Question

Solve the following equations :
$x^2+xy+xz=18,$
$y^2+yz+yx+12=0,$
$z^2+zx+zy=30.$

Answer 1

Given equations can be rewritten as
$x(x+y+z)=18$
$y(x+y+z)=-12$
$z(x+y+z)=30$
Adding $\left(1\right),\left(2\right)$ and $\left(3\right),$ we get
${(x+y+z)}^2=36$ or $x+y+z=\pm 6$
Then from $\left(1\right)$ and $\left(4\right)$, we get $x=\pm 3$
Similarly from $\left(2\right)$ and $\left(4\right)$,$y=\mp 2$,
and from $\left(3\right)$ and $\left(4\right)$, $z=\pm 5$.
Hence the solutions are
$x=3$, $y=-2$, $z=5$
or $x=-3$, $y=2$, $z=-5$.