Question 242912
{{{(x^-2)+(y^-2)=36}=36}}} (1)
{{{(3x^-2)-(3y^-2)=-36}}} (2)

Multiply equation (1) by 3 to give {{{(3x^-2)+(3y^-2)=108}}} (3)

Take (2) away from (3) to give {{{(6y^-2)=144}} so {{{y = sqrt(1/24)}}}

{{{y=0.2041}}} Or {{{y=-0.2041}}}

So plug y back into our equation (1) gives x to be {{{x = (1/12)^(1/2)}}}

so {{{x = 0.2887}}} or {{{x = -0.2887}}}