math
posted by Robyn .
A ship left at 9:00 am going 10 miles per hour on a course. Another ship left at noon on the same course going 15 miles per hour. At what time did the second ship overtake the first ship?

Well write it out, let 1 represent 10:00 in the x coordinant's position, and let y be the miles out on the course the ship is.
Ship 1
10:00 11:00 12:00 1:00 2:00
(1,10);(2,20);(3,30);(4,40);(5,50);
3:00 4:00 5:00 6:00 7:00
(6,60);(7,70);(8,80);(9,90);(10,100)
Ship 2
1:00 2:00 3:00 4:00 5:00
(4,15);(5,30);(6,45);(7,60);(8,75);
6:00 7:00
(9,90);(10,105)
There is probably an easier way to find it out, though. 
Let t hours be the time since noon, when the second ship passes the first.
distance covered by slower ship = 10(t+3) mile
distance covered by faster ship = 15 t
but at that moment they both had gone the same distance, so ....
15t = 10(t+3)
15t = 10t + 30
5t=30
t = 6
so the second ship passed the first 6 hours past noon or 6:00 pm
check:
slower ship sailed for 9 hours at 10 mph = 90 miles
faster ship sailed for 6 hours at 15 mph = 90 miles
Answer is correct.