The system of equations in the image is:

1. $8x - y = -4$
2. $4x + 7y = -32$

Both substitution and elimination methods would work here, but elimination may be more efficient given the structure of the equations. Let's use elimination to solve this.

Elimination Method:

Step 1: Multiply both sides of the first equation by 7 to eliminate $y$ when we add the two equations together.

$7(8x - y) = 7(-4)$ $56x - 7y = -28$

So now, we have the following system:

1. $56x - 7y = -28$
2. $4x + 7y = -32$

Step 2: Add the two equations together to eliminate $y$.

$(56x - 7y) + (4x + 7y) = (-28) + (-32)$ $60x = -60$

Step 3: Solve for $x$.

$x = \frac{-60}{60} = -1$

Step 4: Substitute $x = -1$ into the first equation to solve for $y$.

$8(-1) - y = -4$ $-8 - y = -4$ $-y = -4 + 8$ $-y = 4 \quad \Rightarrow \quad y = -4$

Solution:

The solution to the system of equations is: $x = -1, \quad y = -4$

Would you like further clarification on any step or more details?

Here are some related questions:

1. What is the substitution method and how does it differ from elimination?
2. Can the substitution method also solve this system of equations?
3. How do you check if a solution is correct for a system of equations?
4. Why might elimination be more efficient in some cases than substitution?
5. Can systems of equations sometimes have no solutions or infinitely many solutions?

Tip: When using elimination, it's useful to multiply the equations in a way that cancels out one of the variables immediately.