To use the substitution method to solve the system of equations \( y = x^2 - 4x + 8 \) and \( y = 3x - 4 \), we can set the two equations equal to each other since they both equal \( y \).
-
Set the equations equal: \[ x^2 - 4x + 8 = 3x - 4 \]
-
Rearrange the equation: \[ x^2 - 4x - 3x + 8 + 4 = 0 \] \[ x^2 - 7x + 12 = 0 \]
-
To solve this quadratic equation, we can factor it: \[ (x - 3)(x - 4) = 0 \] So, \( x = 3 \) or \( x = 4 \).
-
Now, we substitute these \( x \) values back into one of the original equations to find the corresponding \( y \) values. We'll use \( y = 3x - 4 \).
-
For \( x = 3 \): \[ y = 3(3) - 4 = 9 - 4 = 5 \] So one solution is \( (3, 5) \).
-
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 \( (3, 5) \).
Among the choices given, the correct answer using the substitution method to solve this linear-quadratic system is:
- (4, 8) and (3, 5).