SOLUTION: Two cruise ships depart from the same harbor. Seabourn Odyssey departs at 8 AM, and travels due south 2 mph faster than the northbound Norwegian Bliss, that left the harbor 1 hour

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Question 1184535: Two cruise ships depart from the same harbor. Seabourn Odyssey departs at 8 AM, and travels due south 2 mph faster than the northbound Norwegian Bliss, that left the harbor 1 hour later than Seabourn Odyssey. At noon, the ships are 176 miles apart. How fast has each ship been travelling
Found 2 solutions by ikleyn, MathTherapy:
Answer by ikleyn(52912)   (Show Source):
You can put this solution on YOUR website!
.

The basic equation is 3*r + 4*(r+2) = 176, where x is the rate of the slower ship in miles per hour.

Having the equation, you do the rest.

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Answer by MathTherapy(10557)   (Show Source):
You can put this solution on YOUR website!

Two cruise ships depart from the same harbor. Seabourn Odyssey departs at 8 AM, and travels due south 2 mph faster than the northbound Norwegian Bliss, that left the harbor 1 hour later than Seabourn Odyssey. At noon, the ships are 176 miles apart. How fast has each ship been travelling

Let Norwegian Bliss' speed be S
Then Seabourn Odyssey's speed = S + 2
Distances covered by Norwegian Bliss and Seabourn Odyssey are: 3S and 4(S + 2), or 4S + 8, respectively
Their respective distances covered, sum to 176 miles, and so, we get the following DISTANCE equation: 3S + 4S + 8 = 176
3S + 4S = 176 - 8
7S = 168
Norwegian Bliss' speed, or
Seabourn Odyssey's speed: