To find the solution to each of the systems of equations, we need to substitute the values of x and y into both equations in the system and check if they satisfy them.
1. y = -2x - 2 and 2x - 2y = 5
Using the values of x = -2 and y = 1:
y = -2(-2) - 2
y = 4 - 2
y = 2
Substitute y = 2 into the second equation:
2(-2) - 2(1) = 5
-4 - 2 = 5
-6 = 5 (This does not hold true)
So, this system of equations does not have the solution (-2, 1).
2. 2x - y = -5 and x + 2y = 10
Using the values of x = -2 and y = 1:
2(-2) - 1 = -5
-4 - 1 = -5
-5 = -5
x + 2(1) = 10
-2 + 2 = 10
0 ≠ 10 (This does not hold true)
So, this system of equations does not have the solution (-2, 1).
3. y = 3/2 * x - 2 and 2x - y = -5
Using the values of x = -2 and y = 1:
1 = 3/2*(-2) - 2
1 = -3 - 2
1 = -5 (This does not hold true)
So, this system of equations does not have the solution (-2, 1).
4. x - 4y = 8 and 3x + 10 = -2y
Using the values of x = -2 and y = 1:
-2 - 4(1) = 8
-2 - 4 = 8
-6 = 8 (This does not hold true)
3(-2) + 10 = -2(1)
-6 + 10 = -2
4 = -2 (This does not hold true)
So, this system of equations does not have the solution (-2, 1).
Therefore, none of the given systems of equations has the solution (-2, 1).
Which of the following systems of equations has the solution (- 2, 1)
This question requires you to show your work.
(1 point)
y = - 2x - 2 and 2x - 2y = 5
2x - y = - 5 and x + 2y = 10
y = 3/2 * x - 2 and 2x - y = - 5
x - 4y = 8 and 3x + 10 = - 2y
1 answer
1 answer