document.write( "Question 106101: A river has a steady current. A boat travels 30 mph over still water. If a trip 85 miles downstream and back takes 10 hours, then what is the speed of the current?
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Algebra.Com's Answer #77325 by ankor@dixie-net.com(22740)\"\" \"About 
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A river has a steady current. A boat travels 30 mph over still water. If a trip 85 miles downstream and back takes 10 hours, then what is the speed of the current?
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\n" ); document.write( "Let x = speed of the current
\n" ); document.write( "then
\n" ); document.write( "(30+x) speed downstream
\n" ); document.write( "and
\n" ); document.write( "(30-x) speed up-stream
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\n" ); document.write( "Write a time equation: Time = Distance/speed
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\n" ); document.write( "Time upstream + Time downstream = 10 hrs
\n" ); document.write( "\"85%2F%28%2830-x%29%29\" + \"85%2F%28%2830%2Bx%29%29\" = 10
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\n" ); document.write( "Multiply equation by (30-x)(30+x) to get rid of the denominators:
\n" ); document.write( "(30-x)(30+x)*\"85%2F%28%2830-x%29%29\" + (30-x)(30+x)*\"85%2F%28%2830%2Bx%29%29\" = 10(30-x)(30+x)
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\n" ); document.write( "Cancel out the denominators, FOIL (30-x)(30+x) and you have;
\n" ); document.write( "85(30-x) + 85(30+x) = 10(900-x^2)
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\n" ); document.write( "2550 - 85x + 2550 + 85x = -10x^2 + 9000
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\n" ); document.write( "The 85x's cancel leaving us with:
\n" ); document.write( "5100 = -10x^2 + 9000
\n" ); document.write( "Or
\n" ); document.write( "10x^2 = 9000 - 5100
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\n" ); document.write( "10x^2 = 3900
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\n" ); document.write( "x^2 = 3900/10
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\n" ); document.write( "x^2 = 390
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\n" ); document.write( "x = \"sqrt%28390%29\"
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\n" ); document.write( "x = 19.75 mph is the current, that seems high so let's check it in the equation
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\n" ); document.write( "\"85%2F%28%2830-19.75%29%29\" + \"85%2F%28%2830%2B19.75%29%29\" = 10
\n" ); document.write( "\"85%2F%28%2810.25%29%29\" + \"85%2F%28%2849.75%29%29\" = 10
\n" ); document.write( " 8.29 + 1.71 = 10 hrs, confirms our solution
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\n" ); document.write( "Could you follow the logic here OK? Any questions?
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