How find this real value $x+y+z $ if such this equation

Question

let $x,y,z>0$ and such $$\begin{cases} \dfrac{x}{xy-z^2}=-\dfrac{1}{7}\\ \dfrac{y}{yz-x^2}=\dfrac{2}{5}\\ \dfrac{z}{zx-y^2}=-3 \end{cases}$$ show that: $$x+y+z=6$$

Answer 1

Ones first impression is that this is a set of three quadratic equations. So if we would proceed by eliminating two of the three variables, we would get a higher order equation that presumably can not be solved. However, in this case the functions are such that we can side-step these complications. With a nice trick we can work with linear equations!

$xy - z^2 = P x$

$yz - x^2 = Q y$

$zx - y^2 = R z$

Subtract these equations from each other (e.g. the second from the first). This yields:

$(x - z) S = P x - Q y$

$(y - x) S = Q y - R z$

$(z - y) S = R z - P x$

where we have defined $S = x + y + z$. We now proceed as if $S$ an adjustable parameter, and solve the set of $3+1$ linear equations (of which one can be omitted, since it depends on the other two). The final result is:

$x = (S^2 - QS + QR) C$

$y = (S^2 - RS + RP) C$

$z = (S^2 - PS + PQ) C$

where $C = \frac S {3S^2 - (P+Q+R)S + QR + RP + PQ}$

Inserting $P = -7, Q = 5/2, R = -1/3$, we find all sets of values $(x, y, z)$ in terms of $S$. In principle $S$ can range from $-\infty$ to $+\infty$. But if we are only interested in positive solutions, than we must restrict $S$ to values equal to or larger than $5/4 + (1/12)\sqrt{345}$. This comes down to $S \implies 2.797848$. There is one solution in terms of natural numbers: $x = 1, y = 2, z = 3$.