\begin{cases} \frac{x - 2y}{x + y} - \frac{3x + 3y}{x - 2y} = 2, \\ x + 2y = -5. \end{cases}

Ответ: x = -1, y = -2

\[x = -5 - 2y\]\[\frac{(-5 - 2y) - 2y}{(-5 - 2y) + y} - \frac{3(-5 - 2y) + 3y}{(-5 - 2y) - 2y} = 2\]\[\frac{-5 - 4y}{-5 - y} - \frac{-15 - 3y}{-5 - 4y} = 2\]\[\frac{-5 - 4y}{-5 - y} - \frac{-15 - 3y}{-5 - 4y} = 2\]\[\frac{(-5 - 4y)(-5 - 4y) - (-15 - 3y)(-5 - y)}{(-5 - y)(-5 - 4y)} = 2\]\[\frac{(25 + 40y + 16y^2) - (75 + 15y + 15y + 3y^2)}{25 + 25y + 4y^2} = 2\]\[\frac{25 + 40y + 16y^2 - 75 - 30y - 3y^2}{25 + 25y + 4y^2} = 2\]\[\frac{13y^2 + 10y - 50}{4y^2 + 25y + 25} = 2\]\[13y^2 + 10y - 50 = 2(4y^2 + 25y + 25)\]\[13y^2 + 10y - 50 = 8y^2 + 50y + 50\]\[5y^2 - 40y - 100 = 0\]\[y^2 - 8y - 20 = 0\]\[y = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(-20)}}{2(1)}\]\[y = \frac{8 \pm \sqrt{64 + 80}}{2}\]\[y = \frac{8 \pm \sqrt{144}}{2}\]\[y = \frac{8 \pm 12}{2}\]\[y_1 = \frac{8 + 12}{2} = 10, \quad y_2 = \frac{8 - 12}{2} = -2\]\[x = -5 - 2(10) = -5 - 20 = -25\]\[x = -5 - 2(-2) = -5 + 4 = -1\]\[\frac{-25 - 20}{-25 + 10} - \frac{3(-25) + 3(10)}{-25 - 20} = \frac{-45}{-15} - \frac{-75 + 30}{-45} = 3 - \frac{-45}{-45} = 3 - 1 = 2\]\[x + 2y = -25 + 2(10) = -25 + 20 = -5\]\[\frac{-1 - 2(-2)}{-1 + (-2)} - \frac{3(-1) + 3(-2)}{-1 - 2(-2)} = \frac{-1 + 4}{-3} - \frac{-3 - 6}{-1 + 4} = \frac{3}{-3} - \frac{-9}{3} = -1 - (-3) = -1 + 3 = 2\]\[x + 2y = -1 + 2(-2) = -1 - 4 = -5\]

Ответ: x = -1, y = -2