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Thread: Simultaneous equations - 3 variables

  1. #1
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    Question Simultaneous equations - 3 variables

    Hi!

    I've got myself into this rather unhelpful question. I have tried many different ways to solve it, but all of them have turned out to fail miserably .The question is:

    Solve,
    $\displaystyle xy+5(x+y)=47$
    $\displaystyle yz+5(y+z)=65$
    $\displaystyle zx+5(z+x)=55$

    I thought it would be reasonably simple to solve, however, (for me at least!) it is harder than it looks!

    Could anybody help me a little here?
    Thanks very much!
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  2. #2
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    Re: Simultaneous equations - 3 variables

    2 integer solutions:
    a = 3, b = 4, c = 5
    a = -13, b = -14, c = -15
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  3. #3
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    Re: Simultaneous equations - 3 variables

    $\displaystyle xy+5(x+y)=47$ (1)
    $\displaystyle yz+5(y+z)=65$ (2)
    $\displaystyle zx+5(z+x)=55$ (3)

    Solving (1) for $\displaystyle x$ yields

    $\displaystyle x=\frac{47-5y}{y+5}$ (4)

    And solving (2) for $\displaystyle z$,

    $\displaystyle z=\frac{65-5y}{y+5}$ (5)

    Then, substituting (4) and (5) into (3):

    $\displaystyle \left(\frac{65-5y}{y+5}\right)\left(\frac{47-5y}{y+5}\right)+5\left(\frac{65-5y}{y+5}+\frac{47-5y}{y+5}\right)=55$

    $\displaystyle \Rightarrow\frac{5\left(5y^2-112y+611\right)}{(y+5)^2}+\frac{10(56-5y)}{y+5}=55$

    $\displaystyle \Rightarrow5\left(5y^2-112y+611\right)+10(56-5y)(y+5)=55(y+5)^2$

    $\displaystyle \Rightarrow80y^2+800y-4480=0$

    $\displaystyle \Rightarrow80(y+14)(y-4)=0$

    $\displaystyle \Rightarrow y=-14\mathrm{\quad or\quad}y=4$

    And substituting both of these into (4) and (5) gives us $\displaystyle x$ and $\displaystyle z$.
    Thanks from BobtheBob
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  4. #4
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    Re: Simultaneous equations - 3 variables

    Oh right thanks! I must have made a mistake when I did a similar thing.

    Thanks very much!
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