document.write( "Question 268391: .A and B can do a piece of work in 12 days,B and C can do it in 15 days and C and A can do
\n" ); document.write( " the same work in 20 days. How long would each take to complete the job?
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Algebra.Com's Answer #196783 by ankor@dixie-net.com(22740)\"\" \"About 
You can put this solution on YOUR website!
Let the piece of work = 1
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\n" ); document.write( "Write a shared work equation for each phrase:
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\n" ); document.write( "A and B can do a piece of work in 12 days,
\n" ); document.write( "\"12%2FA\" + \"12%2FB\" = 1
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\n" ); document.write( "B and C can do it in 15 days and
\n" ); document.write( "\"15%2FB\" + \"15%2FC\" = 1
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\n" ); document.write( " C and A can do the same work in 20 days.
\n" ); document.write( "\"20%2FA\" + \"20%2FC\" = 1
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\n" ); document.write( "Use elimination; Mult the 1st eq by 5, mult the 3nd eq by 3
\n" ); document.write( "\"60%2FA\" + \"60%2FB\" = 5
\n" ); document.write( "\"60%2FA\" + \"60%2FC\" = 3
\n" ); document.write( "------------------------------Subtraction eliminates A, leaving us with
\n" ); document.write( "\"60%2FB\" - \"60%2FC\" = 2
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\n" ); document.write( "Multiply the 2nd equation by 4, add to the above equation
\n" ); document.write( "\"60%2FB\" - \"60%2FC\" = 2
\n" ); document.write( "\"60%2FB\" + \"60%2FC\" = 4
\n" ); document.write( "-------------------------------addition eliminates C, find B
\n" ); document.write( "\"120%2FB\" = 6
\n" ); document.write( "120 = 6B
\n" ); document.write( "B = \"120%2F6\"
\n" ); document.write( "B = 20 hrs working alone
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\n" ); document.write( "Use the 2nd equation to find C, substitute 20 for B, find C
\n" ); document.write( "\"15%2F20\" + \"15%2FC\" = 1
\n" ); document.write( "Multiply by 20C, results:
\n" ); document.write( "15C + 20(15) = 20C
\n" ); document.write( "15C + 300 = 20C
\n" ); document.write( "300 = 20C - 15C
\n" ); document.write( "5C = 300
\n" ); document.write( "C = \"300%2F5\"
\n" ); document.write( "C = 60 hrs working alone
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\n" ); document.write( "Use the 3rd equation to find A
\n" ); document.write( "\"20%2FA\" + \"20%2F60\" = 1
\n" ); document.write( "Multiply by 60A
\n" ); document.write( "60(20) + 20A = 60A
\n" ); document.write( "1200 = 60A - 20A
\n" ); document.write( "40A = 1200
\n" ); document.write( "A = \"1200%2F40\"
\n" ); document.write( "A = 30 hrs working alone
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\n" ); document.write( "Summarize: A=30; B=20; C=60
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\n" ); document.write( ":
\n" ); document.write( "Check it in the 1st equation
\n" ); document.write( "\"12%2FA\" + \"12%2FB\" = 1
\n" ); document.write( "\"12%2F30\" + \"12%2F20\" =
\n" ); document.write( "\"4%2F10\" + \"6%2F10\" = 1
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\n" ); document.write( "YOu can confirm it in the other two equations.
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