SOLUTION: You commute 56 miles one way to work. The trip to work takes 10 minutes longer than the trip home. Your average speed on the trip home is 8 miles per hour faster. What is your aver
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-> SOLUTION: You commute 56 miles one way to work. The trip to work takes 10 minutes longer than the trip home. Your average speed on the trip home is 8 miles per hour faster. What is your aver
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Question 199641This question is from textbook Algebra and Trigonometry
: You commute 56 miles one way to work. The trip to work takes 10 minutes longer than the trip home. Your average speed on the trip home is 8 miles per hour faster. What is your average speed on the trip home? I'm having a hard time figuring out where to place everything, I know it's DxRxT promblem. This question is from textbook Algebra and Trigonometry
You can put this solution on YOUR website! You commute 56 miles one way to work. The trip to work takes 10 minutes longer than the trip home. Your average speed on the trip home is 8 miles per hour faster. What is your average speed on the trip home?
:
Let s = average speed on the trip home
then
(s-8) = average speed to work (It's 8 mph slower than the return trip)
:
Change 10 min to hrs: 10/60 = hr
:
Write a time equation: Time =
Return time + 10 min = To time + =
:
Multiply equation by the common denominator; 6s(s-8)
6s(s-8)* + 6s(s-8)* = 6s(s-8)*
Cancel out the denominators and you have:
6(s-8)*56 + s(s-8) = 6s(56)
:
56(6s-48) + s^2 - 8s = 336s
:
336s - 2688 + s^2 - 8s - 336s = 0
:
Arrange as a quadratic equation; (336s's cancel)
s^2 - 8s - 2688 = 0
With a little sweat, we can factor this or you can use the quadratic formula:
(s-56)(s+48) = 0
Positive solution
s = 56 mph is the return trip av speed
;
:
Check the solution by finding the times (To speed = 48 mph)
56/48 = = 1 hr
56/56 = 1 hr
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difference 1/6 hr which is 10 min
