document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=148508>Question 148508</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Suppose it takes Tom and Dick 2 hours to do a certain job, it takes Tom and Harry 3 hours to do the same job. It takes Dick and Harry 4 hours to do this same job.\r<BR>\n" );
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document.write( "How long will it take Tom, Dick, and Harry to do the same job together? \r<BR>\n" );
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document.write( "So, just for organizing: <BR>\n" );
document.write( "T + D = 2hrs<BR>\n" );
document.write( "T + H = 3hrs<BR>\n" );
document.write( "D + H = 4hrs<BR>\n" );
document.write( "T+D+H = ? hrs\r<BR>\n" );
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document.write( "Therefore: (I think, but I'm sure it's most probably right) <BR>\n" );
document.write( "1/T + 1/D =1/2<BR>\n" );
document.write( "1/T + 1/H =1/3<BR>\n" );
document.write( "1/D + 1/H =1/4<BR>\n" );
document.write( "1/T + 1/D + 1/H =1/x\r<BR>\n" );
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document.write( "So far, I got to T=12/5 and D=12<BR>\n" );
document.write( "using reciprocals and the first equation, the sum is 6/12 or 1/2; so far so good.<BR>\n" );
document.write( "But I've tried using it to solve for the H one.. and it doesn't quite work.. so when I get to the last equation I still have two variables, H and X.\r<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #108835 by </B><A HREF=/tutors/aboutme.mpl?userid=ptaylor><B>ptaylor(2198)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=ptaylor><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=ptaylor><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=108835\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> YOU ARE RIGHT ON TARGET!!!!! EXCEPT YOU MAY BE MAKING IT A LITTLE MORE COMPLICATED THAT IT REALLY IS.   FROM THE THREE EQUATIONS BELOW (THE ONES YOU CAME UP WITH), YOU CAN READILY CALCULATE THE RATE AT WHICH EACH OF THE THREE WORK.  WHEN YOU HAVE DONE THAT, THEN YOU WILL NOT NEED THE 1/x IN YOUR FOURTH EQUATION. INSTEAD, JUST ADD THE THREE RATES TOGETHER AND THEN SOLVE THE FOLLOWING EQUATION:<BR>\n" );
document.write( "(RATE AT WHICH TOM, DICK AND HARRY DOES THE JOB)*X HOURS=1(1 JOB, THAT IS)<BR>\n" );
document.write( "Let x=amount of time required if all three are working together<BR>\n" );
document.write( "Let 1/T=rate at which Tom does the job<BR>\n" );
document.write( "Let 1/D=rate at which Dick does the job<BR>\n" );
document.write( "And Let 1/H=rate at which Harry does the job\r<BR>\n" );
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document.write( "1/T + 1/D =1/2---------------------eq1<BR>\n" );
document.write( "1/T + 1/H =1/3------------------------eq2<BR>\n" );
document.write( "1/D + 1/H =1/4-------------------------------eq3\r<BR>\n" );
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document.write( "subtract eq3 from eq2 and we get:<BR>\n" );
document.write( "1/T - 1/D=4/12-3/12=1/12<BR>\n" );
document.write( "1/T-1/D=1/12-----------------------------eq3a<BR>\n" );
document.write( "next, add eq3a and eq1 and we get:<BR>\n" );
document.write( "2/T=7/12 divide each term by 2<BR>\n" );
document.write( "1/T=7/24----------------------------rate at which Tom works<BR>\n" );
document.write( "substitute  1/T=7/24 into eq1:<BR>\n" );
document.write( "7/24 +1/D=1/2 subtract 7/24 from each side<BR>\n" );
document.write( "1/D=12/24-7/24=5/24-------------------------------rate at which Dick works<BR>\n" );
document.write( "substitute 1/D=5/24 into eq3:<BR>\n" );
document.write( "5/24+1/H=1/4  subtract 5/24 from each side<BR>\n" );
document.write( "1/H=6/24-5/24=1/24---------------------------------rate at which Harry works\r<BR>\n" );
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document.write( "So, together, they work at the rate of <BR>\n" );
document.write( "7/24 +5/24 +1/24=13/24 of the job per hour<BR>\n" );
document.write( "So, (13/24)*x=1  multiply each side by 24<BR>\n" );
document.write( "13x=24  divide each side by 13<BR>\n" );
document.write( "x=24/13 hr\r<BR>\n" );
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document.write( "Hope this helps---ptaylor\r<BR>\n" );
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