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Solving the following simultaneous equations:

$2y+2z=myz$

$2x+z=mxz$

$2x+y=mxy$

$xyz=27$

These are 4 equations in 4 unknowns: $x, y, z, m$ so I think a solution is possible, though I'm unsure. They are non-linear so no matricies :(

Also are there any online applications that may compute this non-linear solution?

Thanks!

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The first equation has $2y + 2z$, while the second and third left-hand-sides are not symmetrical. Is this a typo? –  Arthur Oct 17 '12 at 12:11
    
no typo, it is as is :) –  redrum Oct 17 '12 at 12:12
    
OK, I'm just asking because problems like this are usually very symmetric in nature. –  Arthur Oct 17 '12 at 12:12
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WolframAlpha managed to find a real solution. –  Douglas S. Stones Oct 17 '12 at 12:17
    
wow nice, I didn't know wolf could do that! though I have the solution now, i'd appreciate any working if possible. –  redrum Oct 17 '12 at 12:20
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2 Answers

up vote 1 down vote accepted

You can ask WolframAlpha, if you want, but in this case, you can do it also by hand: Multiply the first three equations by $x$, $y$ resp. $z$, get \[ 27m = 2xy + 2xz = 2xy+yz = 2xz + yz \] So $2xy = 2xz$ and - as $x \ne 0$ since $xyz = 27$ - $y = z$. On the other hand $2xy = yz$, so $x = \frac z2$. Now \[ 27 = xyz = \frac 12z^3 \iff z = 3\cdot 2^{1/3} \] so $y=z =3\cdot 2^{1/3}$, $x = \frac 12z = 3\cdot 2^{-2/3}$. Now \begin{align*} m &= \frac{2y + 2z}{yz}\\ &= \frac{4z}{z^2}\\ &= \frac 4z\\ &= \frac{2^2}{3\cdot 2^{1/3}}\\ &= \frac{2^{5/3}}3. \end{align*}

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How would you feed these kinds of problems to WolframAlpha? I find it tends to break down for problems in lots of variables. I generally have more success with Mathematica. –  Thomas Oct 17 '12 at 12:35
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@Thomas check the comments on OP –  Alex Oct 17 '12 at 12:40
    
@Alex Ah, thanks, I missed the link. –  Thomas Oct 17 '12 at 12:47
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$$27m=(2y+2z)x=(2x+z)y=(2x+y)z$$ $$2xy+2xz=2xy+yz=2xz+yz$$


We get: $$x(y-z)=0$$ $$z(2x-y)=0$$ $$y(2x-z)=0$$


as $x,y,z \ne 0$

we have: $y=z$,$2x=y$ and $2x=z$

so, $xyz=\frac{1}{2}y^3=27 \implies y=z=3\sqrt[3]{2}$ and $x=\frac{3}{2}\sqrt[3]{2}$ and $m=\frac{(2y+2z)x}{27}=\frac{2}{3}\sqrt[3]{4}$

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