Need equation help with this word problem...please!....thx!

Question

An airplane leaves Louisville for Cleveland at a speed of 475 mph. Twenty minutes later; a plane heading for Louisville leaves for Cleveland, which is 350 miles from Louisville at a speed of 500 mph. When they meet, how far are they from Cleveland?

 

Answer 1

distance=rate*time

350=475t+500(t-[1/3])

350=475t+500t-(500/3)

350+(500/3)=975t

350+(166 2/3)=975t

516 2/3=975t

1550/3=975t

(1550/3)(1/975)=t

1550/2925=t

62/117=t

475*(62/117)=29,450/117

29,450/117=251.7094 miles from Louisville when they meet; 350-251.7094=98.2906 miles from Cleveland

check: going the other way from Cleveland...

62/117-1/3=(186-117)/351=69/351=23/117

(23/117)*500=(23*500)/117=11,500/117=98.29059 miles from Cleveland

 

Answer 2

Distance between the planes after 10 min (or 1/3 hr) = 350 - 475/3 = 575/3 miles

Let t hours be the time (after the first 10 min) that the two planes meet.

Distance covered by the first plane = 475t

Distance covered by the second plane = 500t

The sum of the distances covered by each plane = 575/3 miles

475t + 500t = 575/3

hence t = 575/(975*3) * 60 = 11.8 minutes