With a tail wind, a plane traveled 800 mi in 5 hours. With a head wind, the plane . Algebra 2 hlep?

Question

With a tail wind, a plane traveled 800 mi in 5 hours. With a head wind, the plane traveled the same distance in 8 hours. Which system of equations could be used to solve this problem? Please help. I don't understand this.

(s + w)5 = 800 and (s - w)8 = 800

(s - w)5 = 800 and (s + w )8 = 800

5s + 8w = 800 and 5s - 8w = 800

none of the above

Answer 1

Well, it seems that "s" represents the speed of the plane and "w" represents the speed of the wind.

With a tail wind, the plane traveled 800 miles in 5 hours. So it's speed with a tail wind is 800miles/5hours = 160 miles/hour.

With a tail wind, since the wind is blowing in the same direction as the plane's direction, the 160 miles/hour speed is the sum of the plane's speed and the wind speed.

With a head wind, the plane traveled 800 miles in 8 hours. So it's speed with a head wind is 800miles/8hours = 100 miles/hour.

With a head wind, since the wind is blowing in the opposite direction as the plane's direction, the 100 miles/hour speed is the difference between the plane's speed and the wind speed.

Can you figure it out now?

Answer 2

s= plane speed

w = wind

With the wind, plane makes it in 5 hours

Against wind, plane makes it in 8 hours

rate * time = distance

(s+w)5 = 800 and (s-w)8 = 800

The solution if you care...

s= 130

w= 30