Question

A)[\ x + y =1,x2 + y2 = 1;6)Slxl+ly1=3(x²y² = 4;​

Answer 1

Ответ:

Смотри на фото

Пошаговое объяснение:

Answer 2

It seems like you have provided two sets of equations. Let's solve them one by one:

Set A:

1. x + y = 1
2. x^2 + y^2 = 1

To solve this system of equations, we can use substitution or elimination method. Let's use the substitution method:

From equation 1, we can express x in terms of y: x = 1 - y

Substituting this value of x in equation 2: (1 - y)^2 + y^2 = 1 Simplifying the equation: 1 - 2y + y^2 + y^2 = 1 2y^2 - 2y = 0 2y(y - 1) = 0

So we have two possible solutions for y:

1. y = 0
2. y - 1 = 0, which gives y = 1

For y = 0: x = 1 - 0 x = 1

For y = 1: x = 1 - 1 x = 0

Therefore, the solutions to Set A are: Solution 1: (x = 1, y = 0) Solution 2: (x = 0, y = 1)

Set B:

1. 6x + y = 3
2. x^2y^2 = 4

We can start by rearranging equation 1 to solve for y: y = 3 - 6x

Substituting this value of y in equation 2: x^2(3 - 6x)^2 = 4 Expanding the equation: x^2(9 - 36x + 36x^2) = 4 9x^2 - 36x^3 + 36x^4 = 4 9x^4 - 36x^3 + 9x^2 - 4 = 0

Unfortunately, this equation cannot be easily solved analytically. You might need to use numerical methods or approximation techniques to find the values of x that satisfy this equation.

I hope this helps you with the given sets of equations. If you have any further questions, please, let me know!