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Math Help - Tricky Equation Setup

  1. #1
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    Tricky Equation Setup

    Gilbert Greenfield owns a pasture on which the grass has been cut to uniform height of two inches. The grass in the pasture grows uniformly at a constant rate. In addition to owning the pasture, Mr. Greenfield also owns a cow, a horse, and a sheep. The grass existing and growing in the pasture is sufficient to feed all three animals grazing together for 20 days. It would feed the cow and the horse alone for 25 days; the cow and the sheep alone for 33⅓ days; and the horse and the sheep alone for 50 days. How long would it sustain (a) the cow alone? (b) The horse alone? (c) The sheep alone?
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  2. #2
    RJH
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    Ok, so set up 3 simultaneous equations as follows:
    h + c = 25
    c + s = 33.5
    h + s = 50
    then solve them by substitution

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  3. #3
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    Would'nt the 4 equations in this problem be:

    1/H + 1/C = 25
    1/C + 1/S = 33 1/3
    1/H + 1/S = 50
    1/H + 1/C + 1/S = 20
    ???

    If this is true, can someone be so kind as to explain why?

    Thanks.
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  4. #4
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    How would the variables be defined???
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  5. #5
    RJH
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    Hi MATNTRNG!
    I apologize for any confusion caused by my response- it was done without reading the question at all properly - here's my second go:
    X amount of grass in the field
    cow, horse and sheep eat C,H and S amount respectively
    so: C+H+S=X
    so the rate at which it was eaten is:
    C/t + H/t + S/t =X/t
    so the total time is:
    T= X/(C/t + H/t + S/t)
    so the equations are:
    20=
    X/(C/t + H/t + S/t)
    25=X/(C/t + H/t)
    33.3=X/(C/t + S/t)
    50=X/(S/t + H/t)
    the question asks for X/(C/t), X/(H/t) and X/(S/t)
    sorry again.


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