# Find $x,y,z$ satisfying $x(y+z-x)=68-2x^2$, $y(z+x-y)=102-2y^2$, $z(x+y-z)=119-2z^2$

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?

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).

• $(-4,-6,-7)$ is also a solution. Apr 23, 2022 at 20:17
• how do you conclude that $a^2=68+102+119$? Apr 23, 2022 at 20:20
• Thanks @paw88789 for pointing that out. I always forget about the negative roots. Apr 23, 2022 at 20:26
• You can also sum the 3 lines together to get $a^2=68+102+119$
– zwim
Apr 23, 2022 at 21:56
• @zwim thanks that's what I wasn't seeing. Apr 23, 2022 at 22:45

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 $$x=-4$$, $$y=-6$$, $$z=-7$$.

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.

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