To determine which system of equations has the point (0, -2) as a solution, we can substitute \( x = 0 \) and \( y = -2 \) into each of the equations in the systems and check if both equations hold true.
First System
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\(-x + y = 2\) \[-(0) + (-2) = 2\] \[-2 \neq 2\] (Not a solution)
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\(-x + 2y = 16\) \[-(0) + 2(-2) = 16\] \[-4 \neq 16\] (Not a solution)
Second System
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\(-4x + y = 6\) \[-4(0) + (-2) = 6\] \[-2 \neq 6\] (Not a solution)
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\(-5x - y = 21\) \[-5(0) - (-2) = 21\] \[2 \neq 21\] (Not a solution)
Third System
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\(-5x + y = -3\) \[-5(0) + (-2) = -3\] \[-2 \neq -3\] (Not a solution)
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\(3x - 8y = 24\) \[3(0) - 8(-2) = 24\] \[16 \neq 24\] (Not a solution)
Fourth System
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\(-5x + y = -2\) \[-5(0) + (-2) = -2\] \[-2 = -2\] (Is a solution)
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\(-3x + 6y = -12\) \[-3(0) + 6(-2) = -12\] \[-12 = -12\] (Is a solution)
Since both equations in the fourth system hold true with the point (0, -2), the answer is:
-5x + y = -2 and -3x + 6y = -12.