Question 186854
Distance(d) equals Rate(r) times Time(t) or d=rt; r=d/t and t=d/r

Let r=rate (speed) or the plane in still air
Then r+50=rate of plan with the wind (return trip)
And r-50=rate of plane against the wind  (SF to NY)


time to fly to NY:
6=d/(r-50)-----------------eq1
time to return:
5=d/(r+50)-----------------eq2
from eq1, d=6(r-50).  Substitute this into eq2 for d
5=6(r-50)/(r+50)  multiply each side by (r+50)
5(r+50)=6(r-50)  get rid of parens
5r+250=6r-300 add 300 to and subtract 5r from both sides
250+300+5r-5r=6r-5r-300+300  collect like terms
r=550 mph---------------------speed of plane in still air

CK
distance to SF=(550-50)*6=3000 mi
return distance=(550+50)*5=3000 mi
3000=3000

Hope this helps---ptaylor