Two systems of equations are shown:

To determine the value of \( c \) that makes the second system of equations equivalent to the first, we need to first solve the first system and then ensure that the second system produces the same solution.

The first system of equations is:

1. \( 4x + y = -1 \)
2. \( x + y = 2 \)

We can solve this system using substitution or elimination. Let's use substitution:

From the second equation \( x + y = 2 \):
\[
y = 2 - x
\]

Now substitute \( y \) into the first equation:
\[
4x + (2 - x) = -1
\]
\[
4x + 2 - x = -1
\]
\[
3x + 2 = -1
\]
\[
3x = -1 - 2
\]
\[
3x = -3
\]
\[
x = -1
\]

Now substitute \( x = -1 \) back into \( y = 2 - x \):
\[
y = 2 - (-1) = 2 + 1 = 3
\]

So the solution to the first system is:
\[
(x, y) = (-1, 3)
\]

Now, we need to solve the second system:

1. \( 8x + 2y = -2 \)
2. \( -4x + cy = -8 \)

We can simplify the first equation:
\[
8x + 2y = -2 \implies 4x + y = -1 \quad \text{(Divide by 2)}
\]

This is the same as the first equation in the first system.

Next, we can substitute \( x = -1 \) from our solution into the modified second equation:
\[
-4(-1) + cy = -8
\]
\[
4 + cy = -8
\]
Now substitute \( y = 3 \) which we found earlier:
\[
4 + c(3) = -8
\]
\[
4 + 3c = -8
\]
Now, isolate \( c \):
\[
3c = -8 - 4
\]
\[
3c = -12
\]
\[
c = -4
\]

Thus, the value of \( c \) that we need for the two systems to have the same solution is:
\[
\boxed{-4}
\]