document.write( "Question 142297: A boat travels 20 miles upstream in the same time that it would take to travel 30 miles downstream. If the rate of the water is 5 miles per hour, find the speed of the boat alone. \n" ); document.write( "
Algebra.Com's Answer #103613 by ptaylor(2198)\"\" \"About 
You can put this solution on YOUR website!

\n" ); document.write( "Distance(d) equals Rate(r) times Time(t) or d=rt; r=d/t and t=d/r\r
\n" ); document.write( "\n" ); document.write( "Let r= rate of the boat in still water
\n" ); document.write( "r-5=rate of the boat upsteam
\n" ); document.write( "r+5=rate of the boat downstream\r
\n" ); document.write( "\n" ); document.write( "Time to travel upstream=20/(r-5)
\n" ); document.write( "Time to travel downstream=30/(r+5)\r
\n" ); document.write( "\n" ); document.write( "Now we are told that the above two times are equal, so:\r
\n" ); document.write( "\n" ); document.write( "20/(r-5)=30/(r+5) multiply each term by (r-5)(r+5) or cross-multiply and we get:\r
\n" ); document.write( "\n" ); document.write( "20(r+5)=30(r-5) get rid of parens\r
\n" ); document.write( "\n" ); document.write( "20r+100=30r-150 subtract 30r and also 100 from each side
\n" ); document.write( "20r-30r+100-100=30r-30r-150-100 collect like terms\r
\n" ); document.write( "\n" ); document.write( "-10r=-250 divide each side by -10\r
\n" ); document.write( "\n" ); document.write( "r=25 mph-----rate of boat in still water\r
\n" ); document.write( "\n" ); document.write( "CK\r
\n" ); document.write( "\n" ); document.write( "20/(25-5)=30/(25+5)
\n" ); document.write( "20/20=30/30
\n" ); document.write( "1=1\r
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\n" ); document.write( "\n" ); document.write( "Hope this helps---ptaylor
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