To solve for the value of \(c\) that makes both systems of equations have the same solution, we first need to find the solution for the first system:
- The first system is given by: \[ 4x + y = -1 \quad (1) \] \[ x + y = 2 \quad (2) \]
From equation (2), we can express \(y\) in terms of \(x\): \[ y = 2 - x \] Now, substitute this expression for \(y\) into equation (1): \[ 4x + (2 - x) = -1 \] Simplifying: \[ 4x + 2 - x = -1 \] \[ 3x + 2 = -1 \] \[ 3x = -3 \] \[ x = -1 \]
Now substitute \(x = -1\) back into equation (2) to find \(y\): \[ -1 + y = 2 \] \[ y = 3 \]
Thus, the solution for the first system is \( (x, y) = (-1, 3) \).
Next, let's examine the second system: \[ 8x + 2y = -2 \quad (3) \] \[ -4x + cy = -8 \quad (4) \]
We can simplify equation (3) by dividing through by 2: \[ 4x + y = -1 \quad (5) \]
We see that equation (5) is the same as equation (1) from the first system. Hence, the first two equations in both systems are equivalent.
Now we need to ensure that equation (4) gives us the same solution of \( (x, y) = (-1, 3) \).
Substituting \(x = -1\) and \(y = 3\) into equation (4): \[ -4(-1) + c(3) = -8 \] This simplifies to: \[ 4 + 3c = -8 \] Now, solve for \(c\): \[ 3c = -8 - 4 \] \[ 3c = -12 \] \[ c = -4 \]
Thus, the value of \(c\) that would make both systems have the same solution is: \[ \boxed{-4} \]
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