SOLUTION: Two ships leave port, one sailing east and the other south. Some time later they are 17 mi apart, with the eastbound ship 7 mi farther from port than the southbound ship. How far i

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Question 165413This question is from textbook
: Two ships leave port, one sailing east and the other south. Some time later they are 17 mi apart, with the eastbound ship 7 mi farther from port than the southbound ship. How far is each from port? This question is from textbook

Found 2 solutions by checkley77, ankor@dixie-net.com:
Answer by checkley77(12844) (Show Source):
You can put this solution on YOUR website!
22x=32(2.25-x)
22x=72-32x
22x+32x=72
54x=72
x=72/54
x=1.333 is the travel time for the trip to work @ 22 mph.
2.25-1.333=.917 hours for the return trip @ 32 mph.
Proof:
22*1.333=32*.917
29.33~29.34

Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website!
Two ships leave port, one sailing east and the other south. Some time later they are 17 mi apart, with the eastbound ship 7 mi farther from port than the southbound ship. How far is each from port?
:
This is a right triangle problem
:
Let x = distance from port of the southbound ship (one side of a right triangle)
then
(x+7) = distance from port of the eastbound ship (2nd side of a right triangle)
:
Distance between them = 17 mi (hypotenuse of the triangle)
:
x^2 + (x+7)^2 = 17^2
:
x^2 + x^2 + 14x + 49 = 289
:
2x^2 + 14x + 49 - 289 = 0
:
2x^2 + 14x - 240 = 0
Simplify, divide equation by 2
x^2 + 7x - 120 = 0
Factor
(x-8)(x-15) = 0
Positive solution
x = 8 distance from port of the southbound ship
and
8 + 7 = 15 mi distance of the eastbound ship
:
:
Check on calc enter: sqrt(8^2+15^2) = 17