Posted by **megan** on Wednesday, February 25, 2009 at 11:09am.

Solve the problem. A man rode a bicycle for 12 miles and then hiked an additional 8 miles. The total time for the trip was 5 hours. If his rate when he was riding a bicycle was 10 miles per hour faster than his rate walking, what was each rate?

- college algebra -
**Reiny**, Wednesday, February 25, 2009 at 12:44pm
let walking rate be x mph

let biking rate be x+10

time for walking = 12/x

time biking = 8/(x+10)

isn't 12/x + 8(x+10) = 5 ?

multiply each term by x(x+10)

you will get quadratic, solve for x

- college algebra -
**drwls**, Wednesday, February 25, 2009 at 12:49pm
Let the walking speed be v and the bike speed V = v + 10.

12/(v+10) + 8/v = 5 is the sum of the times spent bidking and walking.

Solve that equation for v. You will have to start with a common denominator.

[12v + 8(v+10)]/[v(v+10)] = 5

5 v^2 + 50 v = 20 v + 80

v^2 + 6v - 16 = 0

(v + 8)(v -2) = 0

Choose the positive root, v = 2 miles/hr

The bike speed was then v + 10 = 12 mph.

- college algebra -
**Reiny**, Wednesday, February 25, 2009 at 1:07pm
go with drwls solution, I mixed up the two rates.

- college algebra -
**Tim**, Sunday, July 31, 2011 at 11:19am
Solve. V^2-6v-16=0

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