## Precalculus (6th Edition) Blitzer

$(3,0)$, $(\dfrac{-9}{5}, \dfrac{12}{5})$
Re-arrange the given equations to get $x=-2y+3$ and $(-2y+3)^2 +y^2=9$ $\implies 4y^2-12y+9+y^2=9$ $\implies y(5y-12) =0$ when $y=\dfrac{12}{5}$ then we have $x=\dfrac{-9}{5}$ when $y=0$ then we have $x=-2(0)+3=3$ Hence, our answers are: $(3,0)$, $(\dfrac{-9}{5}, \dfrac{12}{5})$