document.write( "Question 1066350: A boat traveled downstream a distance of 50 mi and then came right back. If the speed of the current was 20 mph and the total trip took 3 hours and 20 minutes, find the average speed of the boat relative to the water. \n" ); document.write( "
Algebra.Com's Answer #681547 by MathTherapy(10557)\"\" \"About 
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\n" ); document.write( "A boat traveled downstream a distance of 50 mi and then came right back. If the speed of the current was 20 mph and the total trip took 3 hours and 20 minutes, find the average speed of the boat relative to the water.
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Let speed in still water be S
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document.write( "Then average speed downstream = S + 20
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document.write( "Average speed upstream: S – 20
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document.write( "We then get the following TIME equation: \"50%2F%28S+%2B+20%29+%2B+50%2F%28S+-+20%29+=+3%2620%2F60\"
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document.write( "\"50%2F%28S+%2B+20%29+%2B+50%2F%28S+-+20%29+=+10%2F3\"
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document.write( "50(3)(S – 20) + 50(3)(S + 20) = 10(S + 20)(S – 20) ------- Multiplying by LCD, 3(S  + 20)(S – 20)
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document.write( "\"150S+-+%223%2C000%22+%2B+150S+%2B+%223%2C000%22+=+10%28S%5E2+-+400%29\"
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document.write( "\"300S+=+10%28S%5E2+-+400%29\"
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document.write( "\"10%2830S%29+=+10%28S%5E2+-+400%29\"
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document.write( "\"30S+=+S%5E2+-+400\"
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document.write( "\"S%5E2+-+30S+-+400+=+0\"
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document.write( "(S - 40)(S + 10) = 0
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document.write( "S, or speed of boat in still water = \"highlight_green%28matrix%281%2C2%2C+40%2C+mph%29%29\"   	OR   	  S = - 10 (ignore) \n" );
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