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March 4, 2015

March 4, 2015

Posted by **Marysvoice** on Saturday, February 9, 2008 at 10:58am.

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/r-5=5hrs

Then I multiply both sides by the LCD?

r(r-5)(47/r+9/r-5)=5

47(r-5)+9r=5r(r-5)

47r-???+9r=3r^2? I am lost here

Maybe-

47x5r=235?

47r-235+9r=5r-4r?

Quadratic-

5r^2-60r+235=0???

- Speed and distance -
**Quidditch**, Saturday, February 9, 2008 at 12:08pmYou have it figured out.

47(r-5)+9r=5r(r-5)

47r-235+9r=5r^2-25r

Rearranging and combining you had an error.

I got 5r^2-81r+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 (r-5) rate.

- Speed and distance -
**domonic**, Saturday, February 9, 2008 at 6:12pmhi quidditch!!! remember me??

- Speed and distance -
- Speed and distance -
**drbob222**, Saturday, February 9, 2008 at 12:11pmstart with 47/r+9/r-5=5hrs

**OK to here. I will rewrite to see it more clearly.**

[47/r] + [9/(r-5)] = 5

multiply both sides by r(r-5) to get

47(r-5) + 9r = 5(r)(r-5)

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 r-5 = 7.4 mph.*I hope this helps.*

Then I multiply both sides by the LCD?

r(r-5)(47/r+9/r-5)=5

47(r-5)+9r=5r(r-5)

47r-???+9r=3r^2? I am lost here

Maybe-

47x5r=235?

47r-235+9r=5r-4r?

Quadratic-

5r^2-60r+235=0???

- Speed and distance -
**Marysvoice**, Saturday, February 9, 2008 at 1:18pmI GOT IT!!!!! THANK YOU!!! THANK YOU!!! THANK YOU!!!

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