1. **State the problem:** Solve the system of equations: $$3x - 6y = 15$$ $$-2x + 4y = -10$$ 2. **Formula and rules:** We can solve this system using substitution or elimination. Here, elimination is convenient. 3. **Elimination method:** Multiply the first equation by 2 and the second by 3 to align coefficients of $x$: $$2(3x - 6y) = 2(15) \Rightarrow 6x - 12y = 30$$ $$3(-2x + 4y) = 3(-10) \Rightarrow -6x + 12y = -30$$ 4. **Add the two equations:** $$ (6x - 12y) + (-6x + 12y) = 30 + (-30)$$ $$0 = 0$$ 5. **Interpretation:** The result $0=0$ means the two equations are dependent (one is a multiple of the other), so there are infinitely many solutions. 6. **Express $x$ in terms of $y$:** From the first equation: $$3x - 6y = 15 \Rightarrow 3x = 15 + 6y \Rightarrow x = 5 + 2y$$ 7. **Final answer:** The system has infinitely many solutions given by: $$x = 5 + 2y, \quad y \in \mathbb{R}$$