SOLUTION: Making a round trip from Fairview to Casterville, a distance of 20 miles, a pilot faces a 30mph head wind one way and a 30mph tail wind on the return trip. The return trip takes 4

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Question 153165: Making a round trip from Fairview to Casterville, a distance of 20 miles, a pilot faces a 30mph head wind one way and a 30mph tail wind on the return trip. The return trip takes 45 minutes less than the outbound journey. Find the speed of the plane in still air
from Fair/Cast distance=10miles; rate = (r-30); time = t
from Cast/Fair distnace= 10miles; rate = (r+30); time = t -45
the total distance is 20 miles
so 10 = t(r-30)
10 = (r+30)(t-45)
the two variables are throwing me off, somethings wrong in my setup??
Thanks for your help.
Answer by orca(409) About Me  (Show Source):
You can put this solution on YOUR website!
There is only a small mistake in the second equation. It is the unit of time.
Note that the unit for the speed of the wind is miles/hour. So you need to convert 45 minutes into 3/4 hours
So the simultaneous equations become:
10 = t(r-30) ................(1)
10 = (r+30)(t-3/4) ...........(2)
From equation (1), we have
t+=+10%2F%28r-30%29
Substituting it into equation (2), we have
10+=+%28r%2B30%29%2810%2F%28r-30%29-3%2F4%29
Solving for the above equation, we obtain
10+=+10%28r%2B30%29%2F%28r-30%29-3%28r%2B30%29%2F4%29
10%2A4%28r-30%29+=+4%2A10%28r%2B30%29-3%28r%2B30%29%28r-30%29 Multiplying both sides by 4(r-30)
40r-1200=40r%2B1200+-3%28r%5E2-30%5E2%29
40r-1200=40r%2B1200+-3%28r%5E2-900%29
40r-1200=40r%2B1200+-3r%5E2%2B2700%29
3r%5E2=2700%2B1200%2B1200
3r%5E2=5100
r%5E2=1700
r+=+10sqrt%2817%29
r+=+41.23+