Question

Решите систему уравненийх/у = 1/3х^2 + у^2 = 40​

Answer 1

$\left \{ {{\frac{x}{y}=\frac{1}{3}} \atop {x^{2}+y^{2}=40}} \right.\\\\\left \{ {{y=3x} \atop {x^{2}+(3x)^{2}=40}} \right. \\\\\left \{ {{y=3x} \atop {x^{2} +9x^{2}=40 }} \right.\\\\\left \{ {{y=3x} \atop {10x^{2}=40}} \right.\\\\\left \{ {{y=3x} \atop {x^{2}=4}} \right. \\\\1)x_{1}=-2\\\\y_{1}=3*(-2)=-6\\\\2)x_{2}=2\\\\y_{2}=3*2=6\\\\Otvet:\boxed{(-2;-6),(2;6)}$

Answer 2

Ответ:

(2; 6); (-2; -6); x = 2, y = 6; x = -2, y = -6.

Объяснение:

x : y = 1 : 3,

x^2 + y^2 = 40.

y = 3x,

x^2 + (3x)^2 = 40.

y = 3x,

x^2 + 9x^2 = 40.

y = 3x,

10x^2 = 40.

y = 3x,

x^2 = 4 (40 : 10).

y = 6,

x = 2.

y = -6,

x = -2.