To determine the value of \( c \) that will yield the same solution for both systems of equations, we first need to analyze the solutions of the first system:
- First System of Equations: \[ 4x + y = -1 \quad \text{(1)} \] \[ x + y = 2 \quad \text{(2)} \]
From equation (2), we can express \( y \) in terms of \( x \): \[ y = 2 - x \]
Now, we substitute this expression for \( y \) into equation (1): \[ 4x + (2 - x) = -1 \] This simplifies to: \[ 4x + 2 - x = -1 \] \[ 3x + 2 = -1 \] \[ 3x = -3 \] \[ x = -1 \]
Now we substitute \( x = -1 \) back into equation (2) to find \( y \): \[ -1 + y = 2 \] \[ y = 3 \]
Thus, the solution for the first system is \( x = -1 \) and \( y = 3 \).
- Second System of Equations: \[ 8x + 2y = -2 \quad \text{(3)} \] \[ -4x + cy = -8 \quad \text{(4)} \]
Let's check if the first equation of this system has the same solution \( (x, y) = (-1, 3) \).
Substituting \( x = -1 \) and \( y = 3 \) into equation (3): \[ 8(-1) + 2(3) = -2 \] \[ -8 + 6 = -2 \] This is valid.
Now for equation (4): \[ -4(-1) + c(3) = -8 \] \[ 4 + 3c = -8 \] Solving for \( c \): \[ 3c = -8 - 4 \] \[ 3c = -12 \] \[ c = -4 \]
Thus, the value of \( c \) that would yield the same solution for both systems of equations is \(-4\).
The answer is -4.