To determine which system the point (0, -2) is a solution for, we will substitute \( x = 0 \) and \( y = -2 \) into each equation provided.
Checking the first system:
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First equation: \[ -5(0) + (-2) = -2 \quad \Rightarrow \quad -2 = -2 \quad \text{(True)} \]
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Second equation: \[ -3(0) + 6(-2) = -12 \quad \Rightarrow \quad 0 - 12 = -12 \quad \Rightarrow \quad -12 = -12 \quad \text{(True)} \]
Both equations are satisfied, so (0, -2) is a solution to the first system.
Checking the second system:
- First equation: \[ (0) + (-2) = 2 \quad \Rightarrow \quad -2 = 2 \quad \text{(False)} \]
Since the first equation is not true, (0, -2) is not a solution for the second system.
Checking the third system:
- First equation: \[ -4(0) + (-2) = 6 \quad \Rightarrow \quad -2 = 6 \quad \text{(False)} \]
Since the first equation is not true, (0, -2) is not a solution for the third system.
Checking the fourth system:
- First equation: \[ -5(0) = -2 - 3 \quad \Rightarrow \quad 0 = -5 \quad \text{(False)} \]
Since the first equation is not true, (0, -2) is not a solution for the fourth system.
Conclusion:
The point (0, -2) is a solution to the system: \[ \begin{align*} -5x + y &= -2 \ -3x + 6y &= -12 \end{align*} \]