Question

Система уравнений, помогите решить {x^2+3xy+9y^2=12 x^2+3xy+2y^2=0

Answer 1

{x²+3xy+9y² =12 ; x²+3xy+2y²=0 .* * * вычитаем от первого урав второе * * *  
 {y² =12/7; x²+3xy+2y²=0 .
 y = ±2√(3/7)
у₁ =-2√(3/7) ⇒x² -6√(3/7)* y +24/7 =0  .

у₁ =2√(3/7  ⇒x² +6√(3/7)* y +14 =0 ..

и т.д.    (поздно, пара спать)

Answer 2

Solving the System of Equations

To solve the system of equations x^2+3xy+9y^2=12 and x^2+3xy+2y^2=0, we can use the method of substitution or elimination. Let's proceed with the method of substitution.

First, let's solve the second equation for x in terms of y: x^2 + 3xy + 2y^2 = 0

x^2 = -3xy - 2y^2 x = -3y - 2y^2/x

Now, we can substitute this expression for x into the first equation: (-3y - 2y^2/x)^2 + 3(-3y - 2y^2/x)y + 9y^2 = 12

Solving this equation will give us the values of y, and then we can find the corresponding values of x.

Let's proceed with solving the equation: (-3y - 2y^2/x)^2 + 3(-3y - 2y^2/x)y + 9y^2 = 12

This will give us the solution for the system of equations.