1. **State the problem:** Solve the simultaneous equations: $$2x + 3y = 6$$ and $$(2x + 1)^2 + y(y - 2)^2 = 49$$ 2. **Use substitution:** From the first equation, express $x$ in terms of $y$: $$2x = 6 - 3y \implies x = \frac{6 - 3y}{2}$$ 3. **Substitute into the second equation:** Replace $x$ with $\frac{6 - 3y}{2}$: $$\left(2\cdot \frac{6 - 3y}{2} + 1\right)^2 + y(y - 2)^2 = 49$$ Simplify inside the first term: $$\left(6 - 3y + 1\right)^2 + y(y - 2)^2 = 49$$ $$\left(7 - 3y\right)^2 + y(y - 2)^2 = 49$$ 4. **Expand the squares:** $$ (7 - 3y)^2 = 49 - 42y + 9y^2 $$ $$ (y - 2)^2 = y^2 - 4y + 4 $$ So, $$ y(y - 2)^2 = y(y^2 - 4y + 4) = y^3 - 4y^2 + 4y $$ 5. **Rewrite the equation:** $$ 49 - 42y + 9y^2 + y^3 - 4y^2 + 4y = 49 $$ 6. **Simplify and bring all terms to one side:** $$ y^3 + (9y^2 - 4y^2) + (-42y + 4y) + 49 - 49 = 0 $$ $$ y^3 + 5y^2 - 38y = 0 $$ 7. **Factor the cubic:** $$ y(y^2 + 5y - 38) = 0 $$ So either $$ y = 0 $$ or solve quadratic $$ y^2 + 5y - 38 = 0 $$ 8. **Solve quadratic using formula:** $$ y = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 1 \cdot (-38)}}{2} = \frac{-5 \pm \sqrt{25 + 152}}{2} = \frac{-5 \pm \sqrt{177}}{2} $$ 9. **Find corresponding $x$ values:** - For $y=0$: $$ x = \frac{6 - 3\cdot 0}{2} = 3 $$ - For $y = \frac{-5 + \sqrt{177}}{2}$: $$ x = \frac{6 - 3y}{2} = \frac{6 - 3 \cdot \frac{-5 + \sqrt{177}}{2}}{2} = \frac{6 - \frac{-15 + 3\sqrt{177}}{2}}{2} = \frac{\frac{12 + 15 - 3\sqrt{177}}{2}}{2} = \frac{27 - 3\sqrt{177}}{4} $$ - For $y = \frac{-5 - \sqrt{177}}{2}$: $$ x = \frac{6 - 3y}{2} = \frac{27 + 3\sqrt{177}}{4} $$ **Final solutions:** $$ (x,y) = (3, 0), \left(\frac{27 - 3\sqrt{177}}{4}, \frac{-5 + \sqrt{177}}{2}\right), \left(\frac{27 + 3\sqrt{177}}{4}, \frac{-5 - \sqrt{177}}{2}\right) $$