Question

Larry's time to travel 357 miles is 3 hours more than Terrell's time to travel 220 miles. Terrell drove 4 miles per hour faster than Larry. How fast did each one travel?

Answer 1

so, if terrell traveled "h" hours, larry traveled "h+3"

terrel drove 4 miles faster
so if larry drove "r" m/h fast, terrell drove "r+4" m/h faster

Answer 2

so
$$
\begin{matrix}
& d & t & r \\
\hline
larry & 357 & h+3 & r \\
terrell & 220 & h & r+4\\
\end{matrix}
$$

Answer 3

so, using the d = rt distance formula
larry traveled "t" distance, or t = d/r or
\(h+3 = \cfrac{357}{r} \implies h = \cfrac{357}{r}-3\)

Answer 4

terrell's rate is r+4 which

\(r+4 =\cfrac{220}{h}\)
but \( h = \cfrac{357}{r}-3\) from the 1st equation so
$$
r+4 =\cfrac{220}{\cfrac{357}{r}-3}
$$

Answer 5

$$
r+4 =\cfrac{220}{\cfrac{357}{r}-3} \implies
r+4 = \cfrac{220r}{357-3r}\\ \implies
(r+4)(357-3r) = 220r\\
\implies
-3r^2-12r+357r+1428-220r = 0\\
\implies
\color{blue}{-3r^2+125r+1428=0}\\
\text{from there you'd need to use the quadratic formula}\\
\text{to get "r" or "larry's rate"}
$$

Answer 6

@inprogress you'd get 2 values, discard the negative one, since "rate" is positive

Answer 7

@inprogress

$$
\text{quadratic formula for it}\\
\cfrac{-125\sqrt{15625+17136}}{-6}\\
\cfrac{-125\pm 181}{-6}
$$

Answer 8

A solution using Mathematica to perform the calculations is attached.