Two systems of equations are shown:

Question

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

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. $$8x+2y=-2\text{(Equation 3)}$$
2. $$-4x+cy=-8\text{(Equation 4)}$$

First, let's check Equation 3 by substituting $x=-1$ and $y=3$:

For Equation 3: $$8(-1)+2(3)=-2-8+6=-2\text{(True)}$$

So the first equation in the second system holds true.

Now we substitute into Equation 4: $$-4(-1)+c(3)=-84+3c=-8$$ $$3c=-8-43c=-12c=-4$$

Thus, the value of $c$ that makes the second system have the same solution as the first system is: $$\boxed{{\displaystyle -4}}$$