Álgebra. 2 problem

Question

Starting from the same place a plane took 1 hr longer to fly 705mi to Atlanta, than to fly 470 miles to New Orleans.

If the plane flew at the same rate for the entire trip, what was the flying time for the shorter leg?

Answer 1

For the shorter leg to New Orleans, let the distance flown be x_n (miles) and the time taken be t_n (hours). (The subscript 'n' is for New Orleans)

The distance flown can be written as:

x_n = v * t_n, where x_n = 470, and v is the speed (miles/hour) of the plane.

We want to solve for this flight time t_n, in hours.

For the longer leg to Atlanta, let the distance flown be x_a and the time taken be t_a. (The subscript 'a' is for Atlanta)

The flight speed v is the same for the 2 legs so the distance flown to Atlanta is:

x_a = v * t_a, where x_a = 705.

Because it takes 1 hour longer to fly to Atlanta, we know that t_a = t_n + 1, in hours.

We can rewrite the New Orleans speed as v = x_n/t_n

We can rewrite the Atlanta speed as v = x_a/t_a = x_a/(t_n + 1).

Because these 2 speeds are equal:

x_n/t_n = x_a/(t_n + 1)

Multiplying by the denominators:

x_n*(t_n + 1 ) = x_a*t_n

... the left hand side can be written as x_n*t_n + x_n, so:

(x_a - x_n) * t_n = x_n

Solving for t_n :

t_n = x_n / (x_a - x_n) = 470/(705-470) = 2 hours

So the time to fly the shorter leg (to New Orleans) is 2 hours.