We are given the system of linear equations:

1. $-4x - 2y = -12$
2. $4x + 8y = -24$

Step 1: Simplify the equations if possible

We can divide both sides of the first equation by $-2$ to simplify it: -4x - 2y = -12 \quad \Rightarrow \quad 2x + y = 6 \tag{1}

The second equation can also be simplified by dividing both sides by $4$: 4x + 8y = -24 \quad \Rightarrow \quad x + 2y = -6 \tag{2}

Now, we have the simplified system:

1. $2x + y = 6$
2. $x + 2y = -6$

Step 2: Solve using substitution or elimination

Let's use the substitution method. From equation (2), solve for $x$: x = -6 - 2y \tag{3}

Step 3: Substitute into equation (1)

Substitute $x = -6 - 2y$ into equation (1): $2(-6 - 2y) + y = 6$ $-12 - 4y + y = 6$ $-12 - 3y = 6$ $-3y = 18$ $y = -6$

Step 4: Solve for $x$

Now, substitute $y = -6$ into equation (3): $x = -6 - 2(-6) = -6 + 12 = 6$

Solution:

The solution to the system is: $x = 6, \quad y = -6$

Would you like more details, or have any questions?

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Here are 5 related questions for further practice:

1. Solve the system: $3x - 2y = 7$ and $5x + y = 13$.
2. Solve using the elimination method: $x + 3y = 12$ and $2x - y = 3$.
3. Find the intersection point of the lines $y = 2x - 5$ and $y = -x + 4$.
4. Solve the system of equations: $4x + 5y = 1$ and $2x - 3y = -7$.
5. Solve for $x$ and $y$: $x - 4y = 10$ and $2x + y = 3$.

Tip: Always check your solutions by substituting the values of $x$ and $y$ back into the original equations!