1. **Stating the problem:** Solve the system of linear equations for each set (a), (b), and (c). 2. **Method:** We will use the substitution or elimination method to find the values of $x$ and $y$. --- ### a) System: $$\begin{cases}-x + 3y = -5 \\ 2x - 5y = 8\end{cases}$$ 3. From the first equation, express $x$: $$-x + 3y = -5 \implies x = 3y + 5$$ 4. Substitute $x$ into the second equation: $$2(3y + 5) - 5y = 8$$ $$6y + 10 - 5y = 8$$ $$y + 10 = 8$$ $$y = 8 - 10 = -2$$ 5. Substitute $y = -2$ back into $x = 3y + 5$: $$x = 3(-2) + 5 = -6 + 5 = -1$$ **Solution for (a):** $x = -1$, $y = -2$ --- ### b) System: $$\begin{cases}x - 2y = -9 \\ \frac{1}{4}x - \frac{1}{2}y = 10\end{cases}$$ 6. Multiply the second equation by 4 to clear fractions: $$x - 2y = 40$$ 7. Now we have: $$\begin{cases}x - 2y = -9 \\ x - 2y = 40\end{cases}$$ 8. These two equations contradict each other (same left side, different right side), so there is **no solution**. **Solution for (b):** No solution (the system is inconsistent). --- ### c) System: $$\begin{cases}5x + 10y = 25 \\ 15x + 30y = 75\end{cases}$$ 9. Notice the second equation is exactly 3 times the first: $$15x + 30y = 3(5x + 10y) = 3 \times 25 = 75$$ 10. This means the two equations represent the same line, so there are infinitely many solutions. **Solution for (c):** Infinitely many solutions (dependent system). --- **Final answers:** - (a) $x = -1$, $y = -2$ - (b) No solution - (c) Infinitely many solutions