X^2 + xy = 4y y^2 + xy = 4x решите СИСТЕМУ уравнений

$x^2 + xy = 4y \\ y^2 + xy = 4x \\ \\ x^2 = 4y - xy \\ y^2 + xy = 4x \\ \\ x^2 = y(4-x) \\ y^2 + xy = 4x \\ \\ y = \frac{x^2}{4-x} \\ \frac{x^4}{(4-x)^2} + x*\frac{x^2}{4-x} = 4x \\ \\ x^4 + x^3(4-x) = 4x(4-x)^2 \\ \\ x^4 +4x^3 -4x^4 = 64x -32x^2 + 4x^3 \\ \\ 64x - 32x^2 =0 \\ \\ x(x-2) = 0 \\ \\ x_1 = 0 \:\:\:\:\:\: y_1 = \frac{0^2}{4-0} =0 \\ \\ x_2 = 2 \:\:\:\:\:\: y_1 = \frac{2^2}{4-2} =2$

Ответ: x = y = 0 и x = y = 2