1. **State the problem:** We are given a pair of simultaneous equations represented by two graphs: a green line and a purple downward-facing parabola. We want to find their solutions, which are the points where the graphs intersect. 2. **Identify the green line:** The green line crosses the y-axis at approximately $y=-2$ and the x-axis near $x=2$. A linear equation crossing the y-axis at $-2$ can be written as $y = mx - 2$. Since it crosses the x-axis at $x=2$, at this point $y=0$: $$0 = m(2) - 2 \implies m = 1.$$ Therefore, the green line is $$y = x - 2.$$ 3. **Identify the purple parabola:** It is downward facing and peaks around $(2,1)$, descending towards the x-axis near $(5,0)$. We can assume the parabola is of the form $$y = -a(x - h)^2 + k,$$ where $(h,k)$ is the vertex. We know $h=2$ and $k=1$, so $$y = -a(x-2)^2 + 1.$$ To find $a$, use the point $(5,0)$ on the parabola: $$0 = -a(5-2)^2 +1 \implies 0 = -a(3)^2 + 1 \implies a = \frac{1}{9}.$$ Thus, the parabola is $$y = -\frac{1}{9}(x-2)^2 + 1.$$ 4. **Find intersection points by solving simultaneously:** Set the two equations equal: $$x - 2 = -\frac{1}{9}(x-2)^2 +1.$$ Bring all terms to one side: $$x - 2 - 1 = -\frac{1}{9}(x-2)^2$$ $$x - 3 = -\frac{1}{9}(x-2)^2.$$ Multiply both sides by 9: $$9(x - 3) = -(x-2)^2$$ $$9x - 27 = -(x-2)^2.$$ Rewrite: $$(x-2)^2 + 9x - 27 = 0.$$ Expand $(x-2)^2 = x^2 -4x +4$: $$x^2 - 4x + 4 + 9x - 27 = 0,$$ Simplify: $$x^2 + 5x - 23 = 0.$$ 5. **Solve the quadratic for $x$: $$x = \frac{-5 \pm \sqrt{5^2 - 4(1)(-23)}}{2} = \frac{-5 \pm \sqrt{25 + 92}}{2} = \frac{-5 \pm \sqrt{117}}{2}.$$ Approximate the roots: $$\sqrt{117} \approx 10.82.$$ So $$x_1 = \frac{-5 + 10.82}{2} = \frac{5.82}{2} = 2.91,$$ $$x_2 = \frac{-5 - 10.82}{2} = \frac{-15.82}{2} = -7.91.$$ 6. **Find corresponding $y$ values using $y = x - 2$: $$y_1 = 2.91 - 2 = 0.91,$$ $$y_2 = -7.91 - 2 = -9.91.$$ 7. **Final solutions:** The solutions to the simultaneous equations (points of intersection) are approximately $$\boxed{(2.91, 0.91)} \text{ and } \boxed{(-7.91, -9.91)}.$$