Arthur D. answered 04/10/14
Tutor
4.9
(306)
Forty Year Educator: Classroom, Summer School, Substitute, Tutor
distance=rate*time
d=rt
let w=speed of wind
330=(220-w)*t
550=(220+w)*t
use the distributive property
330=220t-wt
550=220t+wt
add the two equations
880=440t (-wt+wt=0)
880/440=t
2 hours=t
substitute t=2 into either of the original equations
330=(220-w)*2
165=220-w
w=220-165
w=55 mph
550=(220+w)*2
275=220+w
w=55 mph
another solution:
330=(220-w)*t
330/(220-w)=t
550=(220+w)*t
550/(220+w)=t
330/(220-w)=550/(220+w)
cross multiply
330(220+w)=550(220-w)
72,600+330w=121,000-550w
330w+550w=121,000-72,600
880w=48,400
w=48,400/880
w=55 mph