Răspuns:
Explicație pas cu pas:
Se egaleaza pur si simplu cele doua ecuatii:
[tex]x^2-3x+1=2x^2+x+4\\x^2-4x+3=0\\\Delta=16-4\cdot 3=16-12=4\Rightarrow \sqrt{\Delta}=2\\x_1=\dfrac{4-2}{2}=\dfrac{2}{2}=1\\\\x_2=\dfrac{4+2}{2}=\dfrac{6}{2}=3\\y=1^2-3+1=-1\\\texttt{Solutiile sistemului sunt:}\\S:(x,y)\in\{(1,-1);(3,-1)\}[/tex]
