Solve the system for (x,y)(x, y)(x,y):
y=x2−4x+3y = x^2 - 4x + 3y=x2−4x+3 y=x−1y = x - 1y=x−1
(1,0)(1, 0)(1,0)
(4,3)(4, 3)(4,3)
(2,1)(2, 1)(2,1)
Both a and b