To solve the given linear-quadratic system using the substitution method, we have the following equations:
- \(y = x^2 - 4x + 8\)
- \(y = 3x - 4\)
We will substitute the second equation into the first equation:
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Set the equations equal: \[ 3x - 4 = x^2 - 4x + 8 \]
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Rearrange the equation to bring all terms to one side: \[ x^2 - 4x - 3x + 8 + 4 = 0 \] \[ x^2 - 7x + 12 = 0 \]
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Now, factor the quadratic: \[ (x - 3)(x - 4) = 0 \] This gives us: \[ x = 3 \quad \text{or} \quad x = 4 \]
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Next, substitute these \(x\) values back into one of the original equations to find the corresponding \(y\) values. We can use \(y = 3x - 4\):
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For \(x = 3\): \[ y = 3(3) - 4 = 9 - 4 = 5 \quad \text{(Point: } (3, 5) \text{)} \]
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For \(x = 4\): \[ y = 3(4) - 4 = 12 - 4 = 8 \quad \text{(Point: } (4, 8) \text{)} \]
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Thus, the solution points for the system are \((3, 5)\) and \((4, 8)\).
Therefore, the correct response is: \[ \text{(4, 8) and (3, 5)} \]