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