1. **State the problem:** We are given the simultaneous equations: $$x - 6y = 10$$ $$3y^2 = 4x + 7$$ We need to: a) Show that $$3y^2 - 24y - 47 = 0$$ b) Use part a) to solve the simultaneous equations. 2. **Part a) Show that $$3y^2 - 24y - 47 = 0$$:** Start with the first equation: $$x - 6y = 10 \implies x = 6y + 10$$ Substitute $$x = 6y + 10$$ into the second equation: $$3y^2 = 4x + 7 = 4(6y + 10) + 7 = 24y + 40 + 7 = 24y + 47$$ Rearranging: $$3y^2 - 24y - 47 = 0$$ This matches the required equation. 3. **Part b) Solve the simultaneous equations using part a):** We solve the quadratic equation: $$3y^2 - 24y - 47 = 0$$ Use the quadratic formula: $$y = \frac{24 \pm \sqrt{(-24)^2 - 4 \times 3 \times (-47)}}{2 \times 3} = \frac{24 \pm \sqrt{576 + 564}}{6} = \frac{24 \pm \sqrt{1140}}{6}$$ Calculate $$\sqrt{1140} \approx 33.8$$ So: $$y = \frac{24 \pm 33.8}{6}$$ Two solutions for $$y$$: 1. $$y = \frac{24 + 33.8}{6} = \frac{57.8}{6} \approx 9.6$$ 2. $$y = \frac{24 - 33.8}{6} = \frac{-9.8}{6} \approx -1.6$$ Find corresponding $$x$$ values using $$x = 6y + 10$$: 1. $$x = 6(9.6) + 10 = 57.6 + 10 = 67.6$$ 2. $$x = 6(-1.6) + 10 = -9.6 + 10 = 0.4$$ **Final solutions:** $$(x,y) \approx (67.6, 9.6) \text{ and } (0.4, -1.6)$$