Question

{2(x+y)-x=-6 {3x-(x-y)=0

Answer 1

2x+2y-x= - 6
2x+y=0

x+2y = - 6
2x+y=0

x+2y= - 6
- 4x-2y=0

- 3x= - 6
2x+y=0

x=2
y= - 4

Ответ:(2; - 4)

Answer 2

I can help you with your math problem. Here is the detailed solution:

To solve the system of equations, we can use the substitution method. First, we isolate one variable in one equation. For example, from the second equation, we can get:

$$y = x - \frac{3x}{x-y}$$

Then, we substitute this expression for y into the first equation and simplify:

$$2(x + x - \frac{3x}{x-y}) - x = -6$$ $$x - \frac{6x}{x-y} = -6$$ $$x^2 - 6x - 6xy = -6x + 6y$$ $$x^2 - 12x - 6xy + 6y = 0$$

Next, we can use the quadratic formula to find the possible values of x:

$$x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(-6)(-6y + 6y)}}{2(-6)}$$ $$x = \frac{12 \pm \sqrt{144}}{-12}$$ $$x = \frac{12 \pm 12}{-12}$$ $$x = -1 \text{ or } x = 0$$

Finally, we can plug these values of x into the expression for y and find the corresponding values of y:

$$y = x - \frac{3x}{x-y}$$ $$y = -1 - \frac{3(-1)}{-1-y} \text{ or } y = 0 - \frac{3(0)}{0-y}$$ $$y = -1 + \frac{3}{1+y} \text{ or } y = 0$$

To find the exact values of y, we need to solve the rational equation for y:

$$-1 + \frac{3}{1+y} = y$$ $$-1 + y + \frac{3}{1+y} = 0$$ $$(-1 + y)(1 + y) + 3 = 0$$ $$y^2 - 2y - 4 = 0$$

Using the quadratic formula again, we get:

$$y = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-4)}}{2(1)}$$ $$y = \frac{2 \pm \sqrt{20}}{2}$$ $$y = 1 \pm \sqrt{5}$$

Therefore, the solutions of the system are:

$$(x, y) = (-1, 1 + \sqrt{5}) \text{ or } (0, 0)$$

I hope this helps you understand how to solve systems of equations. If you have any other questions, feel free to ask me.Ответь+подробно.+{2(x+y)-x=-6+{3x-(x-y)=0