SOLUTION: You commute to work a distance of 80 miles and return on the same route at the end of the day. Your average rate on the return trip is 10 miles per hour faster than your average r

Question 148238: You commute to work a distance of 80 miles and return on the same route at the end of the day. Your average rate on the return trip is 10 miles per hour faster than your average rate on the outgoing trip. If the round trip takes 3 hours, what is your average rate on the outgoing trip to work?
Let x be the rate for the outgoing trip.
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
You commute to work a distance of 80 miles and return on the same route at the end of the day. Your average rate on the return trip is 10 miles per hour faster than your average rate on the outgoing trip. If the round trip takes 3 hours, what is your average rate on the outgoing trip to work?
:
Let x = the rate for the outgoing trip.
then
(x+10) = rate for the return trip
:
Write a time equation: Time =
:
Out time + return time = 3 hrs
+ = 3
Multiply equation by x(x+10) the common denominator
80(x+10) + 80x = 3x(x+10)
:
80x + 800 + 80x = 3x^2 + 30x
:
160x + 800 = 3x^2 + 30x
:
arrange as quadratic equation:
3x^2 + 30x - 160x - 800 = 0
:
3x^2 - 130x - 800 = 0
:
We need to solve this using the quadratic formula:

In this equation a=3; b=-130; c= -800

:
We're only interested in the positive solution here

x =
x = 48.8 mph is the outgoing rate.
then
48.8 + 10 = 58.8 is the return rate
:
:
Check solution by finding the total time
80/48.8 + 80/58.8 =
1.64 + 1.36 = 3.00 hrs

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