<s>[INST] how to find the solution to this system of equations Given the system of equations:
$
xy+xz=54+x^2  \\ 
yx+yz=64+y^2 \\ 
zx+zy=70+z^2 $
Need to find all of the solutions of $ x,y$ and $z$.
Tried to sum up all three equations but got stuck with nothing to factorize. 
 [/INST] Adding, $x^{2}$ ,$y^{2}$, & $z^{2}$ to 1st,2nd & 3rd equations respectively we get, 
$x(x+y+z)=2x^{2}+54$
$y(x+y+z)=2y^{2}+64$
$z(x+y+z)=2z^{2}+70$
So, $\dfrac{2x^{2}+54}{x}=\dfrac{2y^{2}+64}{y}=\dfrac{2z^{2}+70}{z}=x+y+z=k(say)$
From these, $2x^{2}-kx+54=0,2y^{2}-ky+64=0,2z^{2}-kz+70=0$.
So, $x=\dfrac{k\pm \sqrt{k^{2}-4.2.54}}{4}, y=\dfrac{k\pm \sqrt{k^{2}-4.2.64}}{4}, z=\dfrac{k\pm \sqrt{k^{2}-4.2.70}}{4}$.
Putting, $k=24$,& taking only positive sign we get, $x=9,y=8,z=7$.
Note that, $x+y+z=24$. So, if we take negative sign then $x=3,y=4,z=5$, in that case $x+y+z\ne 24$. So, it is not admissible.
Again, if we take, $k=-24$ then the solutions are $x=-9,y=-8,z=-7$ 7 in this case sum is also $-24$.
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