To find the value of \( c \) that makes the second system of equations consistent with the first system, we first need to solve for the solution to the first system of equations:
- \[ 4x + y = -1 \quad \text{(Equation 1)} \]
- \[ x + y = 2 \quad \text{(Equation 2)} \]
Let's solve Equation 2 for \( y \): \[ y = 2 - x \]
Now we substitute \( y \) into Equation 1: \[ 4x + (2 - x) = -1 \] \[ 4x + 2 - x = -1 \] \[ 3x + 2 = -1 \] \[ 3x = -3 \] \[ x = -1 \]
Now, substitute \( x = -1 \) back into Equation 2 to find \( y \): \[ y = 2 - (-1) = 2 + 1 = 3 \]
So the solution to the first system is \( (x, y) = (-1, 3) \).
Next, we need the second system of equations to yield the same solution \( (-1, 3) \). The equations are:
- \[ 8x + 2y = -2 \quad \text{(Equation 3)} \]
- \[ -4x + cy = -8 \quad \text{(Equation 4)} \]
First, let's check Equation 3 by substituting \( x = -1 \) and \( y = 3 \):
For Equation 3: \[ 8(-1) + 2(3) = -2 \ -8 + 6 = -2 \quad \text{(True)} \]
So the first equation in the second system holds true.
Now we substitute into Equation 4: \[ -4(-1) + c(3) = -8 \ 4 + 3c = -8 \] \[ 3c = -8 - 4 \ 3c = -12 \ c = -4 \]
Thus, the value of \( c \) that makes the second system have the same solution as the first system is: \[ \boxed{-4} \]
1 answer