Question 724285
Here we are, exactly seven minutes late. This time I averaged 30 miles per hour.
Last time we did the trip, we were five minutes early. 
We started at the same time.  This time I averaged 36 miles per hour.
What was the distance?
:
Since we are using mph:
Change 5 min to 5/60 hr
Change 7 min to 7/60 hr
:
let t = normal "on-time" time for the trip
:
Write a distance equation, dist = speed * time,
36(t-{{{5/60}}}) = 30(t+{{{7/60}}})
divide both sides by 6
6(t-{{{5/60}}}) = 5(t+{{{7/60}}})
6t - {{{30/60}}} = 5t + {{{35/60}}}
reduce the fractions
6t - {{{6/12}}} = 5t + {{{7/12}}}
6t - 5t = {{{7/12}}} + {{{6/12}}}
t = {{{13/12}}} hr is the normal travel time
:
Find the distance when 7min late, 67 min
d = 30 * [{{{13/12)}}} + {{{7/60}}}]
d = 30 * [{{{65/60)}}} + {{{7/60}}}]
d = 30 * {{{72/60}}}
d = {{{72/2}}}
d = 36 miles
:
Check the distance when 5 min early
d = 36 * [{{{65/60}}} - {{{5/60}}}]
d = 36 * {{{60/60}}} 
d = 36  miles,