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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.

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    The symmetry of the problem makes me want to multiply the first by x, the second by y, and the third by z. This gives an xyz in each. Unfortunately, that doesn't seem to help. Maybe somebody can take this and run with it.2012-11-18

2 Answers 2

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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"!

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    @amWhy +1 As Gerry noted Maple gives a complex form.2013-08-15
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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 $z$: $x={Nz-m\over(z-1)(z+1)},\quad y={Mz-N\over(z-1)(z+1)}$