Two systems of equations are shown:

To solve for the value of \( c \) such that both systems of equations have the same solution, we first need to determine the solution of the first system of equations:

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

We can solve for \( y \) in terms of \( x \) by re-arranging Equation 2:

\[
y = 2 - x
\]

Now substitute \( y = 2 - x \) into Equation 1:

\[
4x + (2 - x) = -1
\]

Combine like terms:

\[
4x - x + 2 = -1
\]
\[
3x + 2 = -1
\]
\[
3x = -1 - 2
\]
\[
3x = -3
\]
\[
x = -1
\]

Now substitute \( x = -1 \) back into \( y = 2 - x \):

\[
y = 2 - (-1) = 3
\]

Thus, the solution to the first system of equations is \( (x, y) = (-1, 3) \).

Next, we need to analyze the second system of equations:

1. \( 8x + 2y = -2 \) (Equation 3)
2. \( -4x + cy = -8 \) (Equation 4)

We can simplify Equation 3:

\[
8x + 2y = -2
\]
Dividing through by 2:
\[
4x + y = -1 \quad \text{(which is the same as Equation 1)}
\]

Now we know that Equation 3 is consistent with the first system of equations. Therefore, to find the value of \( c \), we need to set up Equation 4 so that it has the same solution \( (x, y) = (-1, 3) \).

Substituting \( x = -1 \) and \( y = 3 \) into Equation 4:

\[
-4(-1) + c(3) = -8
\]
\[
4 + 3c = -8
\]
\[
3c = -8 - 4
\]
\[
3c = -12
\]
\[
c = -4
\]

Thus, the value of \( c \) needed to make both systems have the same solution is \( \boxed{-4} \).