To determine which system of equations has the solution (4, -3), we can substitute x = 4 and y = -3 into each of the given equations.
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First System:
- \( 7x + 9y = 55 \) \[ 7(4) + 9(-3) = 28 - 27 = 1 \quad \text{(not equal to 55)} \]
- \( 3x + 9y = -15 \) \[ 3(4) + 9(-3) = 12 - 27 = -15 \quad \text{(correct)} \]
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Second System:
- \( 7x + 9y = 55 \) \[ 7(4) + 9(-3) = 28 - 27 = 1 \quad \text{(not equal to 55)} \]
- \( 3x - 9y = -15 \) \[ 3(4) - 9(-3) = 12 + 27 = 39 \quad \text{(not equal to -15)} \]
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Third System:
- \( 7x - 9y = 55 \) \[ 7(4) - 9(-3) = 28 + 27 = 55 \quad \text{(correct)} \]
- \( 3x - 9y = -15 \) \[ 3(4) - 9(-3) = 12 + 27 = 39 \quad \text{(not equal to -15)} \]
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Fourth System:
- \( 7x - 9y = 55 \) \[ 7(4) - 9(-3) = 28 + 27 = 55 \quad \text{(correct)} \]
- \( 3x + 9y = -15 \) \[ 3(4) + 9(-3) = 12 - 27 = -15 \quad \text{(correct)} \]
Since the first and second equations from the fourth system both yield correct results when evaluated at \( (4, -3) \), the correct system for the solution \( (4, -3) \) is:
7x−9y=55 and 3x+9y=−15.