Which of the following systems of equations has the solution (-2, 1)?

Question

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(1 point)
Responses

2x−y=−5
and x+2y=10
2 x minus y is equal to negative 5 and x plus 2 y is equal to 10

x−4y=8
and 3x+10=−2y
x minus 4 y is equal to 8 and 3 x plus 10 is equal to negative 2 y

y=−2x−2
and 2x−2y=5
y is equal to negative 2 x minus 2 and 2 x minus 2 y is equal to 5

y=−32x−2
and 2x−y=−5

Answer 1

Let's substitute the values x = -2 and y = 1 into each system of equations to see which one has the solution (-2, 1).

1. For the first system:
2(-2) - 1 = -5 → -4 - 1 = -5 → -5 = -5
-2 + 2(1) = 10 → -2 + 2 = 10 → 0 ≠ 10
The first system does not have (-2, 1) as the solution.

2. For the second system:
-2 - 4(1) = 8 → -2 - 4 = 8 → -6 ≠ 8
3(-2) + 10 = -2(1) → -6 + 10 = -2 → 4 ≠ -2
The second system does not have (-2, 1) as the solution.

3. For the third system:
1 = -2(-2) - 2 → 1 = 4 - 2 → 1 = 2
2(-2) - 2(1) = 5 → -4 - 2 = 5 → -6 ≠ 5
The third system does not have (-2, 1) as the solution.

4. For the fourth system:
1 = -3(2) - 2 → 1 = -6 - 2 → 1 = -8
2(-2) - 1 = -5 → -4 - 1 = -5 → -5 = -5
The fourth system has (-2, 1) as the solution.

Therefore, the system of equations y = -3/2x - 2 and 2x - y = -5 has the solution (-2, 1).

Answer 2

can u answer easier

Answer 3

Of course!

We can check by directly substituting x = -2 and y = 1 into each system of equations.

1. For 2x - y = -5 and x + 2y = 10:
2(-2) - 1 = -5 → -4 - 1 = -5 → -5 = -5
-2 + 2(1) = 10 → -2 + 2 = 10 → 0 ≠ 10
Since the values satisfy the first equation but not the second equation, this system does not have the solution (-2, 1).

2. For y = -2x - 2 and 2x - y = -5:
1 = -2(-2) - 2 → 1 = 4 - 2 → 1 = 2
2(-2) - 1 = -5 → -4 - 1 = -5 → -5 = -5
Since the values satisfy both equations, this system has the solution (-2, 1).

Therefore, the system of equations y = -2x - 2 and 2x - y = -5 has the solution (-2, 1).