Question

Solve the following equations :
$y^2+z^2-(y+z)x=12,x^2+z^2-(x+z)y=8,$
$x^2+y^2-(x+y)z=2.$

Answer 1

We rewrite the equations as
$(y^2-zx)+(z^2-xy)=12$
$(x^2-yz)+(z^2-xy)=8$
$(x^2-yz)+(y^2-zx)=2$
Adding $\left(1\right),\left(2\right)$ and $\left(3\right)$, we get
$2(y^2-zx)+2(z^2-xy)+2(x^2-yz)=22$
or $(y^2-zx)+(z^2-xy)+(x^2-yz)=11$
From $\left(1\right)$ and $\left(4\right)$,
$x^2-yz=-1$
Similarly from $\left(2\right)$ and $\left(4\right)$,
$y^2-zx=3$
and from $\left(3\right)$ and $\left(4\right)$,
$z^2-xy=9$
Multiplying $\left(5\right),\left(6\right),\left(7\right)$ by $y,z,x$ respectively and adding, we get
$-y+3z+9x=0$ or $9x-y+3z=0$
Again multiplying these equations by $z,x$ and $y$ and adding, we get
$-z+3x+9y=0$
or $3x+9y-z=0$
Solving $\left(8\right)$ and $\left(9\right)$ by cross-multiplication, we get
$\frac{x}{1-27}=\frac{y}{9+9}=\frac{z}{81+3}$
or $\frac{x}{-26}=\frac{y}{18}=\frac{z}{84}$
or $\frac{x}{-13}=\frac{y}{9}=\frac{z}{42}=k,$say
Substituting these values in $\left(5\right)$, we get
$169k^2-378k^2=-1$
or $209k^2=1$ or $k=\pm \frac{1}{\surd \left(209\right)}$
Hence $x=\pm \frac{13}{\surd \left(209\right)},y=\pm \frac{9}{\surd \left(209\right)},z=\pm \frac{42}{\surd \left(209\right)}$.