To solve the linear-quadratic system using the substitution method, we have the equations:
- \( y = x^2 - 4x + 8 \)
- \( y = 3x - 4 \)
We can set these equations equal to each other since they both equal \( y \):
\[ x^2 - 4x + 8 = 3x - 4 \]
Now, rearranging the equation:
\[ x^2 - 4x - 3x + 8 + 4 = 0 \]
This simplifies to:
\[ x^2 - 7x + 12 = 0 \]
Next, we can factor this quadratic equation:
\[ (x - 3)(x - 4) = 0 \]
This gives us the solutions:
\[ x = 3 \quad \text{or} \quad x = 4 \]
Now we can find the corresponding \( y \) values using the linear equation \( y = 3x - 4 \).
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For \( x = 3 \): \[ y = 3(3) - 4 = 9 - 4 = 5. \] So one solution is \( (3, 5) \).
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For \( x = 4 \): \[ y = 3(4) - 4 = 12 - 4 = 8. \] So the other solution is \( (4, 8) \).
Thus, the solutions to the system are \( (3, 5) \) and \( (4, 8) \).
From the options provided, the correct response is:
(4,8) and (3,5).