To determine which system of equations the point (0, -2) is a solution for, we can substitute \( x = 0 \) and \( y = -2 \) into each equation to see if they hold true.
Let's check each system one by one:
-
System:
- \( -4x + y = 6 \)
- \( -5x - y = 21 \)
Substituting (0, -2):
- For \( -4(0) + (-2) = 6 \) → \( -2 \neq 6 \) (not a solution)
- For \( -5(0) - (-2) = 21 \) → \( 2 \neq 21 \) (not a solution)
-
System:
- \( x + y = 2 \)
- \( -x + 2y = 16 \)
Substituting (0, -2):
- For \( 0 + (-2) = 2 \) → \( -2 \neq 2 \) (not a solution)
- For \( -0 + 2(-2) = 16 \) → \( -4 \neq 16 \) (not a solution)
-
System:
- \( -5x + y = -2 \)
- \( -3x + 6y = -12 \)
Substituting (0, -2):
- For \( -5(0) + (-2) = -2 \) → \( -2 = -2 \) (solution)
- For \( -3(0) + 6(-2) = -12 \) → \( -12 = -12 \) (solution)
-
System:
- \( -5x = y - 3 \)
- \( 3x - 8y = 24 \)
Substituting (0, -2):
- For \( -5(0) = -2 - 3 \) → \( 0 = -5 \) (not a solution)
- For \( 3(0) - 8(-2) = 24 \) → \( 16 \neq 24 \) (not a solution)
Based on these checks, the point (0, -2) is a solution to the third system:
\[ -5x + y = -2 \quad \text{and} \quad -3x + 6y = -12 \]