1. **State the problem:** We are given the simultaneous equations: $$\frac{3x}{2y + 5} + 2y = 1$$ and $$4y + x = -6$$ We need to: a) Show that $$4y^2 - 4y - 23 = 0$$ b) Use this quadratic to solve the simultaneous equations. 2. **Show that $$4y^2 - 4y - 23 = 0$$:** From the second equation, express $$x$$ in terms of $$y$$: $$x = -6 - 4y$$ Substitute this into the first equation: $$\frac{3(-6 - 4y)}{2y + 5} + 2y = 1$$ Simplify numerator: $$\frac{-18 - 12y}{2y + 5} + 2y = 1$$ Multiply both sides by $$2y + 5$$ to clear the denominator: $$-18 - 12y + 2y(2y + 5) = 1(2y + 5)$$ Expand terms: $$-18 - 12y + 4y^2 + 10y = 2y + 5$$ Combine like terms on the left: $$4y^2 - 2y - 18 = 2y + 5$$ Bring all terms to one side: $$4y^2 - 2y - 18 - 2y - 5 = 0$$ Simplify: $$4y^2 - 4y - 23 = 0$$ This matches the required quadratic. 3. **Solve the quadratic equation:** Use the quadratic formula: $$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $$a=4$$, $$b=-4$$, $$c=-23$$. Calculate discriminant: $$\Delta = (-4)^2 - 4 \times 4 \times (-23) = 16 + 368 = 384$$ Calculate roots: $$y = \frac{4 \pm \sqrt{384}}{8} = \frac{4 \pm 8\sqrt{6}}{8} = \frac{4}{8} \pm \frac{8\sqrt{6}}{8} = 0.5 \pm \sqrt{6}$$ Approximate values: $$y_1 = 0.5 + 2.45 = 2.95$$ $$y_2 = 0.5 - 2.45 = -1.95$$ 4. **Find corresponding $$x$$ values:** Recall $$x = -6 - 4y$$. For $$y_1 = 2.95$$: $$x_1 = -6 - 4(2.95) = -6 - 11.8 = -17.8$$ For $$y_2 = -1.95$$: $$x_2 = -6 - 4(-1.95) = -6 + 7.8 = 1.8$$ **Final solutions:** $$(x, y) = (-17.8, 2.95) \text{ and } (1.8, -1.95)$$ These satisfy the simultaneous equations.