document.write( "Question 132751: With Current, a motorboat travels 60 mi in 3 h. Against the current, it takes 2h longer to travel 60 mi. Find the rate of the current. \n" ); document.write( "
Algebra.Com's Answer #97013 by solver91311(24713)\"\" \"About 
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Rate of the boat in still water: \"r%5Bb%5D\"
\n" ); document.write( "Rate of the current: \"r%5Bc%5D\"
\n" ); document.write( "Then:
\n" ); document.write( "Rate of the boat with the current: \"r%5Bb%5D%2Br%5Bc%5D\"
\n" ); document.write( "Rate of the boat against the current: \"r%5Bb%5D-r%5Bc%5D\"\r
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\n" ); document.write( "\n" ); document.write( "But we know that \"r=d%2Ft\", so the rate of the boat with the current is \"60%2F3=20\", and the rate of the boat against the current is \"60%2F%283%2B2%29=60%2F5=12\", so:\r
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\n" ); document.write( "\n" ); document.write( "\"r%5Bb%5D%2Br%5Bc%5D=20\"
\n" ); document.write( "\"r%5Bb%5D-r%5Bc%5D=12\"\r
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\n" ); document.write( "\n" ); document.write( "Add the two equations term by term to eliminate \"r%5Bc%5D\"
\n" ); document.write( "\"2r%5Bb%5D=32\"
\n" ); document.write( "\"r%5Bb%5D=16\"\r
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\n" ); document.write( "\n" ); document.write( "Since \"r%5Bb%5D%2Br%5Bc%5D=20\", \"16%2Br%5Bc%5D=20\", and finally \"r%5Bc%5D=4\"\r
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\n" ); document.write( "\n" ); document.write( "Check:
\n" ); document.write( "16 + 4 = 20 mph times 3 hours = 60 miles
\n" ); document.write( "16 - 4 = 12 mph times (3 + 2) hours = 60 miles
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