[tex]\displaystyle Avem \left \{ {{x^2+y^2+3xy+2=0~~(*)} \atop {xy+y^2-4=0}~~(**)}\right.. \\ \\ Din~(**)~rezulta~xy=4-y^2.~Inlocuim~in~(*)~si~obtinem \\ \\ \boxed{x^2=2y^2-14}~. \\ \\ Evident~y \neq 0.~(In~caz~contrar~(**)~ar~deveni~-4=0,~fals!) \\ \\ Din~(**)~rezulta~\boxed{ x=\frac{4-y^2}{y}}~. \\ \\ Din~relatiile~din~chenar~rezulta~ \left( \frac{4-y^2}{y} \right)^2=2y^2-14. \\ \\ \Leftrightarrow y^4-6y^2-16=0.[/tex]
[tex]\displaystyle Notam~y^2=t \Rightarrow t^2-6t-16=0.~Rezolvand~ecuatia~obtinem \\ \\ t \in \{-2,8 \},~dar~t \ge 0,~deci~t=8. \\ \\ Asadar~ \boxed{y^2=8}~.~Deci~y \in \{-2 \sqrt{2},2 \sqrt{2} \}. \\ \\ Cum~x= \frac{4-y^2}{y},~rezulta: \\ \\ \bullet y= -2 \sqrt{2} \Rightarrow x= \sqrt{2}. \\ \\ \bullet y= 2 \sqrt{2} \Rightarrow x=- \sqrt{2}. \\ \\ Verificand~solutii~gasite~(trebuie~verificate!),~constatam~ca \\ \\ sunt~bune. \\ \\ SOLUTIE:~(x,y) \in \{ (- \sqrt{2}, 2 \sqrt{2})~;~( \sqrt{2},-2 \sqrt{2})\}.[/tex]}
