SOLUTION: I'm having trouble setting up the equation for the following problem: A plane flies 450 miles with the wind in 3 hours. Flying back against the wind, the plane takes 5 hours to

Question 1014664: I'm having trouble setting up the equation for the following problem:
A plane flies 450 miles with the wind in 3 hours. Flying back against the wind, the plane takes 5 hours to make the trip What was the rate of the plane in still air? What was the rate of the wind?
Thanks so much for your help!
Found 2 solutions by josgarithmetic, macston:
Answer by josgarithmetic(39621) (Show Source): You can put this solution on YOUR website!
Study this very instructive example on video, and use it to understand your exercise. Maybe you will be able to solve yours on your own:

travel rates equal times different distances, involve wind

Assign variables to unknown quantities.
Make a data table.
Write system of equations.
Solve the system.
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Do you form and simplify a system which ultimately resembles ?
Answer by macston(5194) (Show Source): You can put this solution on YOUR website!
.
Rate going=450 mi/3 hours= 150 mi/hr
Rate returning=450 mi/5 hours= 90 mi/hr
.
R=rate in still air; W=wind speed
.
Rate going =R+W=150mph
Rate back = R-W=90 mph Subtract this from equation above.
2W=60 mph
W=30 mph
ANSWER 1: Wind speed was 30 mph
.
R+W=150 mph
R=150mph-W
R=150mph-30mph=120mph
ANSWER 2: The speed of the plane in still air is 120 mph.
.
CHECK:
R-W=90 mph
120mph-30mph=90mph
90mph=90mph
.

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