Question

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

Answer 1

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

Ответ: x = y = 0    и    x = y = 2