To solve the linear-quadratic system using the substitution method, we have the following equations:
- \( y = x^2 - 4x + 8 \) (quadratic)
- \( y = 3x - 4 \) (linear)
We can substitute the expression for \( y \) from the second equation into the first equation:
- Substitute \( 3x - 4 \) for \( y \) in the first equation:
\[ 3x - 4 = x^2 - 4x + 8 \]
- Rearranging gives us:
\[ x^2 - 4x - 3x + 8 + 4 = 0 \]
This simplifies to:
\[ x^2 - 7x + 12 = 0 \]
- Now, we can factor or use the quadratic formula to solve for \( x \):
\[ (x - 3)(x - 4) = 0 \]
Thus, the solutions are:
\[ x = 3 \quad \text{and} \quad x = 4 \]
- We now find the corresponding \( y \) values by substituting back into \( y = 3x - 4 \):
- For \( x = 3 \):
\[ y = 3(3) - 4 = 9 - 4 = 5 \quad \Rightarrow \quad (3, 5) \]
- For \( x = 4 \):
\[ y = 3(4) - 4 = 12 - 4 = 8 \quad \Rightarrow \quad (4, 8) \]
The solutions to the system are:
\[ (3, 5) \quad \text{and} \quad (4, 8) \]
Therefore, the correct response is:
(4,8) and (3,5).