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!
Best Answer
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*}