Let x = plane speed and w = wind speed
3(x + w) = 1680
4(x - w) = 1680
x + w = 560
x - 4 = 420
You should be able to solve it from here!
Zuheily V.
asked 11/18/18Matt and Anna Killian are frequent fliers on Fast-n-Go Airlines. They often fly between two cities that are a distance of 1680 miles apart. On one particular trip, they flew into the wind and the trip took
4 hours. The return trip with the wind behind them, only took about
3hours.
Let x = plane speed and w = wind speed
3(x + w) = 1680
4(x - w) = 1680
x + w = 560
x - 4 = 420
You should be able to solve it from here!
David M. answered 11/19/18
Dave "The Math Whiz"
We have 2 unknowns, the plane speed and wind speed, therefore, we need 2 equations to solve this problem.
Let x=the planes speed in still air
y=winds speed
Eq. I
When the plane is traveling with the wind you get the planes speed + the wind speed for the total speed of the plane. This gives us 3 hrs X (x + y) = 1680
x + y = 1680/3
x + y = 560
Eq. II
The opposite is true when going into the wind, so this gives us
4hrs X (x - y) = 1680
x - y = 1680/4
x - y = 420
You can use either the substitution or the elimination method to solve for one of the variables and then use that variable to solve for the other one. Let's use the elimination method:
x + y = 560 Eq. I
x - y = 420 Eq. II
Adding the them together and we get
x + y + x - y = 560 + 420
2x = 980
x = 490
Putting this value into Eq. I and solving for y we get
x + y = 560
490 + y = 560
y = 560 - 490
y = 70
Checking the values for x & y in both equations we find that the answers are correct. Therefore, the speed of the plane in still air, x, is 490 mph and the wind speed, y, is 70 mph.
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