To solve the linear-quadratic system of equations given by:
- \( y = x^2 - 4x + 8 \) (quadratic)
- \( y = 3x - 4 \) (linear)
we can use the substitution method.
Since both equations equal \( y \), we can set them equal to each other:
\[ x^2 - 4x + 8 = 3x - 4 \]
Now, rearranging the equation gives:
\[ x^2 - 4x - 3x + 8 + 4 = 0 \]
\[ x^2 - 7x + 12 = 0 \]
Next, we can factor the quadratic:
\[ (x - 3)(x - 4) = 0 \]
Setting each factor to zero gives us the possible \( x \) values:
\[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \] \[ x - 4 = 0 \quad \Rightarrow \quad x = 4 \]
Next, we substitute these \( x \) values back into either of the original equations to find the corresponding \( y \) values. We'll use the linear equation \( y = 3x - 4 \).
- For \( x = 3 \):
\[ y = 3(3) - 4 = 9 - 4 = 5 \quad \Rightarrow \quad (3, 5) \]
- For \( x = 4 \):
\[ y = 3(4) - 4 = 12 - 4 = 8 \quad \Rightarrow \quad (4, 8) \]
Thus, the solutions to the system are:
\[ (3, 5) \quad \text{and} \quad (4, 8) \]
From your options, the correct response is:
(4,8) and (3,5).
1 answer