Two planes left simultaneously from the same airport and headed in the same direction towards another airport 3600 km away. The speed of one of the planes was 200 km/hour slower than the speed of the other plane, and so it arrived at its destination 1.5 hours after the faster plane. Find the speeds of both planes.

Step-by-step explanation:

Equation

Let s represent the speed of the faster plane. Then its travel time is ...

  time = distance/speed = 3600/s

travel time for the slower plane is ...

  time = distance/speed = 3600/(s -200)

The difference in these times is 1.5 hours, so we have ...

  3600/(s -200) -3600/s = 1.5

Solution

Multiplying by (2/3)s(s -200), we get ...

  2400(s) -2400(s -200) = s(s -200)

  s^2 -200s -480000 = 0 . . . . . . . . . . rewrite in standard form

  (s -800)(s +600) = 0 . . . . . . . . . . . . . factor

The positive value of s that makes a factor zero is s = 800.

Conclusion

The speed of the faster plane was 800 km/h; of the slower plane, 600 km/h.

Check

The travel time for the faster plane was 3600/800 = 4.5 hours. For the slower plane, 3600/600 = 6 hours, a difference of 1.5 hours.

the correct answers i think are b. and d.

give me brainliest if it is correct!