Решите систему: x^2+3y^2=7 и xy+y^2=3

Решите задачу:

$x^2+3y^2=7\\ xy+y^2=3\\ \\ y^2=\frac{7-x^2}{3}\\ x\sqrt{\frac{7-x^2}{3}}+\frac{7-x^2}{3}=3\\ x\sqrt{\frac{7-x^2}{3}}=3-\frac{7-x^2}{3}\\ x^2*\frac{7-x^2}{3}=\frac{x^4+4x^2+4}{9}\\ \frac{21x^2-3x^4}{9}=\frac{x^4+4x^2+4}{9}\\ 21x^2-3x^4=x^4+4x^2+4\\ 4x^4-17x^2+4=0\\ x^2=t\\ D=289-4^3=15^2\\ x_{1}=\frac{17+15}{8}=4\\ x_{2}=\frac{17-15}{8}=\frac{1}{4}\\ x=+-2\\ x=+-0.5\\ y=+-1\\ y=+-\frac{3}{2}$