Question 214844
I DON'T THINK YOU HAVE STATED THE PROBLEM CORRECTLY. FLYING WITH THE WIND, EDDIE SHOULD BE ABLE TO TRAVEL A LOT FARTHER IN THE SAME AMOUNT OF TIME THAN HE COULD FLYING AGAINST THE WIND UNLESS THIS PROBLEM IN IMPLYING THAT HE IS FLYING THE 270 MILE TRIP AT THE SPEED OF THE WIND. ANYWAY, HERE'S HOW TO SET UP THE PROBLEM:

Distance(d) equals Rate(r) times Time(t) or d=rt;r=d/t and t=d/r
Let s=speed of the wind
time travelled against the wind=570/(220-s)
time travelled with he wind=270/(220+s)
Now we are told that the above two times are equal, so:
570/(220-s)=270/(220+s)  multiply each side by (220-s)(220+s)
570(220+s)=270(220-s)  divide each side by 10 (just to reduce the size of the numbers)
57(220+s)=27(220-s) get rid of parens
12540+57s=5940-27s subtract 12540 from and add 27s to each side
12540-12540+57s+27s=5940-12540-27s+27s  collect like terms
84s=-6600
s=-78.6 mph----------------WE CAN'T HAVE NEGATIVE WIND SPEED

CHECK YOU PROBLEM

NOW IF THIS PROBLEM IS INDICATING THAT HE FLYS THE 270 MI TRIP AT THE SPEED OF THE WIND, THEN OUR PROBLEM TO SOLVE IS:
570/(220-s)=270/s  multiply each term by s(220-s)
570s=270(220-s)  divide each side by 1-
57s=27(220-s) or
57s=5940-27s  add 27s to each side
57s+27s=5940
84s=5940
s=70.7 mph-------------------wind speed

Hope this helps---ptaylor