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