You are going to San Francisco to visit your Aunt Carly. On the way there, the plane is flying into a headwind; your flight from Boston, Massachusetts to San Francisco, California takes 6 hours and 15 minutes. However, on the return flight, this same 3000 mile distance between the two cities is covered in 5 hours and 30 minutes. Find the airspeed of the plane and the speed of the wind, assuming that both remain constant.

Speed of plane = mph

Speed of wind = mph

Round all answers to one decimal place.

The two unknown quantities are the speeds of the wind and the plane.

If s_{1}=speed
of the plane and s_{2}=speed of the wind, then

s_{1}-s_{2}=
speed of the plane against the wind

s_{1}+s_{2}=speed
of the plane with the wind

Using the formula distance=(rate)(time) for these two speeds, you get the following equations

3000=(s_{1}-s_{2})(6+15/60)
, 3000=6.25s_{1}-6.25s_{2},12000=25s_{1}-25s_{2}

3000=(s_{1}+s_{2})(5+30/60),
3000=5.5s_{1}+5.5s_{2} , 6000=11s_{1}+11s_{2}

Using elimination to solve this system, multiply the first equation by 11 and the second equation by 25.

132000=275s_{1}-275s_{2}

+(150000=275s_{1}+275s_{2})

282000=550s_{1}+0

S_{1}=5640/11512.7 mph

By back substituting s_{1}=5640/11
mph into either equation,

6000=11(5640/11)+11s_{2}

360/11=s_{2}32.7 mph

S_{1}512.7 mph ; s_{2}32.7 mph