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Thread: Solve an equation system with three unknown variables

  1. #1
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    Solve an equation system with three unknown variables

    Please solve this equation system:
    xy+x+y=11
    xz+x+z=14
    yz+y+z=19

    First I factorized everything:
    (x+1)(y+1)=12
    (x+1)(z+1)=15
    (y+1)(z+1)=20

    After that I tried the substitution method, however I fail to solve it.
    I should I countinue?
    Thank you in advance!
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  2. #2
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    Divide the second equation by the first to give $\displaystyle \displaystyle \frac{z+1}{y+1} = \frac{5}{4} \implies z + 1 = \frac{5}{4}(y + 1)$.

    Now substitute this into the third equation and you can solve for $\displaystyle \displaystyle y$, and then use that to solve for $\displaystyle \displaystyle x,z$.
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  3. #3
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    Thank you for your help!
    I substituted it into the third equation. My answer was that y can be 1 or -3, however when I looked in the key, y should be 3 or 5. Can you please show be how to do it?
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  4. #4
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    I think you'll find that $\displaystyle \displaystyle y = 3$ or $\displaystyle \displaystyle y = -5$...
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  5. #5
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    Hello, Anna55!

    $\displaystyle \text{Solve this system: }\;\begin{array}{cccc}xy+x+y&=&11 & [1] \\ xz+x+z&=&14 & [2] \\ yz+y+z&=&19 & [3] \end{array}$

    I tried all kinds of "clever" tricks . . . and failed abysmally.
    . . I was reduced to using simple Substitution . . .

    $\displaystyle \text{In [3], solve for }z\!:\;\;z \;=\;\dfrac{19-y}{1+y}\;\;[4]$

    Substitute into [2]: .$\displaystyle x\left(\dfrac{19-y}{1+y}\right) + x + \dfrac{19-y}{1+y} \:=\:14$

    . . $\displaystyle \text{Solve for }y\!:\;\;y \;=\;\dfrac{4x+1}{3}\;\;[5]$


    $\displaystyle \text{Substutite into [1]: }\;x\left(\dfrac{4x+1}{3}\right) + x + \dfrac{4x+1}{3} \;=\;11$

    This simplifies to: .$\displaystyle x^2 + 2x - 8 \:=\:0 \quad\Rightarrow\quad (x-2)(x+4) \:=\:0$

    . . Hence: .$\displaystyle x \;=\;2,\:\text{-}4$


    $\displaystyle \text{Substitute into [5]: }\;y \:=\:\begin{Bmatrix}\dfrac{4(2)+1}{3} \:=\:3 \\ \\[-3mm] \dfrac{4(\text{-}4)+1}{3} \;=\;\text{-}5 \end{Bmatrix}$


    $\displaystyle \text{Substitute into [4]: }\;z \;=\;\begin{Bmatrix}\dfrac{19-3}{1+3} &=& 4 \\ \\[-3mm] \dfrac{19-(\text{-}5)}{1 - 5} &=& \text{-}6 \end{Bmatrix}$


    $\displaystyle \text{Therefore: }\;(x,y,z) \;=\;(2,3,4),\;(\text{-}4,\text{-}5,\text{-}6)$

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