Speed and distance
posted by Marysvoice on .
I can't figure this one out
During a trip, a canoe travles 47 miles at r speed. The second part of the trip the canoe travels 9 miles, but 5 mph slower. The total trip is 5 hrs. What is the speed of each part of the trip?
I start with 47/r+9/r5=5hrs
Then I multiply both sides by the LCD?
r(r5)(47/r+9/r5)=5
47(r5)+9r=5r(r5)
47r???+9r=3r^2? I am lost here
Maybe
47x5r=235?
47r235+9r=5r4r?
Quadratic
5r^260r+235=0???

You have it figured out.
47(r5)+9r=5r(r5)
47r235+9r=5r^225r
Rearranging and combining you had an error.
I got 5r^281r+235=0
Solve using the quadratic formula. One of the solutions of the quadratic will be less than 5 which would imply that the canoe would change directions for the (r5) rate. 
hi quidditch!!! remember me??

start with 47/r+9/r5=5hrs OK to here. I will rewrite to see it more clearly.
[47/r] + [9/(r5)] = 5
multiply both sides by r(r5) to get
47(r5) + 9r = 5(r)(r5)
47r  235 + 9r = 5r^2  25r
47r  235 + 9r 5r^2 + 25r = 0
5r^2 + 81r 235 = 0
I changed the sign here but it isn't necessary.
5r^2 81r + 235 = 0
and solve by the quadratic formula. I et r = 12.4 and if I remember the problem that is mph. Then r5 = 7.4 mph.
I hope this helps.
Then I multiply both sides by the LCD?
r(r5)(47/r+9/r5)=5
47(r5)+9r=5r(r5)
47r???+9r=3r^2? I am lost here
Maybe
47x5r=235?
47r235+9r=5r4r?
Quadratic
5r^260r+235=0??? 
I GOT IT!!!!! THANK YOU!!! THANK YOU!!! THANK YOU!!!