Question

Two systems of equations are shown:

4x+y=−1


x+y=2


and

8x+2y=−2


−4x+cy=−8


What would the value of c need to be to get the same solution for both systems?

(1 point)
Responses

-4
-4

-2
-2

1
1

4
4

2
2
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Answer 1

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:
1. \( 4x + y = -1 \)
2. \( x + y = 2 \)

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:
1. \( 8x + 2y = -2 \)
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 \( \boxed{-4} \).