Use systems of lineear equations to solve:
An airplane flew for 6 hours with a 22-km/h tail wind. The return flight against the same wind took 8 hours. Find the speed of the plane in still air.
I know the answer is 154 km/hr (back of book resources :) )...I just don't understand how you would get to that answer.
Use the equation d = rt, once for each trip. I will use capitals for the variables and lowercase for subscripts.
D = R * T
Dr = Rr * Tr
where "r" is a subscript used to indicate that values for the return tripe.
We are given the time for each trip: 6 hours and then 8 hours for the return trip.
D = R * 6
Dr = Rr * 8
And the rate for both trips is based on x -- the speed in still air. It is x + 22 in one direction and x - 22 in the other.
D = (x + 22) * 6
Dr = (x - 22) * 8
Finally, we know that the distance is the same for both trips.
D = Dr
Therefore, through transitibity, we know that the right sides of both of equations are equal.
(x + 22) * 6 = (x - 22) * 8
or
6(x + 22) = 8(x - 22)
Solve that and you will get 154.
To solve this problem using systems of linear equations, let's make some assumptions:
Let's assume that the speed of the plane in still air is "x" km/h, and the speed of the wind is "w" km/h.
When the airplane is flying with the tailwind, the total speed is the sum of the plane's speed and the wind's speed. So the equation for the tailwind situation is:
Speed with tailwind = (Speed of plane in still air) + (Speed of wind)
(x + w) km/h
Given that the airplane flew for 6 hours with a 22 km/h tailwind, the distance covered can be calculated as:
Distance = Time × Speed
Distance = 6 × (x + w) km
On the return flight against the wind, the total speed is the difference between the plane's speed and the wind's speed. So the equation for the return flight is:
Speed against wind = (Speed of plane in still air) - (Speed of wind)
(x - w) km/h
Given that the return flight took 8 hours, the distance covered can be calculated as:
Distance = Time × Speed
Distance = 8 × (x - w) km
Since the distance traveled in both directions is the same, we can set the two distance equations equal to each other:
6(x + w) = 8(x - w)
Now, we can solve this equation for "x," which represents the speed of the plane in still air.
Simplifying the equation:
6x + 6w = 8x - 8w
Rearranging the terms:
6w + 8w = 8x - 6x
14w = 2x
Dividing both sides by 2:
7w = x
So, the speed of the plane in still air, denoted by "x," equals 7 times the speed of the wind, denoted by "w."
Now, we can substitute the value of "x" back into one of the original equations. Let's use the first equation:
x = 7w
Substituting this back into the equation:
7w = x
7w = 7w
This equation is true for any value of "w." Therefore, there are infinitely many solutions.
However, if we assume that the speed of the wind is positive (as it is given as a tailwind), we can solve for "x" by assuming a value for "w." In this case, let's assume that "w" (the speed of the wind) is 22 km/h.
Substituting this value into the equation x = 7w:
x = 7(22)
x = 154
So, when the speed of the wind is 22 km/h, the speed of the plane in still air is 154 km/h.