document.write( "Question 1204175: P can complete a task in x days while Q takes (x + 12) days to do the same task.
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Algebra.Com's Answer #840250 by Edwin McCravy(20056)\"\" \"About 
You can put this solution on YOUR website!
P can complete a task in x days while Q takes (x + 12) days to do the same task.
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document.write( "P can complete 1 task in x days, so P can complete 1/x of a task in 1 day.\r\n" );
document.write( "Q can complete 1 task in (x + 12) days, so Q can complete 1/(x + 12) of a task\r\n" );
document.write( "in 1 day.

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document.write( "P can complete 1/x of a task in 1 day, so P can complete 8/x of a task in 8\r\n" );
document.write( "days.\r\n" );
document.write( "Q can complete 1/(x + 12) of a task in 1 day, so Q can complete 8/(x + 12) of a\r\n" );
document.write( "task in 8 days.\r\n" );
document.write( "So together, they can complete 8/x + 8/(x+12) of a task in 8 days\r\n" );
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document.write( "That's 1 task, so \r\n" );
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document.write( "\"8%2Fx%2B8%2F%28x%2B12%29\"\"%22%22=%22%22\"\"1\"\r\n" );
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document.write( "LCD = x(x + 12)\r\n" );
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document.write( "\"8%28x%2B12%29+%2B+8x\"\"%22%22=%22%22\"\"x%28x%2B12%29\"\r\n" );
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document.write( "\"8x%2B96%2B8x\"\"%22%22=%22%22\"\"x%5E2%2B12x\"\r\n" );
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document.write( "\"16x%2B96\"\"%22%22=%22%22\"\"x%5E2%2B12x\"\r\n" );
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document.write( "\"x%5E2%2B12x\"\"%22%22=%22%22\"\"16x%2B96\"\r\n" );
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document.write( "\"x%5E2-4x-96\"\"%22%22=%22%22\"\"0\"\r\n" );
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document.write( "\"%28x%2B8%29%28x-12%29\"\"%22%22=%22%22\"\"0\"\r\n" );
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document.write( "x+8=0; x-12=0\r\n" );
document.write( "  x=-8;   x=12\r\n" );
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document.write( "x cannot be negative.\r\n" );
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document.write( "x = 12.\r\n" );
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document.write( "Now that we have x, we can answer (a) as a numerical fraction:\r\n" );
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document.write( "a) What fraction of the task can each of them complete in a day?
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document.write( "P can complete 1 task in 12 days, so P can complete 1/12 of a task in 1 day.\r\n" );
document.write( "Q can complete 1 task in (12 + 12) = 24 days, so Q can complete 1/(12 + 12) = 1/24 of a task in 1 day.
Edwin
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