Use the elimination method to find all solutions of the system $$\begin{array}{l} x^2+y^2 = 6, \\ x^2-y^2 = 3. \\ \end{array}$$ The four solutions of the system are:
the one with $x<0, y<0$ is
$x=$
$y=$
the one with $x<0, y>0$ is
$x=$
$y=$
the one with $x>0, y<0$ is
$x=$
$y=$
the one with $x>0, y>0$ is
$x=$
$y=$