To solve the system of equations:
First Equation: 3x² - 6x + 5 = 2
Second Equation: y = (x - 1)² + 2
We can start by finding the value of x from the first equation and then substitute it into the second equation to find y.
Let's solve the first equation:
3x² - 6x + 5 = 2
Rearranging the equation:
3x² - 6x + 3 = 0
This quadratic equation can be solved using the quadratic formula:
x = (-b ± √(b² - 4ac)) / (2a)
For our equation a = 3, b = -6, and c = 3:
x = (-(-6) ± √((-6)² - 4(3)(3))) / (2(3))
x = (6 ± √(36 - 36)) / 6
x = (6 ± √0) / 6
x = (6 ± 0) / 6
We have x = 1 for both cases, so x = 1.
Now we can substitute x = 1 into the second equation to find y:
y = (x - 1)² + 2
y = (1 - 1)² + 2
y = 0² + 2
y = 0 + 2
y = 2
Therefore, the solution to the given system of equations is x = 1 and y = 2.