Question

X²-2y²+x=-6 x²-3y²=-11​

Answer 1

превет это очень крута а

Answer 2

To solve this system of equations:

Equation 1: x² - 2y² + x = -6 Equation 2: x² - 3y² = -11

We can use a substitution method to eliminate one variable and solve for the other. Let's solve for x in Equation 1:

x² - 2y² + x = -6 Rearrange: x² + x - 2y² = -6 x² + x = 2y² - 6 x(x + 1) = 2y² - 6 x = (2y² - 6) / (x + 1) -- Equation 3

Now we substitute this value of x in Equation 2:

(2y² - 6)² / (x + 1)² - 3y² = -11

Let's simplify this equation step by step:

(2y² - 6)² - 3y²(x + 1)² = -11(x + 1)²

Expand the squared terms:

(4y⁴ - 24y² + 36) - 3y²(x² + 2x + 1) = -11(x² + 2x + 1)

Distribute -3y² and -11:

4y⁴ - 24y² + 36 - 3y²x² - 6y²x - 3y² - 11x² - 22x - 11 = 0

Rearrange the terms:

4y⁴ - (3x² + 30)y² - (6x + 22)x - (11x² + 22x + 47) = 0 -- Equation 4

We now have a quadratic equation in terms of y². Let's solve it:

To solve for y², we can use the quadratic formula:

y² = (-b ± √(b² - 4ac)) / (2a)

In this case: a = 4 b = -(3x² + 30) c = -(6x + 22)x - (11x² + 22x + 47)

Substituting these values into the quadratic formula, we get:

y² = (-(3x² + 30) ± √((3x² + 30)² - 4 * 4 * (-(6x + 22)x - (11x² + 22x + 47)))) / (2 * 4)

Simplifying further, we have:

y² = (-(3x² + 30) ± √(9x⁴ + 180x² + 900 + 96x² + 384x + 1888)) / 8

y² = (-(3x² + 30) ± √(9x⁴ + 276x² + 384x + 2788)) / 8

Now we can substitute these values of y² back into Equation 3 to solve for x:

x = (2((-(3x² + 30) ± √(9x⁴ + 276x² + 384x + 2788))) / 8) - 1

Simplifying:

x = (-(3x² + 30) ± √(9x⁴ + 276x² + 384x + 2788)) / 4 - 1

This is the equation to solve for x. However, solving it directly is quite complex due to the presence of higher-order terms. In