Question

Помогите пожалуйста решить систему уравнений:

Answer 1

$x-y=4\\ xy=5\\ \\ x=4+y\\ (4+y)y=5\\ 4y+y^2-5=0\\ D=16+4*5=6^2\\ y_{1}=\frac{-4+6}{2}=1\\ y_{2}=\frac{-4-6}{2}=-5\\ x=5\\ x=-1$

Answer 2

System of Equations

To solve the system of equations, we need to find the values of x and y that satisfy both equations. The system of equations is as follows:

Equation 1: x^2 + xy = -4 Equation 2: y^2 + xy = 8

To solve this system, we can use various methods such as substitution, elimination, or matrix methods. Let's use the substitution method to solve this system.

Substitution Method

We can solve Equation 1 for x and substitute it into Equation 2 to eliminate x. Let's solve Equation 1 for x:

x^2 + xy = -4 x(x + y) = -4 x = -4 / (x + y) Now, substitute the value of x from Equation 1 into Equation 2:

(-4 / (x + y))^2 + (-4 / (x + y))y = 8 16 / (x + y)^2 - 4y / (x + y) = 8 16 - 4y(x + y) = 8(x + y)^2 16 - 4xy - 4y^2 = 8x^2 + 16xy + 8y^2 8x^2 + 4xy + 12y^2 - 4xy - 4y^2 - 16 = 0 8x^2 + 8y^2 - 16 = 0 x^2 + y^2 - 2 = 0 x^2 + y^2 = 2 Now we have a new equation, x^2 + y^2 = 2. Let's substitute the value of x from Equation 1 into this equation:

(-4 / (x + y))^2 + y^2 = 2 16 / (x + y)^2 + y^2 = 2 16 + y^2(x + y)^2 = 2(x + y)^2 16 + y^2x^2 + 2y^3x + y^4 = 2x^2 + 4xy + 2y^2 14x^2 + 2y^3x + y^4 - 4xy - 2y^2 - 16 = 0 Now we have a quadratic equation in terms of x and y. We can solve this equation using numerical methods or by factoring. However, since the equation is quite complex, it may not have simple solutions.

Unfortunately, without further information or constraints, it is not possible to determine the exact values of x and y that satisfy the system of equations. Additional information or constraints are needed to find a unique solution.

Please let me know if there is anything else I can help you with!