1. **Stating the problem:** We are given the system of equations involving sequences $U_n$ and $v_n$: $$U_{n+1} + v = \alpha (U_n - 1)$$ $$v_n = U_n - 6$$ $$v_n = U_{n+1} + \alpha$$ $$U_{n+1} = U_n + \alpha - 6$$ We want to find the value of $\alpha$ that satisfies these equations. 2. **Using the given equations:** From the second and third equations, since both equal $v_n$, we set them equal: $$U_n - 6 = U_{n+1} + \alpha$$ 3. **Substitute $U_{n+1}$ from the fourth equation:** $$U_n - 6 = (U_n + \alpha - 6) + \alpha$$ 4. **Simplify the right side:** $$U_n - 6 = U_n + \alpha - 6 + \alpha$$ $$U_n - 6 = U_n + 2\alpha - 6$$ 5. **Cancel $U_n$ and $-6$ on both sides:** $$\cancel{U_n} - 6 = \cancel{U_n} + 2\alpha - 6$$ $$-6 = 2\alpha - 6$$ 6. **Add 6 to both sides:** $$-6 + 6 = 2\alpha - 6 + 6$$ $$0 = 2\alpha$$ 7. **Divide both sides by 2:** $$\frac{0}{\cancel{2}} = \frac{2\alpha}{\cancel{2}}$$ $$0 = \alpha$$ 8. **Check with the first equation:** Substitute $\alpha=0$ into the first equation: $$U_{n+1} + v = 0 \cdot (U_n - 1) = 0$$ $$U_{n+1} + v = 0$$ Since $v_n = U_n - 6$, and $v$ presumably equals $v_n$, then: $$U_{n+1} + (U_n - 6) = 0$$ Using $U_{n+1} = U_n + 0 - 6 = U_n - 6$, substitute: $$(U_n - 6) + (U_n - 6) = 0$$ $$2U_n - 12 = 0$$ $$2U_n = 12$$ $$U_n = 6$$ This is consistent. **Final answer:** $$\boxed{\alpha = 0}$$