find $x,y,z$ such that $x+yz=M$, $y+zx=N$, $z+xy=K$ somebody asked me this. I don't know whether it is interesting but I hope someone can solve it.
find $x,y,z$ such that
$x+yz=M$,
$y+zx=N$,
$z+xy=K$,
where $M,N,K$ are constants.
 A: To solve for $x, y, z$, given constants $M,N,K$ and the following equations:
$$x+yz=M,\tag{1}$$
$$y+zx=N,\tag{2}$$
$$z+xy=K,\tag{3}$$
we can start by solving for $x$ in equation $(1)$, and then substituting $x = M-yz$ into $(2)$ and $(3)$. This gives two equations $(2), (3)$, two unknowns.
$(1)$: $x=M-yz,$ 
$(2)$: $y + z(M-yz) = N,$
$(3)$: $z + y(M-yz) = K.$
You can then solve for $y$, substitute that value into $(3)$, and then get $z$ as an expression in $M, N, K$, then back substitute, solving first for $y$, and then for $x$.

It does indeed get messy, quickly. See WolframAlpha's solution.
(Perhaps trivial) Observation: Note that if we have $M = N = K$, then by symmetry of the equations, we must have $x = y = z$. One thought: try to solve for $M, N, K$ when $M=N=K$, and then see how $M=N, \;M\neq K$ impacts the solution, and finally, what this means for $M\neq N,;N\neq K,\;M\neq K$. But given WolframAlpha's solutions, it isn't going to be "pretty"!
A: Maple says $z$ is a root of $$u^5-Ku^4-2u^3+(MN+2K)u^2+(1-M^2-N^2)u+MN-K=0$$ and then expresses $x$ and $y$ in terms of $z$: $$x={Nz-m\over(z-1)(z+1)},\quad y={Mz-N\over(z-1)(z+1)}$$
