Question

Решите систему уравнений
x²-3xy=-2
3y²-xy=2​

Answer 1

$\left \{ {\bigg{x^{2} - 3xy = -2,} \atop \bigg{3y^{2} - xy = 2. \ \ \ }} \right.$

$\left \{ {\bigg{x(x - 3y) = -2,} \atop \bigg{y(3y - x) = 2. \ \ \ }} \right.$

$\left \{ {\bigg{x(3y - x) = 2,} \atop \bigg{y(3y - x) = 2. \ }} \right.$

ПОДЕЛИМ ПЕРВОЕ УРАВНЕНИЕ НА ВТОРОЕ:

$\dfrac{x(3y - x)}{y(3y - x)} = \dfrac{2}{2}$

$\dfrac{x}{y} = 1$

$x = y$

$3y^{2} - y^{2} = 2\\2y^{2} = 2\\y^{2} = 1\\y = \pm 1\\x = \pm 1$

Answer 2

Ответ: ( -1; -1 ), ( 1; 1 )

Объяснение:

$\left \{ {{x^2-3xy=-2} \atop {3y^2-xy=2}} \right. \\\left \{ {{x*(x-3y)=-2} \atop {y*(3y-x)=2}} \right. \\x*(x-3y)=-y*(3y-x)\\x*(x-3y)=y*(x-3y)\\x=y\\\\3y^2-y^2=2\\2y^2=2\\y^2=1\\y=-1; y=1\\x=-1; x=1$