SOLUTION: A farmer drives a tractor from one town to another, a distance of 180 kilometers. He drives 15 kilometers per hour faster on the return trip, cutting 1 hour off the time. How fast

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: A farmer drives a tractor from one town to another, a distance of 180 kilometers. He drives 15 kilometers per hour faster on the return trip, cutting 1 hour off the time. How fast       Log On


   

Question 401134: A farmer drives a tractor from one town to another, a distance of 180 kilometers. He drives 15 kilometers per hour faster on the return trip, cutting 1 hour off the time. How fast does he drive each way?
I got to the point where the problem reads as followed:
r^2-10r-1800
(r-60)(r ?) and that is where I am stuck
Answer by Tatiana_Stebko(1539) About Me  (Show Source):
You can put this solution on YOUR website!
Let r km/h is tractor's rate from one town to another. For a distance of 180 kilometers it takes time 180%2Fr hours
He drives 15 kilometers per hour faster on the return trip, so it rate is (r+15) km/h. For a distance of 180 kilometers it takes 180%2F%28r%2B15%29 hours
A farmer cuts one hour off the time on the return road, so
180%2Fr=180%2F%28r%2B15%29%2B1

180%28r%2B15%29=180r%2Br%28r%2B15%29
180r%2B2700=180r%2Br%5E2%2B15r
r%5E2%2B15r-2700=0
a quadratic equation , the roots are given by the quadratic formula r=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
r+=+%28-15+%2B-+sqrt%28+15%5E2-4%2A1%2A%28-2700+%29%29%29%2F%282%2A1%29+
r+=+%28-15+%2B+105%29%2F2+ or r+=+%28-15+-+105%29%2F2+
r+=+45 or r+=+-60+<0 extraneous root of the equation, does not satisfy the problem
r+=+45km/h is tractor's rate from one town to another, back trip rate is %28r%2B15%29=45%2B15=60km/h