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Llavinia11
a fost răspuns

Sisteme :
{x+y=3
{x^2+x-y =0

Răspuns :

Renatemambouko
x+y=3
x²+x-y=0

x=3-y
(3-y)²+3-y-y=0
9-6y+y²+3-2y=0
y²-8y+12=0
Δ=64+48=16
y₁=(8+√16)/2=(8+4)/2=6  ⇒ x=3-6=-3
y₂=(8-√16)/2=(8-4)/2=2  ⇒ x=3-2=1

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