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

Let's substitute the solution (-2, 1) into each system of equations:

1) 2(-2) - 1 = -5
-4 - 1 = -5
-5 = -5 ✔️

(-2) + 2(1) = 10
-2 + 2 = 10
0 ≠ 10 ❌

Thus, the solution (-2, 1) does not satisfy the second equation in the first system of equations.

2) (-2) - 4(1) = 8
-2 - 4 = 8
-6 ≠ 8 ❌

3(-2) + 10 = -2(1)
-6 + 10 = -2
4 ≠ -2 ❌

Thus, the solution (-2, 1) does not satisfy either equation in the second system of equations.

3) 1 = -2(-2) - 2
1 = 4 - 2
1 = 2 ❌

2(-2) - 2(1) = 5
-4 - 2 = 5
-6 ≠ 5 ❌

Thus, the solution (-2, 1) does not satisfy either equation in the third system of equations.

4) 1 = -3(2) - 2
1 = -6 - 2
1 = -8 ❌

2(-2) - 1 = -5
-4 - 1 = -5
-5 = -5 ✔️

Thus, the only system of equations that has the solution (-2, 1) is:

y = -32x - 2
2x - y = -5

Show me how to do this promblem

To solve this problem, we are looking for the system of equations where the solution (-2, 1) satisfies both equations.

Let's go through each of the given options and substitute x = -2 and y = 1 to check if they satisfy both equations.

1) 2x - y = -5
2(-2) - 1 = -5
-4 - 1 = -5
-5 = -5 ✔️

x + 2y = 10
-2 + 2(1) = 10
-2 + 2 = 10
0 ≠ 10 ❌

The solution (-2, 1) does not satisfy the second equation in the first system of equations.

Repeat this process for the other systems of equations, and the system where both equations are satisfied by the solution (-2, 1) is the system you are looking for.