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Help me out in these linear equations

  1. Sep 16, 2014 #1
    1. The problem statement, all variables and given/known data
    2 women and 5 men can finish a work a work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 women alone to finish the work, and also that taken by 1 man alone.

    2. Relevant equations
    Pair of linear equations in two variables

    3. The attempt at a solution
    I can't figure how can I do it out:( LOL
    But in the solution, I saw that it had been given that,
    Work done by woman in 1 day=x
    Work done by man in 1 day=y

    =>2/x+ 5/y =1/4

    =>3/x + 6/y =4/3

    Solving them, gave the answers. But, my doubt here is, how did 2/x,5/y,etc. arrive?

    Could anybody please enlighten me:)
  2. jcsd
  3. Sep 16, 2014 #2


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    Staff: Mentor

    Strange, these are not linear equations.

    Whole work is 1 - if a woman does x per day, she will need 1/x days to finish the work. 2 women need 2/x.
  4. Sep 16, 2014 #3

    Ray Vickson

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    Why should 2 women need twice as long to finish the work?
  5. Sep 16, 2014 #4


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    Staff: Mentor

    Good question, no idea what I was thinking :frown:
  6. Sep 16, 2014 #5


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    Because they get yakity yaking??
  7. Sep 16, 2014 #6

    Ray Vickson

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    Think about rates, and that rates add (but times do not).

    Anyway, the RHS of the second equation above should be 1/3, not 4/3.
  8. Sep 16, 2014 #7


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    I think the confusion comes from the units in your problem set up.

    Your setup says x is the work done by a woman in 1 day, so I take units are [work woman##^{-1}## day##^{-1}##].
    If so, then the work per day by 2 women would be: 2 [women] * x [work woman##^{-1}## day##^{-1}##] = 2x [work day##^{-1}##].

    Then the first equation would be 2x + 5y = 1/4 [work / day]. This is the way I'd do the problem.

    However, if you follow the problem statement in choosing your variables:

    Let x = the time taken by 1 woman alone to finish the work. Then x has units [woman days work##^{-1}##].
    Similarly, let y be the time for a 1 man alone [man days] to finish the work.

    It follows that 2 [women] / x [woman days work##^{-1}##] has units [work days##^{-1}##], and you get the equations that you posted (with a 1/3 on the RHS as the previous poster points out).

    As some of the other posters noticed, these aren't linear equations. That's why it seems easier to me to do it the other way, with your original definition: x is work/woman/day, and so on.
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