1. **State the problem:** Solve the system of equations: $$\begin{cases} 2x - 4y = -8 \\ -4x + y = -12 \end{cases}$$ 2. **Use substitution or elimination method.** Here, we use substitution. From the second equation: $$-4x + y = -12 \implies y = 4x - 12$$ 3. **Substitute $y$ into the first equation:** $$2x - 4(4x - 12) = -8$$ 4. **Simplify:** $$2x - 16x + 48 = -8$$ $$-14x + 48 = -8$$ 5. **Isolate $x$:** $$-14x = -8 - 48$$ $$-14x = -56$$ 6. **Divide both sides by $-14$:** $$x = \frac{-56}{-14}$$ $$x = \cancel{\frac{-56}{-14}}4$$ 7. **Find $y$ using $y = 4x - 12$:** $$y = 4(4) - 12 = 16 - 12 = 4$$ **Answer:** $$x = 4, \quad y = 4$$ --- 1. **State the problem:** Solve the system: $$\begin{cases} 4x + 2y = 14 \\ 3x - 5y = 4 \end{cases}$$ 2. **Use elimination method.** Multiply the first equation by 5 and the second by 2 to align $y$ coefficients: $$\begin{cases} 20x + 10y = 70 \\ 6x - 10y = 8 \end{cases}$$ 3. **Add the two equations:** $$20x + 10y + 6x - 10y = 70 + 8$$ $$26x = 78$$ 4. **Solve for $x$:** $$x = \frac{78}{26}$$ $$x = \cancel{\frac{78}{26}}3$$ 5. **Substitute $x=3$ into the first original equation:** $$4(3) + 2y = 14$$ $$12 + 2y = 14$$ 6. **Isolate $y$:** $$2y = 14 - 12 = 2$$ $$y = \frac{2}{2} = 1$$ **Answer:** $$x = 3, \quad y = 1$$