document.write( "Question 964015: Jan and Tariq took a canoeing trip, traveling 6 mi upstream against a 2-mi/h current. They then returned to the same point downstream. If their entire trip took 4 hr, how fast could they paddle in still water? \n" ); document.write( "
Algebra.Com's Answer #589036 by LinnW(1048)\"\" \"About 
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Use the formula, rate * time = distance
\n" ); document.write( "Set x = speed in still water
\n" ); document.write( "(x-2) * t1 = 6 miles
\n" ); document.write( "(x+2) * t2 = 6 miles
\n" ); document.write( "t1, the time paddling upstream = distance 6 miles/ rate (x-2) or
\n" ); document.write( "6/(x-2)
\n" ); document.write( "t2, the time paddling downstream = distance 6 miles/ rate (x+2) or
\n" ); document.write( "6/(x+2)
\n" ); document.write( "t1 + t2 = 4 hours, so
\n" ); document.write( "6/(x-2) + 6/(x+2) = 4
\n" ); document.write( "A common denominator is (x-2)(x+2) so we want
\n" ); document.write( "\"6%2F%28x-2%29%2A%28x%2B2%29%2F%28x%2B2%29+%2B+6%2F%28x%2B2%29%2A%28x-2%29%2F%28x-2%29+=+4\"
\n" ); document.write( "expanding
\n" ); document.write( "\"%286x+%2B+12+%2B+6x+-12%29%2F%28x%5E2-4%29+=+4%2F1\"
\n" ); document.write( "simplify
\n" ); document.write( "\"%2812x%29%2F%28x%5E2-4%29+=+4%2F1\"
\n" ); document.write( "do cross products
\n" ); document.write( "\"12x=4%28x%5E2-4%29\"
\n" ); document.write( "\"12x=4x%5E2-16\"
\n" ); document.write( "\"0=4x%5E2-12x-16\"
\n" ); document.write( "divide by 4
\n" ); document.write( "\"0=x%5E2-3x-4\"
\n" ); document.write( "factoring
\n" ); document.write( "\"0+=+%28x-4%29%28x%2B1%29\"
\n" ); document.write( "So possible values for x are 4 and -1
\n" ); document.write( "Since we need a positive speed, the speed in
\n" ); document.write( "still water is 4 mph
\n" ); document.write( "
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