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

2
2

-2
-2

-4
-4

4
4

1
1
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Answer 1

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:

1.
\[
4x + y = -1 \quad \text{(Equation 1)}
\]
2.
\[
x + y = 2 \quad \text{(Equation 2)}
\]

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:

1.
\[
8x + 2y = -2 \quad \text{(Equation 3)}
\]
2.
\[
-4x + cy = -8 \quad \text{(Equation 4)}
\]

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:
\[
\boxed{-4}
\]