To find the value of \( c \) such that both systems of equations have the same solution, we first need to determine the solution to the first system of equations.
The first system is:
- \( 4x + y = -1 \)
- \( x + y = 2 \)
We can solve for \( y \) in terms of \( x \) using the second equation: \[ y = 2 - x. \]
Next, we substitute this expression for \( y \) into the first equation: \[ 4x + (2 - x) = -1 \] \[ 4x + 2 - x = -1 \] Combine like terms: \[ 3x + 2 = -1 \] Subtract 2 from both sides: \[ 3x = -3 \] Divide by 3: \[ x = -1. \]
Now we can substitute \( x = -1 \) back into the equation for \( y \): \[ y = 2 - (-1) = 2 + 1 = 3. \]
So the solution to the first system is \( (x, y) = (-1, 3) \).
Now we will use the second system of equations to find the value of \( c \) that will yield the same solution \( (x, y) = (-1, 3) \).
The second system is:
- \( 8x + 2y = -2 \)
- \( -4x + cy = -8 \)
First, we substitute \( x = -1 \) and \( y = 3 \) into the first equation of the second system: \[ 8(-1) + 2(3) = -2 \] \[ -8 + 6 = -2 \] This is true, so the first equation in the second system has the same solution.
Next, we need to substitute \( x = -1 \) and \( y = 3 \) into the second equation to find \( c \): \[ -4(-1) + c(3) = -8 \] \[ 4 + 3c = -8 \] Subtract 4 from both sides: \[ 3c = -12 \] Divide by 3: \[ c = -4. \]
Thus, the value of \( c \) needed to get the same solution for both systems is \( -4 \).
So the answer is: \[-4\]