SOLUTION: An airplane flying against the wind travels 430 miles in the same amount of time it would take the same plane to travel 470 miles with the wind. If the wind speed is a constant 20

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: An airplane flying against the wind travels 430 miles in the same amount of time it would take the same plane to travel 470 miles with the wind. If the wind speed is a constant 20       Log On


   

Question 1035988: An airplane flying against the wind travels 430 miles in the same amount of time it would take the same plane to travel 470 miles with the wind. If the wind speed is a constant 20 miles per hour, how fast would the plane travel in still air?
In still air, the plane would travel
miles per hour.
Just need to know how to compute this problem
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +s+ mi/hr = the plane's speed in still air
+20+ mi/hr = constant wind speed
Let +t+ = time in hrs traveling both
with the wind and against the wind
--------------------------------------
Equation for traveling against the wind:
(1) +430+=+%28+s+-+20+%29%2At+
Equation for traveling with the wind:
(2) +470+=+%28+s+%2B+20+%29%2At+
---------------------------
(1) +t+=+430+%2F+%28+s+-+20+%29+
Plug this into (2)
(2) +470+=+%28+s+%2B+20+%29%2A%28+430+%2F+%28+s+-+20+%29+%29+
(2) +470%2A%28+s+-+20+%29+=+430%2A%28+s+%2B+20+%29+
(2) +470s+-+9400+=+430s+%2B+8600+
(2) +40s+=+18000+
(2) +s+=+450+
---------------
The speed of the plane in still air is 450 mi/hr
---------------
check:
(1) +430+=+%28+s+-+20+%29%2At+
(1) +430+=+%28+450+-+20+%29%2At+
(1) +430+=+430t+
(1) +t+=+1+ hr
and
(2) +470+=+%28+s+%2B+20+%29%2At+
(2) +470+=+%28+450+%2B+20+%29%2At+
(2) +470+=+470t+
(2) +t+=+1+ hr
OK
Sorry I did the whole thing -can't resist