Homework Help: algebra

Posted by becca on Tuesday, January 17, 2012 at 9:40pm.

a two part canoe trip,the first part of the trip the canoist traveled 58 miles.Second part of the trip he traveled 18 miles at 5mph slower.the total time for trip was 4 hours.What was the speed of both parts of the trip?

algebra - Reiny, Tuesday, January 17, 2012 at 9:53pm
faster speed --- x mph
slower speed ---- x-5 mph

time at slower speed = 58/x
time at faster speed = 18/(x-5)
58/x + 18/(x-5) = 4
times x(x-5)

58(x-5) + 18x = 4x(x-5)
58x - 290 + 18x = 4x^2 - 20x
4x^2 -96x + 290 = 0
2x^2 - 48x + 145 = 0

x = (48 ą √1144)/4
= appr. 20.45 mph or 3.544 mph

if x = 20.45 , then x-5 = 15.45 mph
if x = 3.544 mph , then x-5 is a negative, so NO GOOD

Ridiculous answer!
I have done my share of canoeing, there is no way you can paddle a canoe at 20 mph!!!!
Absurd to canoe 76 miles in 4 hours.
algebra - becca, Tuesday, January 17, 2012 at 10:00pm
Thanks

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