Adunand ecuatiile, avem:
(x+y)²-2xy=2xy+x+y⇔4xy=(x+y)²-(x+y) (1)
Scazand ecuatiile, obtinem:
(x+y)(x-y)=3(x-y)⇔(x-y)(x+y-3)=0. De aici avem doua variante:
1) x-y=0⇒x=y si dupa ce inlocuim in una dintre ecuatiile sistemului ⇒x=y=0
2) x+y=3 care inlocuit in (1), obtinem xy=3/2.
Deci avem S=x+y=3 si P=xy=3/2.
Ecuatia ale carei solutii dau suma S si produsul P, este z²-Sz+P=0.
Deci z²-3z+3/2=0
Δ=9-6=3; [tex]z_{1;2}=\dfrac{3\pm\sqrt3}{2};\ deci\ [/tex]
[tex](x;y)\in\left\{\left(\dfrac{3+\sqrt3}{2};\dfrac{3-\sqrt3}{2}\right);\ \left(\dfrac{3-\sqrt3}{2};\dfrac{3+\sqrt3}{2}\right);\ (0;0);\right\}[/tex]