[tex]\text{Mai intai sa observam ca este un sistem omogen(daca admite solutia }\\
\text{(a,b) atunci admite si solutie (b,a))}\\
a){ \left \{ {{xy+x+y=5|\cdot 2} \atop {x^2+y^2=5}} \right.\Leftrightarrow \left \{ {{2xy+2x+2y=10} \atop {x^2+y^2=5} \right.} \\
\text{Adunand relatiile obtinem :} x^2+y^2+2xy+2x+2y=15|+1\\
x^2+y^2+2xy+2x+2y+1=16\\
(x+y+1)^2=16\\
|x+y+1|=4\\
\text{Deci avem de analizat doua cazuri:}\\
i)x+y+1=4,\text{ adica }x+y=3\text{ de unde x=3-y}\\
(3-y)^2+y^2=5\\
9-6y+y^2+y^2=5[/tex][tex]2y^2-6y+4=0\\
y^2-3y+2=0\\
(y-1)(y-2)=0\Rightarrow y\in \{1,2\},\text{de unde obtinem solutiile :}(1,2),(2,1)\\
ii)x+y+1=-4\Rightarrow x+y=-5,\text{ deci x=-5-y}\\
(-5-y)^2+y^2=5\\
25+10y+y^2+y^2=5\\
2y^2+10y+20=0\\
y^2+5y+10=0\\
\Delta =25-40=-20\ \textless \ 0 \text{deci nu are solutii}\\
\text{Prin urmare }S:(x,y)\in \{(1,2),(2,1)\}[/tex][tex]b)\text{Observam ca si acesta este un sistem omogen si }x,y\neq 0\\
\dfrac{x}{y}+\dfrac{y}{x}=\dfrac{5}{2}|\cdot 2xy\\
2x^2+2y^2=5xy\\
2x^2-5xy+2y^2=0\\
2x^2-4xy-xy+2y^2=0\\
2x(x-2y)-y(x-2y)=0\\
(x-2y)(2x-y)=0\\
\text{Iar distingem doua cazuri posibile }\\
i)x-2y=0\Rightarrow x=2y\\
\text{Inlocuim in a doua ecuatie si obtinem:}\\
(2y)^2\cdot y+2y(y^2)=6\\
4y^3+2y^3=6\\
6y^3=6\\
y^3=1\Rightarrow y=1\text{ si }x=2
[/tex][tex]ii)2x-y=0\Rightarrow x=\dfrac{y}{2}\\
\dfrac{y^2}{4}\cdot y+y^2\cdot \dfrac{y}{2}=6|\cdot 4\\
y^3+2y^3=24\\
3y^3=24\\
y^3=8\Rightarrow y=2\text{ si }x=1\\
\text{Prin urmare }S:(x,y)\in \{(1,2),(2,1)\}[/tex]
