-y\\\\
344=x-y\implies \boxed{y}=x-344\\
---------------\\
1720=(x+y)4\implies \frac{1720}{4}=x+y\\\\
430=x+y\implies 430=x+\left( \boxed{x-344} \right)
\end{cases}
\\\\\\
430=2x-344\implies 774=2x\implies \cfrac{774}{2}=x\implies 387=x[/tex]
