1. **State the problem:** We are given two equations: $$y = x^2 + 6x - 18$$ and $$y = 2x + 3$$ We need to find which of the points $(-5, -23)$, $(1, -11)$, $(2, 7)$, and $(3, 9)$ satisfy both equations simultaneously, i.e., are solutions to the system. 2. **Method:** To find the solution(s), substitute the $y$ from the linear equation into the quadratic equation and solve for $x$: $$2x + 3 = x^2 + 6x - 18$$ 3. **Rearrange the equation:** $$0 = x^2 + 6x - 18 - 2x - 3$$ Simplify: $$0 = x^2 + (6x - 2x) + (-18 - 3)$$ $$0 = x^2 + 4x - 21$$ 4. **Solve the quadratic equation:** $$x^2 + 4x - 21 = 0$$ Use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=1$, $b=4$, $c=-21$. Calculate the discriminant: $$\Delta = 4^2 - 4 \times 1 \times (-21) = 16 + 84 = 100$$ So, $$x = \frac{-4 \pm \sqrt{100}}{2} = \frac{-4 \pm 10}{2}$$ 5. **Find the two roots:** $$x_1 = \frac{-4 + 10}{2} = \frac{6}{2} = 3$$ $$x_2 = \frac{-4 - 10}{2} = \frac{-14}{2} = -7$$ 6. **Find corresponding $y$ values using $y = 2x + 3$:** For $x=3$: $$y = 2(3) + 3 = 6 + 3 = 9$$ For $x=-7$: $$y = 2(-7) + 3 = -14 + 3 = -11$$ 7. **Check which points match these solutions:** - $(3, 9)$ matches the first solution. - $(-7, -11)$ is not among the given points. - Check if $(1, -11)$ is a solution: For $x=1$, $y$ from linear equation is $2(1)+3=5$, but given $y=-11$, so no. - Check $(-5, -23)$: For $x=-5$, linear $y=2(-5)+3=-10+3=-7$, given $y=-23$, no. - Check $(2, 7)$: For $x=2$, linear $y=2(2)+3=4+3=7$, matches $y=7$. Check if $(2,7)$ satisfies quadratic: $$7 = 2^2 + 6(2) - 18 = 4 + 12 - 18 = -2$$ No, so $(2,7)$ is not a solution to the system. 8. **Final answer:** The only point from the list that satisfies both equations is **$(3, 9)$**.