Question

Solve the following equations :
$x^2+y^2+xy=9,z^2+x^2+xz=4,y^2+z^2+yx=1.$

Answer 1

$x^2+y^2+xy=9$
$z^2+x^2+xz=4$
$y^2+z^2+yz=1$
Subtracting $\left(2\right)$ from $\left(1\right)$ and factorizing, we get
$(y-z)(x+y+z)=5$
Similarly, $(x-y)(x+y+z)=3$
$(z-x)(x+y+z)=-8$
Again subtracting $\left(5\right)$ from $\left(4\right)$, we get
$(2y-x-z)(x+y+z)=2$
Similarly, $(2x-y-2)(x+y+z)=11.$
Now putting $x+y+z=t,$ we get from $\left(7\right)$ and $\left(8\right)$,$(3y-t)t=2$ or $y=\frac{t^2+2}{3t}$ and $(3x-t)t=11$ or $x=\frac{t^2+11}{3t}.$
Substituting these values of $x$ and $y$ in $\left(1\right)$,$\frac{(t^2+{11}^2)}{{9t}^2}+\frac{(t^2+2^2)}{{9t}^2}+\frac{(t^2+11)(t^2+2)}{{9t}^2}=9$ or $3t^4-42t^2+147=0$ or$t^4-14t^2+49=0$ or ${(t^2-7)}^2=0$.
$\therefore {(x+y+z)}^2=t^2=7$ or $x+y+z=t=\pm \sqrt{7}.$
Hence $x=\pm \frac{7+11}{3\surd \left(7\right)}=\pm \frac{6}{\surd \left(7\right)},$ and $y=\pm \frac{7+2}{3\surd \left(7\right)}=\pm \frac{3}{\surd \left(7\right)}$.
Then from $\left(4\right),$ we get
$(\frac{3}{\surd \left(7\right)}-z)\sqrt{7}=5$ or $(-\frac{3}{\surd \left(7\right)}-z)(-\sqrt{7})=5.$
$\therefore z=-2\frac{2}{\surd \left(7\right)}$ or $z=2\frac{2}{\surd \left(7\right)}.$
Hence the solution sets are:
$x=\frac{6}{\surd \left(7\right)},y=\frac{3}{\surd \left(7\right)},z=\frac{2}{\surd \left(7\right)}$ or $x=-\frac{6}{\surd \left(7\right)},y=-\frac{3}{\surd \left(7\right)},z=-\frac{2}{\surd \left(7\right)}.$