Răspuns :

Miky93id
x+y=3  =>y=3-x 

x² +x =y
x²+x=3-x

x²+x+x=3

x²+2x-3=0

x²+2x+1 -4=0

[(x+1)-2][(x+1)+2]=0

(x+1-2)(x+1+2)=0

(x-1)(x+3)=0

x-1=0      ∧  x+3=0
x=1                x= -3

1) x=1  =>y= 3-1=2

2)x=-3  =>y= 3+3=6

S(x;y)={(1;2);(-3;6)}

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