To solve the linear-quadratic system using the substitution method, we can follow these steps:
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We have the equations: \[ y = x^2 - 4x - 41 \] \[ y = 4 \]
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Substitute \( y = 4 \) into the quadratic equation: \[ 4 = x^2 - 4x - 41 \]
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Rearrange the equation: \[ x^2 - 4x - 41 - 4 = 0 \ x^2 - 4x - 45 = 0 \]
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Factor the quadratic equation: \[ (x - 9)(x + 5) = 0 \]
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Therefore, we find the solutions: \[ x - 9 = 0 \quad \Rightarrow \quad x = 9 \ x + 5 = 0 \quad \Rightarrow \quad x = -5 \]
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Now, substitute these values back into \( y = 4 \):
- For \( x = 9 \), \( (9, 4) \)
- For \( x = -5 \), \( (-5, 4) \)
Thus, the solution to the system is: \[ (9, 4) \text{ and } (-5, 4) \]
The correct response is: (−5, 4) and (9, 4).