To solve for the value of \( c \) that will make both systems of equations have the same solution, we first need to find the solution to the first system of equations.
The first system is:
- \( 4x + y = -1 \)
- \( x + y = 2 \)
We can solve for \( y \) in the second equation: \[ y = 2 - x \]
Now, substitute this expression for \( y \) into the first equation: \[ 4x + (2 - x) = -1 \] Simplifying this, we get: \[ 4x + 2 - x = -1 \] \[ 3x + 2 = -1 \] \[ 3x = -1 - 2 \] \[ 3x = -3 \] \[ x = -1 \]
Now substitute \( x = -1 \) back into the equation for \( y \): \[ y = 2 - (-1) = 2 + 1 = 3 \]
So the solution to the first system of equations is \( (x, y) = (-1, 3) \).
Now we analyze the second system of equations:
- \( 8x + 2y = -2 \)
- \( -4x + cy = -8 \)
We can simplify the first equation by dividing all terms by 2: \[ 4x + y = -1 \] This is the same as the first equation in the first system, confirming that both systems share one equation.
Next, we need the second equation \( -4x + cy = -8 \) to also have the same solution, \( (-1, 3) \). We substitute \( x = -1 \) and \( y = 3 \) into this equation: \[ -4(-1) + c(3) = -8 \] This simplifies to: \[ 4 + 3c = -8 \] Now, we solve for \( c \): \[ 3c = -8 - 4 \] \[ 3c = -12 \] \[ c = -4 \]
Thus, the value of \( c \) that needs to be in place for both systems of equations to have the same solution is \( \boxed{-4} \).
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