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:
- \( 4x + y = -1 \) (Equation 1)
- \( x + y = 2 \) (Equation 2)
To solve this system, we can use the substitution or elimination method. Let's use substitution.
From Equation 2, we isolate \( y \):
\[ y = 2 - x \]
Now, substitute \( y \) in Equation 1:
\[ 4x + (2 - x) = -1 \]
This simplifies to:
\[ 4x + 2 - x = -1 \]
Combining like terms, we get:
\[ 3x + 2 = -1 \]
Subtracting 2 from both sides:
\[ 3x = -3 \]
Dividing both sides by 3:
\[ x = -1 \]
Now that we have \( x \), we can find \( y \) by substituting \( x = -1 \) back into Equation 2:
\[ y = 2 - (-1) = 2 + 1 = 3 \]
So the solution to the first system is:
\[ (x, y) = (-1, 3) \]
Now we will explore the second system:
- \( 8x + 2y = -2 \) (Equation 3)
- \( -4x + cy = -8 \) (Equation 4)
Notice that Equation 3 can be simplified. Dividing the entire equation by 2 gives us:
\[ 4x + y = -1 \]
This is the same as Equation 1 from the first system, which we used to find that:
\[ y = -1 - 4x \]
For the second equation (Equation 4) to have the same solution, we already found that \( x = -1 \) and \( y = 3 \). Let's substitute these values into Equation 4:
\[ -4(-1) + c(3) = -8 \]
This simplifies to:
\[ 4 + 3c = -8 \]
Subtracting 4 from both sides gives:
\[ 3c = -12 \]
Now, divide by 3:
\[ c = -4 \]
Therefore, the necessary value for \( c \) to obtain the same solution as the first system is:
\[ \boxed{-4} \]
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