Question

Solve $y^2+yz+z^2=49.....\left(1\right)$,
$z^2+zx+x^2=19....\left(2\right)$,
$x^2+xy+y^2=39....\left(3\right)$.

Answer 1

$x^2+yz+z^2=49\dots \dots \left(1\right)$

$z^2+zx+x^2=19\dots \dots \left(2\right)$

$x^2+xy+y^2=39\dots \dots \left(3\right)$

Subtracting (2) from (1), we get

$y^2-x^2+z(y-x)=30⟹(y-x)(x+y+z)=30\dots \dots \left(4\right)$

Similarly from (1) and (3), we get

$(z-x)(x+y+z)=10\dots \dots \left(5\right)$

Dividing (4) and (5), we get

$\frac{y-x}{z-x}=3⟹y=3z-2x$

Substituting the in equation (3), we get

$x^2-3xz+3z^2=13\dots \dots \left(6\right)$

We have equation (2) and (6) as homogenous in $x$ and $z$, hence substituting $z=mx$ in them, we get

$x^2(m^2+m+1)=19\text{and}{x}^2(3m^2-3m+1)=13$

$\therefore \frac{m^2+m+1}{3m^2-3m+1}=\frac{19}{13}$

Solving the above, we get $m=11$ and $m=\frac{2}{3}$

Substituting the value of $m$ in the above equations and solving for $x,y,z$ we get

$x=\pm 2,y=\pm 5,z=\pm 3\text{and}x=\pm \frac{11}{\sqrt{7}},y=\pm \frac{19}{\sqrt{7}},z=\pm \frac{1}{\sqrt{7}}$