Two systems of equations are shown:%0D%0A%0D%0A4x+y=−1%0D%0A4%0D%0A%0D%0A+%0D%0A%0D%0A=%0D%0A−%0D%0A1%0D%0Ax+y=2%0D%0A%0D%0A+%0D%0A%0D%0A=%0D%0A2%0D%0Aand%0D%0A%0D%0A8x+2y=−2%0D%0A8%0D%0A%0D%0A+%0D%0A2%0D%0A%0D%0A=%0D%0A−%0D%0A2%0D%0A−4x+cy=−8%0D%0A−%0D%0A4%0D%0A%0D%0A+%0D%0A%0D%0A%0D%0A=%0D%0A−%0D%0A8%0D%0AWhat would the value of c need to be to get the same solution for both systems?%0D%0A%0D%0A(1 point)%0D%0AResponses%0D%0A%0D%0A2%0D%0A2%0D%0A%0D%0A1%0D%0A1%0D%0A%0D%0A-4%0D%0A-4%0D%0A%0D%0A4%0D%0A4%0D%0A%0D%0A-2
In both systems, we can use the method of elimination to find the value of c that would give the same solution.
For the first system:
Multiply the first equation by 2:
8x + 2y = -2
This matches the second equation of the second system, so c would need to be 2 to get the same solution for both systems.
Therefore, the value of c that would give the same solution for both systems is 2.
To find the value of c that would result in the same solution for both systems of equations, we need to find the values of x and y that satisfy both systems.
System 1:
4x + y = -1 (Equation 1)
x + y = 2 (Equation 2)
System 2:
8x + 2y = -2 (Equation 3)
-4x + cy = -8 (Equation 4)
To solve the first system, we can use the method of substitution.
From Equation 2, we can isolate x:
x = 2 - y
Substituting this value of x into Equation 1:
4(2 - y) + y = -1
8 - 4y + y = -1
-3y = -9
y = 3
Substituting the value of y into Equation 2:
x + 3 = 2
x = -1
So, the solution to the first system is x = -1 and y = 3.
To find the value of c that makes both systems have the same solution, we need to substitute the values of x and y into the equations of the second system.
From Equation 3:
8x + 2y = -2
Substituting x = -1 and y = 3:
8(-1) + 2(3) = -2
-8 + 6 = -2
-2 = -2
From Equation 4:
-4x + cy = -8
Substituting x = -1 and y = 3:
-4(-1) + c(3) = -8
4 + 3c = -8
3c = -12
c = -4
Therefore, the value of c that would result in the same solution for both systems is -4.
To get the same solution for both systems, we need to find the value of c that makes the two equations equivalent.
Let's solve the first system of equations:
Equation 1: 4x + y = -1
Equation 2: x + y = 2
To solve this system, we can use either substitution or elimination method. Let's use the elimination method.
Multiply Equation 2 by 4 to have the same coefficient of x:
4(x + y) = 4(2)
4x + 4y = 8
Now we have the following system of equations:
Equation 1: 4x + y = -1
Equation 3: 4x + 4y = 8
Subtract Equation 1 from Equation 3 to eliminate x:
(4x + 4y) - (4x + y) = 8 - (-1)
3y = 9
y = 3
Substitute the value of y back into Equation 1 to solve for x:
4x + y = -1
4x + 3 = -1
4x = -4
x = -1
Therefore, the solution to the first system of equations is x = -1, y = 3.
Now let's solve the second system of equations:
Equation 4: 8x + 2y = -2
Equation 5: -4x + cy = -8
To get the same solution for both systems, we need the solutions x = -1 and y = 3 to satisfy Equation 4 and Equation 5.
Substitute the values x = -1 and y = 3 into Equation 4:
8(-1) + 2(3) = -2
-8 + 6 = -2
-2 = -2
Now substitute the values x = -1 and y = 3 into Equation 5:
-4(-1) + c(3) = -8
4 + 3c = -8
3c = -12
c = -4
Therefore, the value of c needed to get the same solution for both systems is c = -4.