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?

options:

2
2

1
1

4
4

-2
-2

-4

Answer 1

To solve for the value of $c$ such that both systems of equations have the same solution, we first need to determine the solution of the first system of equations:

1. $4x+y=-1$ (Equation 1)
2. $x+y=2$ (Equation 2)

We can solve for $y$ in terms of $x$ by re-arranging Equation 2:

$$y=2-x$$

Now substitute $y=2-x$ into Equation 1:

$$4x+(2-x)=-1$$

Combine like terms:

$$4x-x+2=-1$$ $$3x+2=-1$$ $$3x=-1-2$$ $$3x=-3$$ $$x=-1$$

Now substitute $x=-1$ back into $y=2-x$:

$$y=2-(-1)=3$$

Thus, the solution to the first system of equations is $(x,y)=(-1,3)$.

Next, we need to analyze the second system of equations:

1. $8x+2y=-2$ (Equation 3)
2. $-4x+cy=-8$ (Equation 4)

We can simplify Equation 3:

$$8x+2y=-2$$ Dividing through by 2: $$4x+y=-1\text{(which is the same as Equation 1)}$$

Now we know that Equation 3 is consistent with the first system of equations. Therefore, to find the value of $c$, we need to set up Equation 4 so that it has the same solution $(x,y)=(-1,3)$.

Substituting $x=-1$ and $y=3$ into Equation 4:

$$-4(-1)+c(3)=-8$$ $$4+3c=-8$$ $$3c=-8-4$$ $$3c=-12$$ $$c=-4$$

Thus, the value of $c$ needed to make both systems have the same solution is $\boxed{{\displaystyle -4}}$.