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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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Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. If this is homework, please add the homework tag; people will still help, so don't worry. Also, many find the use of imperative ("Prove", "Solve", "Find", etc.) to be rude when asking for help; please consider rewriting your post. –  amWhy Nov 18 '12 at 2:22
    
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. –  Ross Millikan Nov 18 '12 at 5:08

2 Answers 2

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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Chan: there's no way around what may turn out to look like a mess. Just keep in mind that the variables you are solving for are $x, y, z$, and the constants are $M, N, K$. Feel free to update your post if you want to check your work. –  amWhy Nov 18 '12 at 2:37
    
I did it. Then I got a quintic equation which I can't solve. –  Chan Yat-Fay Nov 18 '12 at 2:41
    
Chan Why don't you post what you've got (i.e., edit your question and include what you've got)? –  amWhy Nov 18 '12 at 2:46
    
Somebody asked me years ago, and I don't know where he got it. –  Chan Yat-Fay Nov 18 '12 at 2:48
    
The suggestion was that you post your quintic, and how you arrived at that quintic. –  Gerry Myerson Nov 18 '12 at 4:34

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 $z$: $$x={Nz-m\over(z-1)(z+1)},\quad y={Mz-N\over(z-1)(z+1)}$$

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