document.write( "Question 674181: A swimming pool can be emptied in 6 hours using a 10-horsepower pump along with a 6-horsepower pump. The 6-horsepower pump requires 5 hours more than the 10horsepower pump to empty the pool when working by itself. How long would it take to empty the pool using just the 10-horsepower pump?\r
\n" ); document.write( "\n" ); document.write( "How do I go about this? I have tried just about everything I can think of. My book does not provide an example, so I can't work backwards. \r
\n" ); document.write( "\n" ); document.write( "I was thinking this,
\n" ); document.write( "let 6hp pump =x
\n" ); document.write( " 10hp pump =y
\n" ); document.write( " 10hp + 6hp = c\r
\n" ); document.write( "\n" ); document.write( "so the rate of emptying the pool with pump c= -(1/6)
\n" ); document.write( " pump x= -(1)/(5+y)
\n" ); document.write( " pump y= -(1)/(x-5)\r
\n" ); document.write( "\n" ); document.write( "Assuming I applied these rates correctly, where do I go from here?
\n" ); document.write( "Thanks.
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Algebra.Com's Answer #419078 by ewatrrr(24785) $\"\"$ $\"About$

You can put this solution on YOUR website!

 
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document.write( "Hi,
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document.write( "The 6-horsepower pump requires 5 hours more than the 10horsepower pump to empty the pool when working by itself
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document.write( "10horsepower pump 
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document.write( "6-horsepower pump 
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document.write( "PER hr KEY: emptied in 6 hours using both
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document.write( "Using both Algebraically
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document.write( " |Multiplying thru by 6x(x+5) so as all denominators = 1
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document.write( " 6(x+5) + 6x = x(x+5)
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document.write( " 6x + 30 + 6x = x^2 + 5x
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document.write( "   x^2 - 7x - 30 = 0
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document.write( "factoring
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document.write( "(x+3)(x-10) = 0   ||x = 10hr, time it would take the 10hp by itself
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document.write( "Note: Negative solution is tossed out for unit measrement
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