recall your d = rt, distance = rate * time.
bearing in mind that by the time the Jet matches the Propeller, the distance "d" miles is the same for each.
when that happens, if the Jet has travelled for "t" hours, we know the Propeller had 2 hours headstart, so by then the Propeller has been travelling "t + 2" hours.
[tex]\bf \begin{array}{lcccl}
&\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\
&------&------&------\\
Jet&d&500&t\\
Propeller&d&200&t+2
\end{array}
\\\\\\
\begin{cases}
d=500t\implies \cfrac{d}{500}=\boxed{t}\\\\
d=200(t+2)\\
-----------\\
d=200\left(\boxed{\frac{d}{500}}+2 \right)
\end{cases}[/tex]
[tex]\bf d=200\left( \cfrac{d+1000}{500} \right)\implies \stackrel{\textit{multiplying both sides by the }\stackrel{LCD}{500}}{500d=200(d+1000)}
\\\\\\
500d=200d+200000\implies 300d=200000\implies d=\cfrac{200000}{300}
\\\\\\
d=\cfrac{2000}{3}\implies d=\stackrel{miles}{666\frac{2}{3}}\qquad \textit{or 3520000 feet}[/tex]