To solve the linear-quadratic system using the substitution method, we will substitute one equation into the other. The given equations are:
- \( y = x^2 - 4x + 8 \) (quadratic equation)
- \( y = 3x - 4 \) (linear equation)
We can substitute the expression for \( y \) from the linear equation into the quadratic equation. This gives us:
\[ 3x - 4 = x^2 - 4x + 8 \]
Next, we'll rearrange the equation to one side:
\[ 0 = x^2 - 4x - 3x + 8 + 4 \]
Combining like terms results in:
\[ 0 = x^2 - 7x + 12 \]
Now, we can rewrite this as:
\[ x^2 - 7x + 12 = 0 \]
Next, we can factor the quadratic equation:
\[ (x - 3)(x - 4) = 0 \]
Setting each factor to zero gives us the solutions for \( x \):
- \( x - 3 = 0 \) ⟹ \( x = 3 \)
- \( x - 4 = 0 \) ⟹ \( x = 4 \)
Now we can find the corresponding \( y \) values by substituting these \( x \) values back into the linear equation \( y = 3x - 4 \).
For \( x = 3 \):
\[ y = 3(3) - 4 = 9 - 4 = 5 \]
For \( x = 4 \):
\[ y = 3(4) - 4 = 12 - 4 = 8 \]
Thus, the solutions to the system of equations are:
- \( (3, 5) \)
- \( (4, 8) \)
So the solution set of the system is \( (3, 5) \) and \( (4, 8) \).