Two planes leave an airport at the same time, one flying east, the other flying west.The eastbound plane travels 100 mph slower. They are 3040mi apart after 4hrs. find the speed of each plane

Answer 1

let speed of slower plane be x mph
then speed of faster plane is x+100 mph

solve for x

4x + 4(x+100) = 3040

Answer 2

I got x=330 what do i do from there. How do i get the speed of each plane from there?

Answer 3

I defined x as the speed of the slower speed
so the slower plane has speed of 330 mph
I defined x+100 as the speed of the faster plane, so
faster plane goes (330+100) or 430 mph

check:
what is 4(330) + 4(430) ?

Answer 4

3040...okay thanks

Answer 5

Let the speed of the westbound plane be...
x mph, then the speed of the other plane would be (x-100)mph. since the planes are going away in oposite direction so we add the distance they travelled in order to get the distance far apart,
4x +4(x-100)=3040
x is 330mph therefore the speed of the westbound plane is (330-100) which is 230 mph