5) {4xy - y = -40, 5x - 4xy = 27;

Сложим два уравнения: $$(4xy - y) + (5x - 4xy) = -40 + 27$$ $$5x - y = -13$$ $$y = 5x + 13$$ Подставим в первое уравнение: $$4x(5x + 13) - (5x + 13) = -40$$ $$20x^2 + 52x - 5x - 13 = -40$$ $$20x^2 + 47x + 27 = 0$$ $$D = 47^2 - 4 \cdot 20 \cdot 27 = 2209 - 2160 = 49$$ $$x_1 = \frac{-47 + \sqrt{49}}{2 \cdot 20} = \frac{-47 + 7}{40} = \frac{-40}{40} = -1$$ $$x_2 = \frac{-47 - \sqrt{49}}{2 \cdot 20} = \frac{-47 - 7}{40} = \frac{-54}{40} = -\frac{27}{20}$$ Тогда: $$y_1 = 5(-1) + 13 = -5 + 13 = 8$$ $$y_2 = 5(-\frac{27}{20}) + 13 = -\frac{27}{4} + \frac{52}{4} = \frac{25}{4}$$ Ответ: (-1; 8), (-27/20; 25/4)