Answer
$(1,2),(1,-2),(-1,2),(-1,-2)$
Work Step by Step
$2x^{2}+3y^{2} =14$ Equation $(1)$
$-x^{2}+y^{2} =3$ Equation $(2)$
Multiply Equation $(2)$ by $2$ and add with Equation $(1)$
$2(-x^{2}+y^{2} )+2x^{2}+3y^{2}=2(3)+14$
$-2x^{2}+2y^{2} +2x^{2}+3y^{2}=6+14$
$5y^{2}=20$
$y^{2}=4$
$y= ±2$
$y=2$ or $y=-2$
Substitute $y$ values in Equation $(1)$
Let $y=2$
$2x^{2}+3y^{2} =14$
$2x^{2}+3(2)^{2} =14$
$2x^{2}+12 =14$
$2x^{2} =14-12$
$2x^{2} =2$
$x^{2} =1$
$x= ±1$
Let $y=-2$
$2x^{2}+3y^{2} =14$
$2x^{2}+3(-2)^{2} =14$
$2x^{2}+3(4) =14$
$2x^{2}+12 =14$
$2x^{2} =14-12$
$2x^{2} =2$
$x^{2} =1$
$x= ±1$
$(1,2),(1,-2),(-1,2),(-1,-2)$ satisfy both the equations. So the solution set is $\{(1,2),(1,-2),(-1,2),(-1,-2)\}$
