Two planes left simultaneously from the same airport and headed in the same direction towards another airport 1800 km away. The speed of one of the planes was 100 km/hour slower than the speed of the other plane, and so it arrived at its destination 36 minutes after the faster plane. Find the speeds of both planes.

2 answers

speed of faster plane --- x km/h
speed of slower plane ---- x - 100 km/h

time for faster plane = 1800/x
time for slower plane = 180/(x-100)

1800/(x-100) - 1800/x = 36/60 = 3/5
multiply each term by 5x(x-100) , the LCD

9000x - 9000(x-100) = 3x(x-100)
9000x - 9000x + 900000 = 3x^2 - 300x
3x^2 - 300x - 900000 = 0
x^2 - 100x - 300000 = 0
(x - 1000)(x +
x^2 - 100x = 300000
x^2 - 100x + 2500 = 300000+2500
(I completed the square)
(x - 50)^2 = 302500
x - 50 = ±√302500 = ± 550

x = 600 or a negative

the faster plane went 600 km/h
the slower plane went 500 km/h

check:
time of faster = 1800/600 = 3 hrs = 180 minutes
time of slower plane = 1800/500 = 3.6 hrs = 216 minutes

difference = 216-180= 36 minutes
My answer is correct
your answer is not correct