To determine the system of equations for which (4, -3) is a solution, we need to formulate two equations that, when solved simultaneously, yield x = 4 and y = -3. Here are a couple of examples:
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Example System: \[ \text{Equation 1: } x + y = 1 \quad \text{(4 + (-3) = 1)} \] \[ \text{Equation 2: } 2x - 3y = 18 \quad \text{(2(4) - 3(-3) = 8 + 9 = 17)} \]
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Another Example System: \[ \text{Equation 1: } 2x + y = 5 \quad \text{(2(4) + (-3) = 8 - 3 = 5)} \] \[ \text{Equation 2: } x - 2y = 10 \quad \text{(4 - 2(-3) = 4 + 6 = 10)} \]
You can create similar systems by choosing different coefficients for x and y in any linear equations. The key is ensuring that substituting x = 4 and y = -3 satisfies both equations.