[tex] \left \{ {{x+y=3} \atop { x^{2} +x-y=0}} \right. [/tex]
y=3-x
Se inlocuieste valoarea lui y in cea de-a doua ecuatie:
[tex] x^{2} +x-(3-x)=0[/tex]
[tex] x^{2} +x-3+x=0[/tex]
[tex] x^{2} +2x-3=0[/tex]
Δ=4+12=16
x1=1 => y1=3-1=2
x2=-3 => y2=3+3=6
(x,y)∈{(1,2);(-3,6)}
