SOLUTION: Hello! I'm having a hard time figuring out how to set up the equations for this problem: A speed boat traveled two hours with a 4km/h current before turning around. The return tri

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Question 1015062: Hello! I'm having a hard time figuring out how to set up the equations for this problem:
A speed boat traveled two hours with a 4km/h current before turning around. The return trip against the same current took 3 hours. Find the speed of the boat in still water.
Thanks for your help!
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +d+ = distance of the one-way trip in km
Let +s+ = the speed of the boat in still water in km/hr
--------------------
Equation for traveling with the current:
(1) +d+=+%28+s+%2B+4+%29%2A2+
Equation for traveling against the current:
(2) +d+=+%28+s+-+4+%29%2A3+
---------------------
(1) +d+=+2s+%2B+8+
(1) +d+-+2s+=+8+
and
(2) +d+=+3s+-+12+
(2) +-d+%2B+3s+=+12+
--------------------
Add (1) and (2)
(1) +d+-+2s+=+8+
(2) +-d+%2B+3s+=+12+
-------------------
+s+=+20+
and
(1) +d+-+2s+=+8+
(1) +d+-+2%2A20+=+8+
(1) +d+=+48+
The speed of the boat in still water is 20 km/hr
-----------------
check:
(1) +d+=+%28+s+%2B+4+%29%2A2+
(1) +48+=+%2820+%2B+4+%29%2A2+
(1) +48+=+24%2A2+
(1) +48+=+48+
OK
(2) +d+=+%28+s+-+4+%29%2A3+
(2) +48+=+%28+20+-+4+%29%2A3+
(2) +48+=+16%2A3+
(2) +48+=+48+
OK