let he drove for x hours the first day at y miles per hour...
so in total

xy + (x+3)(y+15) = 1020

but x + x + 3 =21

solving we get y = 40

Mandy12's solution is right on... a clean and easy way to get there algebraically.

Another way to get there that avoids using algebra and that might be faster/easier depending on how your brain is wired up:

He drove for 21 hours, 3 more on the second day than on the first. So that means the breakdown is 9 hours on day 1 and 12 on day 2.

Given he averaged 15 extra miles per hour on day 2, that means he drove 12 x 15 = 180 "extra miles" on day 2. Subtracting 180 from 1020 gives 840. Had he not averaged 15 extra miles per hour on the second day, his two day total would have been 840 miles. This would also mean that he averaged the same mileage per hour on both days.

In order to calculate his average rate: D/T = R --> 840/21 = 40.

Rey

You just made things much easier for me. This is EXACTLY how I am wired and I fully understand now! I can do the other ways but this way is crystal clear and I can back it up and even teach others how to do it this way. The moral or takeaway that I am learning is that you have to account for the EXTRA part and try to make the two things equal. THen the computations are fairly straightforward. I had a similiar problem and it looked like this:

Train A left centerville , heading toward dale city station, at 3pm. Train b left dale heading toward centerville at 3:20 on the same day. The trains rode on straight tracks that were parallel to each other. If train a traveled at 30 mph and train b at 10 mph and the distance between stations were 90 miles, when did the trains pass eachother?

a. 4:45
b. 5:00
c. 5:20
d.5:35
e.6:00


qa
is c


and your easy solution was great. How can I learn more of ways to approach this in the rey.fernandez way.