Question 66678
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?

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

Hope this helps------ptaylor