Chapter 9 Quadratic Relations and Conic Sections - 9.7 Solve Quadratic Systems - 9.7 Exercises - Skill Practice - Page 662: 28

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

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 $x=5$ Plugging this into the second equation we get: if $x=3$: $9-y^2-9=0$, thus $y=0$. If $x=5$: $25-y^2-9=0$ thus $y=\pm4$. Thus the solutions are: $(3,0),(5,4),(5,-4)$