Question

Solve the following equation :
$3x^2-2y^2+5z^2=0,7x^2-3y^2-15z^2=0,5x-4y+7z=6.$

Answer 1

$3x^2-2y^2+5z^2=0{7x}^2-3y^2-15z^2=0$

Applying cross multiplication

$\frac{x^2}{30+15}=\frac{{-y}^2}{-45-35}=\frac{z^2}{-9+14}=k\frac{x^2}{9}=\frac{y^2}{16}=\frac{z^2}{1}=k\Rightarrow x^2=9k,y^2=16k,x^2=k\Rightarrow x=\pm 3\sqrt{k},y=\pm 4\sqrt{k},z=\pm \sqrt{k}5x-4y+7z=6$

substituting $x,y$ and $z$

$\Rightarrow \pm 6\sqrt{k}=6\Rightarrow k=1\Rightarrow x=\pm 3,y=\pm 4,z=\pm 1$