## Intermediate Algebra (6th Edition)

$\{ (-6,0), (0,6) \}$
Substituting $y=x+6$ into the first equation results to \begin{array}{l} x^2+(x+6)^2=36 \\\\ x^2+x^2+12x+36=36 \\\\ 2x^2+12x=0 \\\\ x^2+6x=0 \\\\ x(x+6)=0 \\\\ x=\{-6,0\} .\end{array} Substituting $x=-6$ into the 2nd equation results to \begin{array}{l} y=-6+6\\\\ y=0 .\end{array} Substituting $x=0$ into the 2nd equation results to \begin{array}{l} y=0+6\\\\ y=6 .\end{array} Hence, the solution set is $\{ (-6,0), (0,6) \}$.