if they pass each other in 3hrs, that means, that plane one has been flying for 3hrs and plane two has also been flying for 3hrs, their flying time is the same, by the time they meet, in reference to those 2235 miles they're in
now... one is faster than the other, by 45mph... so... let's say hmm plane one is the slower one, and is flying at a rate of "r"
then plane two has to be flying at a rate of "r + 45" in mph
by the time plane one has been flying for 3hrs at "r" speed, it covered distance "d", whatever that is
now, plane two has also been flying for 3hrs, but at "r+45" speed, and it covered, the slack from the 2235 miles, or 2235-d
their time is the same, so hmmm recall that d = r*t
thus [tex]\bf \begin{array}{lcccllll}
&distance(miles)&rate(mph)&time(hrs)\\
&------&------&------\\
\textit{plane one}&d&r&3\\
\textit{plane two}&2235-d&r+45&3
\end{array}\\\\
-----------------------------\\\\
\begin{cases}
d=(r)(3)\to \boxed{d}=3r\\\\
2235-d=(r+45)(3)\\
----------\\
2235-\left( \boxed{3r} \right)=3(r+45)
\end{cases}[/tex]
solve for "r", to find the speed of plane one
what's the speed of plane two? well, r + 45
