Solve for the solutions

−4x − 2y = −12
4x + 8y = −24

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!

Learn how to solve the system of linear equations -4x - 2y = -12 and 4x + 8y = -24 with a detailed step-by-step solution. This problem covers key algebraic concepts, including equation simplification and the substitution method, suitable for students in Grades 8-10.

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