document.write( "Question 679139: Please help -
\n" ); document.write( "Alicia can row 7 miles downstream in the same time it takes her to row 4 miles upstream. She rows downstream 5 miles/hour faster than she rows upstream. Find Alicia’s rowing rate each way.
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Algebra.Com's Answer #421874 by ptaylor(2198)\"\" \"About 
You can put this solution on YOUR website!
Distance(d) equals Rate(r) times Time(t) or d=rt; r=d/t and t=d/r\r
\n" ); document.write( "\n" ); document.write( "Let r=Alicia's rate upstream
\n" ); document.write( "Then r+5=her rate downstream
\n" ); document.write( "Time to row 4 miles upstream=4/r
\n" ); document.write( "Time to row 7 miles downstream=7/(r+5)
\n" ); document.write( "And we are told that these times are the same, sooooo
\n" ); document.write( "4/r=7/(r+5) multiply each term by r(r+5) or simply cross multiply
\n" ); document.write( "4(r+5)=7r
\n" ); document.write( "4r+20=7r
\n" ); document.write( "3r=20
\n" ); document.write( "r=6 2/3 hr-----------rowing rate upstream
\n" ); document.write( "r+5=6 2/3 +5=11 2/3 hr---rowing rate downstream
\n" ); document.write( "CK
\n" ); document.write( "t=d/r
\n" ); document.write( "time upstream=4/(6 2/3)=4/(20/3)---complex fraction; make it a simple fraction by multiplying the numerator and denominator by (3/20) and we get (4*(3/20))/(20/3*3/20)=12/20=3/5 hr
\n" ); document.write( "time downstream=7/(11 2/3)=7/(35/3);(7*(3/35))/(35/3*3/35)=21/35=3/5 hr
\n" ); document.write( "OK\r
\n" ); document.write( "\n" ); document.write( "Hope this helps---ptaylor
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