document.write( "Question 107977: ok, here we go.\r
\n" ); document.write( "\n" ); document.write( "John and Harry can paint a house in 10 days. when working alone John can paint the house in 30 days. How many days will it take Harry to paint the house.\r
\n" ); document.write( "\n" ); document.write( "I have no idea how to do this probolem.
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Algebra.Com's Answer #78689 by ankor@dixie-net.com(22740)\"\" \"About 
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John and Harry can paint a house in 10 days. when working alone John can paint the house in 30 days. How many days will it take Harry to paint the house.
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\n" ); document.write( "Here's a method you can use on all these kind of problems
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\n" ); document.write( "Let the completed = 1, each will do a fraction that will add up to 1
\n" ); document.write( "It's obvious that John will complete 1/3 of the job in 10 days (10/30)
\n" ); document.write( "Therefore John has to do 2/3 of the job. What denominator of 10 = 2/3?
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\n" ); document.write( "Let x = Harry's time by himself
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\n" ); document.write( "\"10%2F30\" + \"10%2Fx\" = 1
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\n" ); document.write( "Get rid of the denominators, multiply equation by 30x
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\n" ); document.write( "30x*\"10%2F30\" + 30x*\"10%2Fx\" = 30x(1)
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\n" ); document.write( "Cancel out the denominators and you have;
\n" ); document.write( "10x + 30(10) = 30x
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\n" ); document.write( "10x - 30x = -300
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\n" ); document.write( "-20x = - 300
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\n" ); document.write( "x = -300/-20
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\n" ); document.write( "x = 15 days required by John by himself
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\n" ); document.write( "Check the solution by adding up the fractions
\n" ); document.write( "\"10%2F30\" + \"10%2F15\" =
\n" ); document.write( "\"1%2F3\" + \"2%2F3\" = 1
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\n" ); document.write( "Did this make sense to you? Any questions?
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