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## Equation Word Problem

A twin-engined aircraft can fly 1216 miles from city A to city B in 4 hours with the wind and make the return trip in 8 hours against the wind. What is the speed of the wind?

1) Calculate the speed on the way from A to B;

2) Calculate the speed on the way back, from B to A;

3) Subtract the latter from the former, this will be twice the speed of the wind (why?).

Ronica:

Lets convert this word problem into algebraic terms:

Let D=the distance between cities, (note that the distance traveled is the same in both directions)

Let Sw= the speed of the wind; Note that this is what we're trying to solve for.

Let Sp= the speed of the plane

The basic formula to be applied is Distance=Speed*Time

Since the first trip is "with the wind", the total net speed is (Sp+Sw) and Distance=1216=(Sp+Sw)*4

Since the return trip is "against the wind", the total net speed is (Sp-Sw) and Distance=1216=(Sp-Sw)*8

We now have two simultaneous equations that can be solved for Sand Sw:

From the 1st equation, 1216=4Sp+4Sw  or Sp=(1216-4Sw)/4

Plugging this into the second equation for Sp, 1216=8(1216-4Sw)/4 - 8Sw

or 1216=2432-8Sw-8Sw

or -1216=-16Sw

or Sw=76 miles per hour; This is the answer.

You can now also solve for the speed of the plane to get Sp=228 miles per hour

You can Check your work, plugging Sw and Sp back into the 2 simultaneous equations:

1216=(Sp+Sw)*4

1216=(228+76)*4

1216=1216 (check!)

1216=(Sp-Sw)*8

1216=(228-76)*8

1216=1216 (check!)