recall your d = rt, distance = rate * time
let's say we have two cars, A and B, B has a speed rate of "r", so A is the faster one and has a speed rate of " r + 10 "
they both are 300 miles apart after 3hrs, after 3hrs, car has been running for 3hrs, and car B has also been running for 3hrs, so their time is the same
we know their distances added up, is 300miles, so if car B covered say, "d" distance on those 3hrs, car A covered the slack, " 300 - d "
[tex]\bf \begin{array}{lccclll}
&distance&rate&time\\
&-----&-----&-----\\
\textit{car A}&300-d&r+10&3\\
\textit{car B}&d&r&3
\end{array}\\\\
-------------------------------\\\\
\begin{cases}
300-d=(r+10)3\\
\boxed{d}=3r\\
----------\\
300-\boxed{3r}=3(r+10)
\end{cases}[/tex]
solve for "r"
