Chapter 9 Quadratic Relations and Conic Sections - 9.7 Solve Quadratic Systems - Guided Practice for Examples 3 and 4 - Page 661: 8

$(3,0),(5,4),(5,-4)$

Adding the first and the second equation we get: $2x^2-16x+30=0\\x^2-8x+15=0\\(x-3)(x-5)=0$ Thus $x=3$ or $x=5$. When $x=3$ then $y=\pm\sqrt{3^2-9}=\pm0=0$ When $x=5$ then $y=\pm\sqrt{5^2-9}=\pm\sqrt{16}=\pm4$ Thus the solutions are: $(3,0),(5,4),(5,-4)$