+1622 So... I encountered this hard speed-time-distance rate problem...
Richard needs to fly from San Diego to Halifax, Nova Scotia and back in order to give an important talk about mathematics. On the way to Halifax, he will get a speed boost from the wind which blows at 50 miles per hour (mph). On the way back, the plane must, unfortunately, fight this wind speed. If the talk lasts 2 hours, and if the distance between the two cities is 3000 miles, how fast must the plane fly in mph if the entire trip is to take 13 hours?
Not sure what to do with the equations... Please help thanks 
+23246 Let R = the rate of the plane.
With the wind, the rate will be R + 50.
Against the wind, the rate will be R - 50.
The total flight time will be 11 hours.
Let T = the time going (with the wind)
11 - T = the time returning (against the wind.
Since Distance = Rate x Time:
Going: (R + 50)(T) = 3000 ---> T = 3000 / (R + 50)
Returning: (R - 50)(11 - T) = 3000 ---> 11 - T = 3000 / (R - 50)
Combining: 11 - T = 3000 / (R - 50) ---> 11 - 3000 / (R + 50) = 3000 / (R - 50)
---> 11 = 3000 / (R + 50) + 3000 / (R - 50)
Multiply both sides the the common denominator [ (R + 50)(R - 50) ].
This will give you a quadratic; since it has a "nice" answer, either factor or use the quadratic formula.
+36916 11 hours flying time
3000/(s+50) + 3000/ (s-50) = 11 Now multipy both sides of the equation by (s+50)(s-50) to get:
6000s = 11s^2 - 27500
11s^2 -6000s -27500 = 0 Use Quadratic formula to find s = 550 m/hr
