document.write( "Question 11412: The Smith family took a canoe 10 miles down the Allegheny River in 1 hour and 15 minutes. After lunch it took them 4 hours to return. Find the rate of the current.\r
\n" ); document.write( "\n" ); document.write( "I have set up a table with:
\n" ); document.write( " Distance = 10 both upstream and downstream
\n" ); document.write( " Canoe speed downstream = x + y
\n" ); document.write( " Canoe speed upstream = x - y
\n" ); document.write( " Time going downstream = 1 1/4 hour
\n" ); document.write( " Time going upstream = 4 hours\r
\n" ); document.write( "\n" ); document.write( "I just keep getting bogged down with the fractions when trying to solve by addition method or substitution method.
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Algebra.Com's Answer #5940 by Earlsdon(6294)\"\" \"About 
You can put this solution on YOUR website!
Using the distance formuls:
\n" ); document.write( " \"d+=+rt\" where: d = distance (10 miles), r = speed (rate) and t = time.\r
\n" ); document.write( "\n" ); document.write( "Downstream trip:\r
\n" ); document.write( "\n" ); document.write( " \"10+=+r1%281.25%29\"
\n" ); document.write( "\"r1+=+8\"mph\r
\n" ); document.write( "\n" ); document.write( "Upstream trip:\r
\n" ); document.write( "\n" ); document.write( "\"10+=+r2%284%29\"
\n" ); document.write( "\"r2+=+2.5\"mph\r
\n" ); document.write( "\n" ); document.write( "Let x = speed of canoe and y = rate of the current.\r
\n" ); document.write( "\n" ); document.write( "Downstream:
\n" ); document.write( "\"x%2By+=+8\"\r
\n" ); document.write( "\n" ); document.write( "Upstream:
\n" ); document.write( "\"x-y+=+2.5\" Rewrite as: \"x+=+y+%2B+2.5\" and substitute into the above.\r
\n" ); document.write( "\n" ); document.write( "\"%28y%2B2.5%29+%2B+y+=+8\" Simplify and solve for y.\r
\n" ); document.write( "\n" ); document.write( "\"2y+%2B+2.5+=+8\" Subtract 2.5 from both sides.\r
\n" ); document.write( "\n" ); document.write( "\"2y+=+5.5\" Divide both sides by 2.\r
\n" ); document.write( "\n" ); document.write( "\"y+=+2.75\" mph is the rate of the current.
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