recall your d = rt, or distance = rate * time
the two planes, are 1180 miles apart
say plane A and plane B
two hours later, they pass each other
how long have both been flying by the time they passed each other?
well, 2 hrs
so the time for both is the same, 2
now, one is faster than the other, by 4 miles, say hmmm plane A, doesn't really matter, it could be either... but say A, so if A has a rate of "r",
B will have a rate of r-4, whatever "r" is
now, the distances travelled, say A travelled distance "d", whatever that is,
B, travelled the difference from 1180 and that, or 1180 - d
1180 is the total amount of miles apart, take away "d", and B travelled
1180 - d
thus [tex]\bf \begin{array}{rccclll}
&distance&rate&time(hrs)\\
&-----&-----&-----\\
\textit{first plane}&d&r&2\\
\textit{second plane}&1180-d&r-4&2
\end{array}
\\\\\\
thus\implies
\begin{cases}
\boxed{d}=(r)(2)\\\\
1180-d=(r-4)(2)\\
--------------\\\\
1180-\boxed{(r)(2)}=(r-4)(2)
\end{cases}[/tex]
solve for "r", to see A's rate
how fast was B going? well, r-4
