A plane took 1 hour longer to travel 560 miles on the first

A plane took 1 hour longer to travel 560 miles on the first portion of a flight than it took to fly 480 miles on the second portion. If the speed was the same for each portion, what was the flying time for the second part of the trip?

Solution

Let t=time it takes to fly the second portion of the trip (480 mi)
Then t+1=time it takes to fly the first portion of the trip (560 mi)
We know that distance (d) equals rate (r) times time (t) or
d=rt ; divide both sides by t and we get:
d/t=r or r=d/t
Now we are told that the speed (or rate) was the same for each portion of the trip
Rate (1st portion)=d/t (first portion) =560/(t+1)
Rate (2nd portion)=d/t (second portion)=480/t

Rate (1st portion)=Rate (2nd portion) So our equation to solve is:
560/(t+1)=480/t Multiply both sides by t(t+1)
(560(t)(t+1))/(t+1)=(480(t)(t+1))/t and this reduces to:
560t=480(t+1)
560t=480t+480 subtract 480t from both sides
560t-480t=480t-480t+480 collect like terms
80t=480
t=6 hours; time it takes to fly second portion of trip
t+1=7 hours; time it takes to fly first portion of trip
ck
d=rt and both rates are the same
d(2nd portion)=rate times Time (2nd portion)
480=6r divide both sides by 6
80=r
d(1st portion)=rate times Time (1st portion)
560=7r divide both sides by 7
80=r