Two planes leave an airport at the same time, one flying east, the other flying west. the eastbound plane travels 20 mph slower. they are 1600 mi apart after 2 hr. find the speed of each plane.

Let the speed of faster plane be x miles/hour

so,
the speed of the other plane will be (x-20) Mph

The distance covered by both of them in 2hours:

Faster plane= 2 h × (x )Mph = 2x miles

Slower plane = 2(x-20) miles

Now according to question,

2x + 2(x-20) = 1600
2x + 2x -40 = 1600
4x = 1560
x = 440miles

Therefore the speed of faster plane = 400Mph

and the slower one will be (400-20)Mph = 380Mph

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