document.write( "Question 291097: 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? \n" ); document.write( "
Algebra.Com's Answer #210589 by Grinnell(63)\"\" \"About 
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As in all Speed (Rate) Distance problems we start out with the formula rate Times time =distance--
\n" ); document.write( "Rate x Time = Distance
\n" ); document.write( "the speed in still water is 15mph
\n" ); document.write( "BUT we are given a current which will add to the speed in still water as the boat goes DOWNSTREAM.
\n" ); document.write( "Let current be x (keeping our variable simple, ok?)
\n" ); document.write( "The current takes away from the speed in still water by the same variable x.
\n" ); document.write( "So, (15+x) TIMES time =12
\n" ); document.write( "(15-x) TIMES time =9
\n" ); document.write( "Since we are given that the time is the same--T
\n" ); document.write( "we can divide by 15+x (in the first equation) and 15-x (in the second equation) and get 12/15+x =9/15-x
\n" ); document.write( "STOP. Understand what I just did. Both sides of the above equations equal T, I substituted.
\n" ); document.write( "Now we solve for x!
\n" ); document.write( "We multiply thru by (15+x) and (15-x)
\n" ); document.write( "We get 180-12x=135+9x
\n" ); document.write( "-21x=-45
\n" ); document.write( "x=2 and 1/7mph. Let's check it.
\n" ); document.write( "2 and 1/7=roughly 2.14
\n" ); document.write( "Let's put it in -- 17.14---our time should come out to be equal, or the SAME time!
\n" ); document.write( "12/17.14=.700
\n" ); document.write( "9/12.86=.699 This is close enough. The speed is then 2.14 miles per hour. Remember this is 45/21. I rounded off. \n" ); document.write( "
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