Answer
{$(1,\dfrac{1}{2}),(1,-\dfrac{1}{2}),(-1,\dfrac{1}{2}),(-1,-\dfrac{1}{2})$}
Work Step by Step
After adding the given two equations, we get $\dfrac{11}{x^2}=11$ when we multiply the first equation by $2$.
or, $x^2=1$;
or $x=\pm 1$
From first equation $\dfrac{6}{x^2}+\dfrac{2}{y^2}=14$ when $x=1$, we have $y=\pm \dfrac{1}{2}$
From first equation $\dfrac{6}{x^2}+\dfrac{2}{y^2}=14$ when $x=-1$, we have $y=\pm \dfrac{1}{2}$
Thus, solution set is {$(1,\dfrac{1}{2}),(1,-\dfrac{1}{2}),(-1,\dfrac{1}{2}),(-1,-\dfrac{1}{2})$}
