Question

(2x+y/x-2y)-(3(x-2y)/2x+y)=2 x^2+3xy-y^2=23

Answer 1

Ответ:

Объяснение:

------------------------------------------

Answer 2

To solve the system of equations:

(2x + y)/(x - 2y) - (3(x - 2y))/(2x + y) = 2

x^2 + 3xy - y^2 = 23

We can start by simplifying the first equation:

[(2x + y)(2x + y) - 3(x - 2y)(x - 2y)] / [(x - 2y)(2x + y)] = 2

Simplifying further:

[(4x^2 + 4xy + y^2) - (3x^2 - 6xy - 3y^2)] / [(x - 2y)(2x + y)] = 2

Combining like terms:

(4x^2 + 4xy + y^2 - 3x^2 + 6xy + 3y^2) / [(x - 2y)(2x + y)] = 2

Simplifying the numerator:

(x^2 + 10xy + 4y^2) / [(x - 2y)(2x + y)] = 2

Now, we can cross-multiply:

(x^2 + 10xy + 4y^2) = 2(x - 2y)(2x + y)

Expanding the right side:

x^2 + 10xy + 4y^2 = 4x^2 - 8xy + 2xy - 4y^2

Combining like terms:

x^2 + 10xy + 4y^2 = 4x^2 - 6xy - 4y^2

Bringing all terms to one side:

3x^2 - 16xy + 8y^2 = 0

Now, we have a quadratic equation in two variables. We can try to factor it:

(3x - 4y)(x - 2y) = 0

This equation holds true if either (3x - 4y) = 0 or (x - 2y) = 0.

Case 1: (3x - 4y) = 0

Solving this equation, we get:

3x = 4y x = (4/3)y

Substituting this into the second equation:

(4/3)y^2 + 3(4/3)y^2 - y^2 = 23

(4/3)y^2 + 4y^2 - y^2 = 23

(4/3 + 4 - 1)y^2 = 23

(7/3)y^2 = 23

y^2 = (3/7) * 23

y^2 = 69/7

y = ± √(69/7)

Using these values of y, we can find the corresponding values of x using x = (4/3)y.

Case 2: (x - 2y) = 0

Solving this equation, we get:

x = 2y

Substituting this into the second equation:

(2y)^2 + 3(2y)(y) - y^2 = 23

4y^2 + 6y^2 - y^2 = 23

9y^2 = 23

y^2 = 23/9

y = ± √(23/9)

Using these values of y, we can find the corresponding values of x using x = 2y.

In summary, the solution to the system of equations is given by the values of (x, y) obtained from the cases described above:

Case 1: x = (4/3)y, y = ± √(69/7) Case 2: x = 2y, y = ± √(23/9)