Question

Пожалуйста,решите данную функцию: {x+xy=4 {xy-6y=1 Смотреть ответ И у меня D<0... в действительных числах решений нет dobra888 dobra888 Ответ:      х  і  у Є∅ .Объяснение:{ x + xy = 4 ,{ -6y + xy = 1 ;    віднімаємо почленно рівняння :{ x + 6y = 3 ,{ x + xy = 4 ;{ x = 3 - 6y ,{ 3 - 6y + ( 3 - 6y )*y = 4;{ x = 3 - 6y ,{ 3 - 6y + 3y - 6y² = 4 ;{ x = 3 - 6y ,{ 6y²+ 3y + 1 = 0 ;  рішаємо ІІ -е рівняння :      6y²+ 3y + 1 = 0 ;    D = - 15 < 0 , тому це рівняння , а разом із нимі початкова система рівнянь не мають розв"язків .

Answer 1

Solution to the Given System of Equations

The given system of equations is:

x + xy = 4
xy - 6y = 1

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

From the first equation, we can express `x` in terms of `y` as:

x = 4 - xy

Substituting this value of `x` into the second equation, we get:

(4 - xy)y - 6y = 1

Simplifying this equation, we have:

4y - xy^2 - 6y = 1

Rearranging the terms, we get:

-xy^2 + 4y - 6y = 1

Combining like terms, we have:

-xy^2 - 2y = 1

Now, let's solve this quadratic equation for `y`. However, you mentioned that `D < 0`, which means the discriminant is negative. In real numbers, a quadratic equation with a negative discriminant does not have any solutions. Therefore, the given system of equations does not have any solutions in real numbers.

In summary, the solution to the given system of equations is empty set (no solutions exist in real numbers).

Please let me know if you have any further questions!