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The system of equations you provided is:

1. \( 8x + 5y = -70 \)
2. \( 4x - 3y = -2 \)

To solve this system using the addition (elimination) method, we need to eliminate one variable by adding or subtracting the equations.

### Step 1: Make the coefficients of \(x\) or \(y\) the same.

We can make the coefficients of \(x\) the same by multiplying the second equation by 2:
\[
2(4x - 3y) = 2(-2)
\]
\[
8x - 6y = -4
\]

Now, we have:
\[
8x + 5y = -70 \quad \text{(Equation 1)}
\]
\[
8x - 6y = -4 \quad \text{(Equation 2, modified)}
\]

### Step 2: Subtract the second equation from the first to eliminate \(x\):
\[
(8x + 5y) - (8x - 6y) = -70 - (-4)
\]
\[
8x + 5y - 8x + 6y = -70 + 4
\]
\[
11y = -66
\]

### Step 3: Solve for \(y\):
\[
y = \frac{-66}{11} = -6
\]

### Step 4: Substitute \(y = -6\) into one of the original equations (e.g., \( 4x - 3y = -2 \)):
\[
4x - 3(-6) = -2
\]
\[
4x + 18 = -2
\]
\[
4x = -2 - 18
\]
\[
4x = -20
\]
\[
x = \frac{-20}{4} = -5
\]

### Final Solution:
\[
x = -5, \quad y = -6
\]

Let me know if you'd like any further details or explanations!

Here are 5 related questions:
1. What if we chose to eliminate \(y\) first instead of \(x\)?
2. Can you verify the solution by substituting the values back into both original equations?
3. How would this method differ from solving using substitution?
4. What happens if both equations have no solution or infinite solutions?
5. How can you use matrix methods to solve systems of equations like this?

**Tip:** Always double-check your solutions by plugging the values back into the original equations to ensure they satisfy both.

Solve the system of equations using the addition method: 8x + 5y = -70 and 4x - 3y = -2.

Learn how to solve the system of equations 8x + 5y = -70 and 4x - 3y = -2 using the addition method. This detailed step-by-step solution involves eliminating variables, solving for y, and substituting to find x. Ideal for students in Grades 8-10.

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