document.write( "Question 201087: Upstream, downstream. Junior’s boat will go 15 miles per
\n" ); document.write( "hour in still water. If he can go 12 miles downstream in the
\n" ); document.write( "same amount of time as it takes to go 9 miles upstream,
\n" ); document.write( "then what is the speed of the current? \n" ); document.write( "

Algebra.Com's Answer #151357 by josmiceli(19441) $\"\"$ $\"About$

You can put this solution on YOUR website!
Let $\"c\"$ = speed of current
\n" ); document.write( "Let $\"b\"$ = speed of boat in still water
\n" ); document.write( "Let $\"t\"$ = time for both trips in hrs
\n" ); document.write( "given:
\n" ); document.write( " $\"b+=+15\"$ mi/hr
\n" ); document.write( " $\"12+=+%2815+%2B+c%29%2At\"$
\n" ); document.write( " $\"9+=+%2815+-+c%29%2At\"$
\n" ); document.write( "-------------------
\n" ); document.write( " $\"12+=+15t+%2B+ct\"$
\n" ); document.write( " $\"9+=+15t+-+ct\"$
\n" ); document.write( "Add the equations
\n" ); document.write( " $\"21+=+30t\"$
\n" ); document.write( " $\"t+=+7%2F10\"$ hrs
\n" ); document.write( "And, since
\n" ); document.write( " $\"12+=+%2815+%2B+c%29%2At\"$
\n" ); document.write( " $\"12+=+%2815+%2B+c%29%2A%287%2F10%29\"$
\n" ); document.write( " $\"120+=+105+%2B+7c\"$
\n" ); document.write( " $\"7c+=+15\"$
\n" ); document.write( " $\"c+=+2.143\"$ mi/hr answer
\n" ); document.write( "check:
\n" ); document.write( " $\"12+=+15t+%2B+ct\"$
\n" ); document.write( " $\"12+=+15%2A.7+%2B+2.143%2A.7\"$
\n" ); document.write( " $\"12+=+10.5+%2B+1.5\"$
\n" ); document.write( " $\"12+=+12\"$
\n" ); document.write( "and
\n" ); document.write( " $\"9+=+%2815+-+c%29%2At\"$
\n" ); document.write( " $\"9+=+%2815+-+2.143%29%2A.7\"$
\n" ); document.write( " $\"9+=+12.857%2A.7\"$
\n" ); document.write( " $\"9+=+8.9999\"$ close enough\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "

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