Two jets leave an air base at the same time and travel in opposite directions. One jet travels 75 mi/h faster than the other. If the two jets are 4383 miles apart after 3 hours, what is the rate of each jet?Solve by distance,rate,time using system of linear equations.

Respuesta :

We need to use the next formula:

distance = rate * time

d = r * t

If one jet travels 75 mi/h faster than the other one:

Let r for the rate of the slower jet

Then, the rate of the faster jet is r+75mi/h

Where t = 3

d=4383

Then, for both jets:

4383 = (r*3)+ (r+75)*3

4383 = 3r + 3r+225

4383 = 6r + 225

Solve for r:

4383 - 225 = 6r

4158 = 6r

r = 4168/6

r = 693

Then, we can replace using the rate for both jets:

- Rate of faster jet = r + 75 = 693 + 75 = 768 mph

- Rate of the slower jet = r = 693 mph

We can confirm this if the sum of both distances is equal to 4383 miles.

d total = d1 +d2 = (r*3)+((r+75)*3) = (693*3)(768*3) = 2079 + 2304 = 4083 miles.

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