To solve the linear-quadratic system using the substitution method, we have the following equations:
- \( y = x^2 - 4x + 8 \)
- \( y = 3x - 4 \)
We can substitute the second equation into the first equation. This gives us:
\[ 3x - 4 = x^2 - 4x + 8 \]
Next, we rearrange the equation:
\[ x^2 - 4x - 3x + 8 + 4 = 0 \]
This simplifies to:
\[ x^2 - 7x + 12 = 0 \]
Now, we factor the quadratic equation:
\[ (x - 3)(x - 4) = 0 \]
Thus, we have two solutions for \( x \):
- \( x = 3 \)
- \( x = 4 \)
Now, we can find the corresponding \( y \) values using either of the original equations (we'll use 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 \( (4, 8) \) and \( (3, 5) \).
The correct response is: (4,8) and (3,5).