document.write( "Question 304186: Juniors boat will go 15 miles per hour in still water. If he can go 12 miles downstream in the same amount of time as it takes to go 9 miles upstream, then what is the speed of the current. \r
\n" ); document.write( "\n" ); document.write( "Word problems are definitely my weakness, they just don't make sense in my head. I have a hard time figuring out which formula I need and what I need to do.
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Algebra.Com's Answer #217898 by ankor@dixie-net.com(22740)\"\" \"About 
You can put this solution on YOUR website!
Juniors boat will go 15 miles per hour in still water.
\n" ); document.write( " If he can go 12 miles downstream in the same amount of time as it takes to go 9 miles upstream, then what is the speed of the current.
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\n" ); document.write( "See if this makes sense to you.
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\n" ); document.write( "Let c = speed of the current
\n" ); document.write( "then
\n" ); document.write( "(15+c) = speed down stream
\n" ); document.write( "and
\n" ); document.write( "(15-c) = speed upstream
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\n" ); document.write( "Write a time equation: Time = dist/speed
\n" ); document.write( "Time up = time down
\n" ); document.write( "\"9%2F%2815-c%29\" = \"12%2F%2815%2Bc%29\"
\n" ); document.write( "cross multiply
\n" ); document.write( "9(15+c) = 12(15-c)
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\n" ); document.write( "135 + 9c = 180 - 12c
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\n" ); document.write( "9c + 12c = 180 - 135
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\n" ); document.write( "21c = 45
\n" ); document.write( "c = \"45%2F21\"
\n" ); document.write( "c = 2.143 mph is the current
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\n" ); document.write( "Check solution, by finding the times (they should be equal)
\n" ); document.write( "9/12.857 = .7 hrs (subtracted the speed of the current from 15)
\n" ); document.write( "12/17.143= .7 hrs (added the speed of the current to 15)
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\n" ); document.write( "\n" ); document.write( "How about this, did I explain this so you can see it?
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