Rozwiązać algebraicznie i graficznie poniższe układy: a) { y=x2-4x+3 b) {x+2y=0 {y=2x-6 {y=-1/4x2+x

a) y = x² - 4x + 3 y = 2x - 6 ---------------------- x² - 4x +3 = 2x - 6 x² - 6x + 9 = 0 (x - 3)² = 0 x = 3 y = 2*3 - 6 = 6 - 6 = 0 Odp. x = 3 , y = 0 b) x + 2y = 0 ---> 2y = - x ----> y = (-1/2) x y =( -1/2)x² + x ----------------------- (-1/2) x = (-1/4) x² +x (1/4) x² - (1/2) x - x = 0 (1/4) x² - (3/2) x = 0 (1/4) x *( x - 6) = 0 x = 0 lub x = 6 y = (-1/2)*0 = 0 lub y = (-1/2)*6 = - 3 Odp.x = 0 i y = 0 oraz x = 6 i y = -3