document.write( "Question 1024241: Traveling upstream against the current, it took a motorboat 2 hours to make a 60-mile trip. The return trip traveling downstream with the current took 1 hour. Find the rate of the current.\r
\n" ); document.write( "\n" ); document.write( "Please show the work
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Algebra.Com's Answer #639681 by josmiceli(19441)\"\" \"About 
You can put this solution on YOUR website!
Let \"+c+\" = the rate of the current
\n" ); document.write( "Let \"+s+\" = the speed of the boat in still water
\n" ); document.write( "\"+s+%2B+c+\" = the speed of the boat going downstream
\n" ); document.write( "\"+s+-+c+\" = the speed of the boat going upstream
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\n" ); document.write( "Equation for going upstream:
\n" ); document.write( "(1) \"+60+=+%28+s+-+c+%29%2A2+\"
\n" ); document.write( "Equation for going downstream:
\n" ); document.write( "(2) \"+60+=+%28+s+%2B+c+%29%2A1+\"
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\n" ); document.write( "(1) \"+s+-+c+=+30+\"
\n" ); document.write( "and
\n" ); document.write( "(2) \"+s+%2B+c+=+60+\"
\n" ); document.write( "Add the equations:
\n" ); document.write( "\"+2s+=+90+\"
\n" ); document.write( "\"+s+=+45+\"
\n" ); document.write( "-------------
\n" ); document.write( "(2) \"+60+=+%28+s+%2B+c+%29%2A1+\"
\n" ); document.write( "(2) \"+60+=+%28+45+%2B+c+%29%2A1+\"
\n" ); document.write( "(2) \"+60+=+45+%2B+c+\"
\n" ); document.write( "(2) \"+c+=+15+\"
\n" ); document.write( "The rate of the current is 15 mi/hr
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\n" ); document.write( "check:
\n" ); document.write( "(1) \"+60+=+%28+s+-+c+%29%2A2+\"
\n" ); document.write( "(1) \"+60+=+%28+45+-+15+%29%2A2+\"
\n" ); document.write( "(1) \"+60+=+30%2A2+\"
\n" ); document.write( "(1) \"+60+=+60+\"
\n" ); document.write( "OK
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