document.write( "Question 672824: A and B do a work together in five days If A do the work with double speed and B with half the speed then they do the work in 4 days together find A alone do the work in how much time. \n" ); document.write( "
Algebra.Com's Answer #418335 by ankor@dixie-net.com(22740)\"\" \"About 
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A and B do a work together in five days
\n" ); document.write( " If A do the work with double speed and
\n" ); document.write( " B with half the speed then they do the work in 4 days together
\n" ); document.write( " find A alone do the work in how much time.
\n" ); document.write( ":
\n" ); document.write( "let a = A's time alone
\n" ); document.write( "let b = B's time alone
\n" ); document.write( "then
\n" ); document.write( ".5a = A's time when working twice as fast
\n" ); document.write( "and
\n" ); document.write( "2b = B's time when working half as fast
\n" ); document.write( ":
\n" ); document.write( "Let the completed job = 1
\n" ); document.write( ":
\n" ); document.write( "Write a shared work equation for each scenario
\n" ); document.write( ":
\n" ); document.write( "\"5%2Fa\" + \"5%2Fb\" = 1
\n" ); document.write( "get rid of the denominators, mult by ab
\n" ); document.write( "5b + 5a = ab
\n" ); document.write( "and
\n" ); document.write( "\"4%2F%28.5a%29\" + \"4%2F%282b%29\" = 1
\n" ); document.write( "multiply by 2ab to get rid of the denominators
\n" ); document.write( "4b(4) + 4a = 2ab
\n" ); document.write( "16b + 4a = 2ab
\n" ); document.write( ":
\n" ); document.write( "Use elimination here, mult the 1st equation by 4, mult the 2nd eq by 5
\n" ); document.write( "then subtract the 1st from the 2nd
\n" ); document.write( "80b + 20a = 10ab
\n" ); document.write( "20b + 20a = 4ab
\n" ); document.write( "------------------subtraction eliminates a
\n" ); document.write( "60b + 0 = 6ab
\n" ); document.write( "divide both sides by b
\n" ); document.write( "60 = 6a
\n" ); document.write( "a = 60/6
\n" ); document.write( "a = 10 days working alone
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