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Mathematics 9 Online
OpenStudy (anonymous):

Kala drove to the mountains last weekend. There was heavy traffic on the way there, and the trip took 7 hours. When Kala drove home, there was no traffic and the trip only took 5 hours. If her average rate was 18 miles per hour faster on the trip home, how far away does Kala live from the mountains?

OpenStudy (wwwhhhaaattt?):

@TheSmartOne

OpenStudy (anonymous):

Giving out Medals and Fans to whoever helps!!

OpenStudy (anonymous):

@satellite73

OpenStudy (anonymous):

again it is distance is rate times time the the rate going is say \(R\) then the rate returning is \(R+18\)

OpenStudy (anonymous):

so one expression for the distance is \[D=7R\] another is \[D=5(R+18)\]

OpenStudy (anonymous):

I'm not very good with the distance, rate and time equations

OpenStudy (anonymous):

since evidently the distances are the same, you can set them equal and solve \[7R=5(R+18)\]

OpenStudy (anonymous):

it is really one idea distance is rate times time \[D=RT\] or \[T=\frac{D}{R}\] or \[R=\frac{D}{T}\]

OpenStudy (anonymous):

i am sure you can solve \[7R=5(R+18)\] right?

OpenStudy (anonymous):

Yes. Thank you!

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