Explicație pas cu pas:
b)
[tex]\left \{ {{y= {x}^{2} - 4x + 3 } \atop {x - y - 3 = 0}}\right. \\ \left \{ {{y= {x}^{2} - 4x + 3 } \atop {y = x - 3}}\right.[/tex]
[tex]x - 3 = {x}^{2} - 4x + 3 \\ {x}^{2} - 5x + 6 = 0 \\ (x - 2)(x - 3) = 0[/tex]
=>
[tex]\left \{ {{x = 2} \atop {y = - 1}} si\left \{ {{x = 3} \atop {y = 0}} \right.\right. \\ [/tex]
