Question

Help! Please!

Answer 1

Ryan drove to the mountains last weekend. There was heavy traffic on the way there, and the trip took
6
hours. When Ryan drove home, there was no traffic and the trip only took
4
hours. If his average rate was
22
miles per hour faster on the trip home, how far away does Ryan live from the mountains?
Do not do any rounding.

Answer 2

In problems like this always start with a diagram showing what you know. I left out the rate for the heavy traffic leg of the trip. Fill it in and make two equations of the form d = .

The d is the same for both legs, so simply equate the right sides of the equations, solve for r, and then d.

Answer 3

\(\large {
\begin{array}{rcccllll}
&(d)istance&(r)ate&(t)ime
\\\hline\\
drive\ over&d&r&6\\
drive\ back&d&r+22&4
\end{array}\implies
\begin{array}{llll}
d=rt\\
\hline\\
d=r\cdot 6\\
d=(r+22)\cdot 4
\end{array}
}\)

as NoelGreco suggested
any ideas?

Answer 4

bear in mind that "his average rate was 22 miles per hour faster on the trip home"
thus
if we went over with rate of say "r"
his way back rate was " r + 22"