how to find the solution to this system of equations

Question

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.

Answer 1

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 $-24$.

Answer 2

First assume $x,y,z\ne 0$.

From equation (1), $y+z=\dfrac {54}{x}+x\Rightarrow y+z-x=\dfrac {54}{x}$ ..........(A1)

From equation (2), $z+x=\dfrac {64}{y}+y\Rightarrow z+x-y=\dfrac {64}{y}$ ..........(A2)

From equation (3), $x+y=\dfrac {70}{z}+z\Rightarrow x+y-z=\dfrac {70}{z}$............(A3)

Now, adding (A1) & (A2), $2xyz=54y+64x$......(B1)

Adding (A2) & (A3), $2xyz=64z+70y$........(B2)

Adding (A3) & (A1), $2xyz=70x+54z$...........(B3)

Equating (B1) & (B2), $64x-16y-64z=0$

Equating (B2) & (B3), $70x-70y-16z=0$

Equating (B3) & (B1), $16x-54y-54z=0$.

Now solve these equations.

Answer 3

with the method above we get $(x,y,z)=\pm (9, 8, 7)$

Answer 4

HINT:

$ x ( x+y+z - 2 x ) =54 \, ; y (x-y+z- 2 y )=64 \,; z (x+y+z- 2 z)=70 $