Question 878504
A good method is solve one equation either for {{{x^2}}} or for {{{y^2}}}, and substitute into the other equation, and solve for this single variable.  Return to the first variable and solve.


{{{y^2=9-x^2}}}.
Now substitute ...
{{{x^2/9-(9-x^2)/9=1}}}
{{{x^2-(9-x^2)=9}}}
{{{2x^2-9=9}}}
{{{2x^2=9+9=18}}}
{{{x^2=9}}}
{{{x=-3}}} OR {{{x=3}}}.


Resuming with {{{y^2=9-x^2}}}, either x solution will give the same y.
{{{y^2=9-9}}}
{{{y^2=0}}}
{{{y=0}}}


<b>SOLUTIONS:  (-3,0) and (3,0).</b>