1. **Problem 15(a):** Solve the system $$\begin{cases} 2x - 3y = 1 \\ 6x - 9y = 3 \end{cases}$$ 2. Notice the second equation is exactly 3 times the first, so both represent the same line. This means infinitely many solutions lie on this line. 3. To find parametric form, solve the first equation for $x$: $$2x - 3y = 1 \implies 2x = 1 + 3y \implies x = \frac{1}{2} + \frac{3}{2}y$$ 4. Let parameter $t = y$. Then the solution set is: $$x = \frac{1}{2} + \frac{3}{2}t, \quad y = t$$ 5. **Answer 15(a):** $$\boxed{(x,y) = \left(\frac{1}{2} + \frac{3}{2}t, t\right), \quad t \in \mathbb{R}}$$ --- 6. **Problem 15(b):** Solve the system $$\begin{cases} x_1 + 3x_2 - x_3 = -4 \\ 3x_1 + 9x_2 - 3x_3 = -12 \\ -x_1 - 3x_2 + x_3 = 4 \end{cases}$$ 7. The second equation is 3 times the first, and the third is the negative of the first, so all three are dependent. 8. From the first equation: $$x_1 = -4 - 3x_2 + x_3$$ 9. Let parameters $s = x_2$ and $t = x_3$. Then: $$x_1 = -4 - 3s + t, \quad x_2 = s, \quad x_3 = t$$ 10. **Answer 15(b):** $$\boxed{(x_1,x_2,x_3) = (-4 - 3s + t, s, t), \quad s,t \in \mathbb{R}}$$ --- 11. **Problem 16(a):** Solve the system $$\begin{cases} 6x_1 + 2x_2 = -8 \\ 3x_1 + x_2 = -4 \end{cases}$$ 12. The second equation is exactly half the first, so infinitely many solutions lie on this line. 13. Solve the second equation for $x_2$: $$x_2 = -4 - 3x_1$$ 14. Let parameter $t = x_1$. Then: $$x_1 = t, \quad x_2 = -4 - 3t$$ 15. **Answer 16(a):** $$\boxed{(x_1,x_2) = (t, -4 - 3t), \quad t \in \mathbb{R}}$$ --- 16. **Problem 16(b):** Solve the system $$\begin{cases} 2x - y + 2z = -4 \\ 6x - 3y + 6z = -12 \\ -4x + 2y - 4z = 8 \end{cases}$$ 17. The second equation is 3 times the first, and the third is -2 times the first, so all are dependent. 18. From the first equation: $$2x - y + 2z = -4 \implies y = 2x + 2z + 4$$ 19. Let parameters $s = x$ and $t = z$. Then: $$x = s, \quad y = 2s + 2t + 4, \quad z = t$$ 20. **Answer 16(b):** $$\boxed{(x,y,z) = (s, 2s + 2t + 4, t), \quad s,t \in \mathbb{R}}$$