Answer
$(\sqrt 3,0),(-\sqrt 3,0)$
Work Step by Step
$3x^{2}+y^{2} =9$ Equation $(1)$
$3x^{2}-y^{2} =9$ Equation $(2)$
Adding Equation $(1)$ and Equation $(2)$
$3x^{2}+y^{2}+ 3x^{2}-y^{2} =9+9$
$6x^{2}=18 $
$x^{2}=3 $
$x=±\sqrt 3$
$x=\sqrt 3$ or $x=-\sqrt 3$
Substitute $x$ values in Equation $(1)$
Let $x=\sqrt 3$
$3x^{2}+y^{2} =9$
$3(\sqrt 3)^{2}+y^{2} =9$
$3(3)+y^{2} =9$
$9+y^{2} =9$
$y^{2} =9-9$
$y^{2} =0$
$y=0$
Let $x=-\sqrt 3$
$3x^{2}+y^{2} =9$
$3(-\sqrt 3)^{2}+y^{2} =9$
$3(3)+y^{2} =9$
$9+y^{2} =9$
$y^{2} =9-9$
$y^{2} =0$
$y=0$
$(\sqrt 3,0),(-\sqrt 3,0)$ satisfy both equations.
Solution set is $(\sqrt 3,0),(-\sqrt 3,0)$
