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Solve the following equation system:

$$x+y+xy=19$$ $$y+z+yz=11$$ $$z+x+zx=14$$

I've tried substituting, adding, subtracting, multiplying... Nothing works. Could anyone drop me a few hints without actually solving it? Thanks!

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    $\begingroup$ This is NOT, strictly speaking "linear algebra" because these equations are not linear. $\endgroup$ – user247327 Feb 23 '16 at 21:42
  • $\begingroup$ Here is a more recent question about the same system. $\endgroup$ – Martin Sleziak Jan 5 '20 at 16:30
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Add $1$ to both sides of all the equations. To get \begin{align*} (x+1)(y+1) & = 20\\ (y+1)(z+1) & = 12\\ (z+1)(x+1) & = 15\\ \end{align*} Now let $u=x+1,v=y+1,w=z+1$. And you have \begin{align*} uv&=20\\ vw&=12\\ wu&=15 \end{align*} From this you can get $$(uvw)=\pm 60.$$ Now use the above equations to compute $u=\pm 5$ and so on.

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Well, $x = (19-y)/(1+y),$ $y = (11-z)/(1+z),$ $z= (14-x)/(1+z).$ If you substitute for $z$ in the second equation, and then for $y$ in the first, you should get something of the form $x = (ax+b)/(cx+d),$ which is a quadratic equation...

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