document.write( "Question 203370: Debbie traveled by boat 5 miles upstream to fish in her favorite spot. Because of the 4-mph current, it took her 20 minuets longer to get there than to return. How fast will her boat go in still water? \n" ); document.write( "
Algebra.Com's Answer #153437 by ankor@dixie-net.com(22740)\"\" \"About 
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Debbie traveled by boat 5 miles upstream to fish in her favorite spot.
\n" ); document.write( " Because of the 4-mph current, it took her 20 minutes longer to get there than to return.
\n" ); document.write( "How fast will her boat go in still water?
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\n" ); document.write( "Let s = boat speed in still water
\n" ); document.write( "then
\n" ); document.write( "(s-4) = speed upstream
\n" ); document.write( "and
\n" ); document.write( "(s+4) = speed down stream
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\n" ); document.write( "Convert 20 min to hrs: 20/60 = \"1%2F3\"hr
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\n" ); document.write( "Write a time equation: time = \"distance%2Fspeed\"\r
\n" ); document.write( "\n" ); document.write( "Upstream time = downstream time + 20 minutes (1/3 hr)
\n" ); document.write( "\"5%2F%28%28s-4%29%29\" = \"5%2F%28%28s%2B4%29%29\" + \"1%2F3\"
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\n" ); document.write( "To clear out these denominators, multiply equation by 3(s+4)(s-4)
\n" ); document.write( "3(s+4)(s-4)*\"5%2F%28%28s-4%29%29\" = 3(s+4)(s-4)*\"5%2F%28%28s%2B4%29%29\" + \"1%2F3\"*3(s+4)(s-4)
\n" ); document.write( "Results in
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\n" ); document.write( "15(s+4) = 15(s-4) + (s+4)(s-4)
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\n" ); document.write( "15s + 60 = 15s - 60 + s^2 - 16
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\n" ); document.write( "0 = 15s - 15s - 60 - 60 + s^2 - 16
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\n" ); document.write( "0 = s^2 - 136
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\n" ); document.write( "s^2 = 136
\n" ); document.write( "s = \"sqrt%28136%29\"
\n" ); document.write( "s = 11.66 mph speed in still water
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\n" ); document.write( "Check solution on calc: (speed up = 7.66 and speed down = 15.66)
\n" ); document.write( "5/7.66 - 5/15.66 = .3333 which is 1/3 of an hr\r
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