6) 
zpět
substituce: 
y2
– 4y + 3 = 0
(y – 3)(y – 1) = 0
y1
= 3 → 
→ x1 = 43 = 64
y2
= 1 → 
→ x2 = 41 = 4
6) zpět
substituce:
y2
– 4y + 3 = 0
(y – 3)(y – 1) = 0
y1
= 3 → → x1 = 43 = 64
y2
= 1 → → x2 = 41 = 4