1. **Problem statement:** We need to determine how many solution pairs each system in problem 9 has by solving the systems algebraically. 2. **General approach:** For each system, solve the two equations simultaneously. If the system has one unique solution, it has exactly one solution pair. If the system is dependent (infinite solutions), it has infinitely many solution pairs. If inconsistent, it has no solution. --- ### 9a) System: I: $0.5s - 3t = 1$ II: $6t = s - 2$ 3. From II, express $s$: $$s = 6t + 2$$ 4. Substitute into I: $$0.5(6t + 2) - 3t = 1$$ $$3t + 1 - 3t = 1$$ $$1 = 1$$ 5. This is always true, so the system is dependent. 6. **Conclusion:** Infinitely many solutions. --- ### 9b) System: I: $m = v + 1$ II: $2m - 2v = 0.5$ 7. Substitute $m$ from I into II: $$2(v + 1) - 2v = 0.5$$ $$2v + 2 - 2v = 0.5$$ $$2 = 0.5$$ 8. This is false, so the system is inconsistent. 9. **Conclusion:** No solution. --- ### 9c) System: I: $2y = 3x + 1$ II: $2x = 3y + 1$ 10. From I: $$y = \frac{3x + 1}{2}$$ 11. Substitute into II: $$2x = 3\left(\frac{3x + 1}{2}\right) + 1$$ $$2x = \frac{9x + 3}{2} + 1$$ $$2x = \frac{9x + 3 + 2}{2}$$ $$2x = \frac{9x + 5}{2}$$ 12. Multiply both sides by 2: $$4x = 9x + 5$$ 13. Rearrange: $$4x - 9x = 5$$ $$-5x = 5$$ $$x = -1$$ 14. Substitute back to find $y$: $$y = \frac{3(-1) + 1}{2} = \frac{-3 + 1}{2} = \frac{-2}{2} = -1$$ 15. **Conclusion:** One unique solution $(x,y) = (-1,-1)$. --- ### 9d) System: I: $2x = 3 - y$ II: $\frac{2}{3}x + \frac{1}{3}y = 1$ 16. From I: $$y = 3 - 2x$$ 17. Substitute into II: $$\frac{2}{3}x + \frac{1}{3}(3 - 2x) = 1$$ $$\frac{2}{3}x + 1 - \frac{2}{3}x = 1$$ $$1 = 1$$ 18. This is always true, so the system is dependent. 19. **Conclusion:** Infinitely many solutions. --- **Final answers:** - 9a) Infinitely many solutions - 9b) No solution - 9c) One unique solution $(x,y) = (-1,-1)$ - 9d) Infinitely many solutions