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.

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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!

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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