document.write( "Question 692018: please help me with this problem. a motorboat with the current went 54mi in 3 hours. against the current, it took 3.6 hours to travel the same distance. find the rate of the boat in calm water and the rate of the current. \n" ); document.write( "
Algebra.Com's Answer #426921 by KMST(5328)\"\" \"About 
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\"b\" = rate of the boat in calm water (in miles per hour)
\n" ); document.write( "\"c\" = rate of the current (in miles per hour)
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\n" ); document.write( "With the current, the boat moves at \"b%2Bc\" miles per hour.
\n" ); document.write( "In \"3\" hours it covers a distance (in miles) of
\n" ); document.write( "\"3%28b%2Bc%29=54\"
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\n" ); document.write( "Against the current, the boat moves at \"b-c\" miles per hour.
\n" ); document.write( "In \"3.6\" hours it covers a distance (in miles) of
\n" ); document.write( "\"3.6%28b-c%29=54\"
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\n" ); document.write( "Those two equations form a system of linear equations that needs to be solved for \"b\" and \"c\",
\n" ); document.write( "but they can be simplified a bit.
\n" ); document.write( "\"3%28b%2Bc%29=54\" --> \"3%28b%2Bc%29%2F3=54%2F3\" --> \"b%2Bc=18\"
\n" ); document.write( "\"3.6%28b-c%29=54\" --> \"3.6%28b-c%29%2F3.6=54%2F3.6\" --> \"b-c=15\"
\n" ); document.write( "Now we can write the system as
\n" ); document.write( "\"system%28b%2Bc=18%2Cb-c=15%29\"
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\n" ); document.write( "You can easily solve the system by substitution, or by elimination.
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\n" ); document.write( "BY ELIMINATION:
\n" ); document.write( "Adding both equations, we get
\n" ); document.write( "\"2b=18%2B15\" --> \"2b=33\" --> \"2b%2F2=33%2F2\" --> \"highlight%28b=16.5%29\"
\n" ); document.write( "Substituting that into \"b%2Bc=18%7D%7D%2C+we+get%0D%0A%7B%7B%7B16.5%2Bc=18\" --> \"16.5%2Bc-16.5=18-16.5\" --> \"highlight%28c=1.5%29\"
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\n" ); document.write( "BY SUBSTITUTION:
\n" ); document.write( "\"b-c=15\" --> \"b-c%2Bc=15%2Bc\" --> \"b=15%2Bc\"
\n" ); document.write( "Substituting \"15%2Bc\" for \"b\" in \"b%2Bc=18\", we get
\n" ); document.write( "\"15%2Bc%2Bc=18\" --> \"15%2B2c=18\" --> \"15%2B2c-18=18-15\" --> \"2c=3\" --> \"2c%2F2=3%2F2\" --> \"highlight%28c=1.5%29\"
\n" ); document.write( "Substituting that into \"b=15%2Bc\", we get
\n" ); document.write( "\"b=15%2B1.5\" --> \"highlight%28b=16.5%29\"
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