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Solve for $x,y,z$:

$$x(y+z-x)=68-2x^2$$ $$y(z+x-y)=102-2y^2$$ $$z(x+y-z)=119-2z^2$$

After some manipulation, I obtain

$$xy+xz=68-x^2$$ $$yz+xy=102-y^2$$ $$xz+yz=119-z^2$$

After combining equations, I get

$$y=\frac{-51-x^2+z^2}{x-z}$$

This seems too tedious. Is there a simpler way?

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3 Answers 3

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Your system is equivalent to \begin{align} x(y+z+x)&=68\\ y(z+x+y)&=102\\ z(x+y+z)&=119. \end{align} Therefore, in effect, you have \begin{align} xa&=68\\ ya&=102\\ za&=119\\ a&=x+y+z. \end{align} This means that $$a=\frac{68}{a}+\frac{102}{a}+\frac{119}{a}$$ or $$a^2=68+102+119\implies a=\pm 17,$$ which results in either $x= 4, y=6, z=7$ or $x=-4, y=-6, z=-7.$ (Thanks to paw88789 for pointing this out).

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    $\begingroup$ $(-4,-6,-7)$ is also a solution. $\endgroup$
    – paw88789
    Commented Apr 23, 2022 at 20:17
  • $\begingroup$ how do you conclude that $a^2=68+102+119$? $\endgroup$
    – homosapien
    Commented Apr 23, 2022 at 20:20
  • $\begingroup$ Thanks @paw88789 for pointing that out. I always forget about the negative roots. $\endgroup$
    – Patricio
    Commented Apr 23, 2022 at 20:26
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    $\begingroup$ You can also sum the 3 lines together to get $a^2=68+102+119$ $\endgroup$
    – zwim
    Commented Apr 23, 2022 at 21:56
  • $\begingroup$ @zwim thanks that's what I wasn't seeing. $\endgroup$
    – homosapien
    Commented Apr 23, 2022 at 22:45
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If you add $$xy+xz=68-x^2$$ $$yz+xy=102-y^2$$ $$xz+yz=119-z^2$$ you get $(x+y+z)^2=17^2$, and so $x+y+z=\pm 17$. If you substitute $x+y=\pm 17-z$ in the third of your equation, you get $\pm 17 z=119$, and so $z=7$ or $z=-7$; now it is easy to get the two solution $x=4$, $y=6$, $z=7$ and $x=-4$, $y=-6$, $z=-7$.

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Solve for $x,y,z$:

$$x(y+z-x)=68-2x^2$$ $$y(z+x-y)=102-2y^2$$ $$z(x+y-z)=119-2z^2$$

$$x(x + y + z) = 68 - 2x^2 + 2x^2 = 68. \tag1$$

Similarly, $$y(x + y + z) = 102. \tag2 $$

$$z(x + y + z) = 119. \tag3 $$

Therefore, $$x::y::z = 68::102::119.\tag4 $$

You can use (4) above to eliminate (for example) the $x,y$ variables, expressing both of them in terms of $z$. Then, if the system is solvable, routine methods should conquer the original equations.

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    $\begingroup$ Or eliminate them all, having $x=68c, y=102c, z=119c,$ getting a simpler equation. $\endgroup$ Commented Apr 23, 2022 at 20:18
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    $\begingroup$ @ThomasAndrews Got me. Originally, I didn't think of that. Then, once I saw the other answer, I didn't want to plagiarize. $\endgroup$ Commented Apr 23, 2022 at 20:23

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