Answer
$\{\}$ or No solution.
Work Step by Step
$x^{2}+3y^{2} = 6$ Equation $(1)$
$x^{2}-3y^{2} = 10$ Equation $(2)$
Adding Equation $(1)$ and Equation $(2)$
$x^{2}+3y^{2}+x^{2}-3y^{2} = 6+10$
$2x^{2}=16 $
$x^{2}=8 $
$x=±\sqrt 8 $
$x=±\sqrt (4 \times 2) $
$x=±2\sqrt 2 $
$x=2\sqrt 2 $ or $x=-2\sqrt 2 $
Substitute $x$ values in Equation $(1)$
Let $x=2\sqrt 2 $
$x^{2}+3y^{2} = 6$
$(2\sqrt 2 )^{2}+3y^{2} = 6$
$8+3y^{2} = 6$
$3y^{2} = 6-8$
$3y^{2} = -2$
$y^{2} = -\frac{2}{3}$
$y=\sqrt ( -\frac{2}{3}) $
$y=\sqrt ( -\frac{2}{3}) $ is not a real number. So, solution set is empty.
