To determine the number of solutions for the system of equations
- \( y = x^2 - 4x + 8 \) (quadratic equation)
- \( y = 3x - 4 \) (linear equation)
we can set the two equations equal to each other since they both equal \( y \):
\[ x^2 - 4x + 8 = 3x - 4 \]
Now, let's rearrange the equation:
\[ x^2 - 4x - 3x + 8 + 4 = 0 \]
This simplifies to:
\[ x^2 - 7x + 12 = 0 \]
Next, we can factor the quadratic equation:
\[ (x - 3)(x - 4) = 0 \]
Setting each factor to zero gives us the potential solutions for \( x \):
- \( x - 3 = 0 \) → \( x = 3 \)
- \( x - 4 = 0 \) → \( x = 4 \)
Since we found two distinct values for \( x \), we will have two corresponding \( y \) values by substituting \( x \) back into either of the original equations.
Thus, the system has 2 solutions.
The answer is:
2 solution(s).