[tex]\Big\{x^2+y^2 = 74 \\ \Big\{xy+x+y = 47 \\ \\ (x+y)^2 = x^2+2xy+y^2 \Rightarrow x^2+y^2 = (x+y)^2-2xy \\ \\ $Avem:$ \\ \\ \Big\{(x+y)^2-2xy = 74 \\ \Big\{x+y+xy = 47\Big|\cdot 2 \\ \\ \Rightarrow \\ \\ \Big\{(x+y)^2-2xy = 74 \\ \Big\{2(x+y)+2xy = 94 \\ ----------($adunam$) \\ (x+y)^2+2(x+y) = 168 \Rightarrow (x+y)^2+2(x+y) - 168 = 0 \\ \\ $Notam x+y = t [/tex]
[tex]\Rightarrow t^2+2t = 168 \Rightarrow t(t+2) = 168 \\ \\ 12\cdot 14 = 168 \Rightarrow 12\cdot(12+2) 168\\ (-12)\cdot(-14) = 168 \Rightarrow -14\cdot(-14+2) = 168 \\ \\ \Rightarrow t_1 = 12,\quad t = -14 \\ \\ \boxed{1}\quad x+y = 12 \\ \\ \quad \Big\{x+y = 12 \\\Big\{xy+x+y = 47 \\ \\ \Big\{x+y = 12 \\\Big\{xy+12= 47\\ \\ \Big\{x+y = 12 \\\Big\{xy= 35 \\ \\ \Rightarrow x = 5, \quad y=7 \quad sau \quad y = 5, \quad x = 7\quad ($se observa direct$) \\ (x,y) = \Big\{(5,7); (7,5)\Big\} [/tex]
[tex]\boxed{2} \quad x+y = -14 \\ \\ \Big\{x+y = -14 \\ \Big\{xy -14 = 47 \\ \\ \Big\{x+y = -14 \\ \Big\{xy = 61 \\ \\ $Aici totusi sa il rezolvam fara sa ghicim solutiile$ \\ \\ $Consideram$: t^2-St+P = 0 \Rightarrow t^2-\cdot(-14)t +61 = 0 \Rightarrow \\ \Rightarrow t^2+14t+61 = 0 \\ \Delta = 14\cdot 14- 4\cdot 61 = 196-244 = -48 \ \textless \ 0 \Rightarrow t\notin \mathbb_{R} $ \\ $ \ $Deci, aici nu avem solutii reale.$ [/tex]
[tex]
\\ \\ $Din \boxed{1} \cup $ $ \boxed{2}$\Rightarrow \boxed{(x,y) = \Big\{(5,7); (7,5)\Big\}}[/tex]
