Question

x+y=3
x^2+x-y=0      sistem de ecuatii

Answer 1

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