To determine to which system the point (0, -2) is a solution, we need to substitute \(x = 0\) and \(y = -2\) into each system and check if the equations are satisfied.
Let's examine each system one by one.
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System 1: \[ -5x + y = -2 \quad \text{(1)} \] \[ -3x + 6y = -12 \quad \text{(2)} \] Substituting \(x = 0\) and \(y = -2\) in (1): \[ -5(0) + (-2) = -2 \Rightarrow -2 = -2 \quad \text{(True)} \] Substituting in (2): \[ -3(0) + 6(-2) = -12 \Rightarrow -12 = -12 \quad \text{(True)} \] Both equations are satisfied.
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System 2: \[ -5x = y - 3 \quad \text{(1)} \] \[ 3x - 8y = 24 \quad \text{(2)} \] Substituting \(x = 0\) and \(y = -2\) in (1): \[ -5(0) = -2 - 3 \Rightarrow 0 = -5 \quad \text{(False)} \] (We don't need to check the second equation because the first one is false.)
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System 3: \[ -4x + y = 6 \quad \text{(1)} \] \[ -5x - y = 21 \quad \text{(2)} \] Substituting \(x = 0\) and \(y = -2\) in (1): \[ -4(0) + (-2) = 6 \Rightarrow -2 = 6 \quad \text{(False)} \] (We don't need to check the second equation because the first one is false.)
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System 4: \[ x + y = 2 \quad \text{(1)} \] \[ -x + 2y = 16 \quad \text{(2)} \] Substituting \(x = 0\) and \(y = -2\) in (1): \[ 0 + (-2) = 2 \Rightarrow -2 = 2 \quad \text{(False)} \] (We don't need to check the second equation because the first one is false.)
After evaluating all systems, we find that the point (0, -2) satisfies only the first system.
Conclusion: The point (0, -2) is a solution to the system:
\[ -5x + y = -2 \] \[ -3x + 6y = -12 \]