SOLUTION: I need help with trying to figure out this problem: OCEAN TRAVEL At noon, a passenger ship and a freighter left a port traveling in opposite directions. By midnight, the passeng

Algebra ->  Customizable Word Problem Solvers  -> Travel -> SOLUTION: I need help with trying to figure out this problem: OCEAN TRAVEL At noon, a passenger ship and a freighter left a port traveling in opposite directions. By midnight, the passeng      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 157419This question is from textbook Pre-Algebra
: I need help with trying to figure out this problem:
OCEAN TRAVEL
At noon, a passenger ship and a freighter left a port traveling in opposite directions. By midnight, the passenger ship was 3 times farther from the port than the frieghter was. How far was the freightner from port if the distanc between the ships was 84 miles? This question is from textbook Pre-Algebra

Answer by gonzo(654) About Me  (Show Source):
You can put this solution on YOUR website!
let x be the distance of the freighter from the port.
then 3*x = distance of the passenger ship from the port because the passenger ship was 3 times farther from the port than the freighter was.
the distance between the passenger ship and the freighter is given as 84 miles.
since they went in opposite directions, this must be the distance between the freighter and the port and the distance between the passenger ship and the port.
this would then equal x + 3x = 4x.
4x=84
x=21 miles.
since x is the distance of the freighter from the port, then the distance of the freighter from the port is 21 miles and the distance of the passenger ship from the port is 63 miles.
this satisfied the distance between the ships at 84 miles.
it also satisfied the fact that the passenger ship is 3 times the distance to the port than the freighter is.