Question

Решите систему уравнений

Answer 1

Ответ:

из первый уравнений найдем у

$\frac{1}{2x - 3y} + x = 1 \\ 1 + 2 {x}^{2} - 3xy = 2x - 3y \\ 3xy - 3y = 2 {x}^{2} - 2x + 1 \\ y = \frac{2 {x}^{2} - 2x + 1 }{3x - 3}$

и постави на второе уравнение

$\frac{x}{2x - 3( \frac{2 {x}^{2} - 2x + 1}{3x - 3} )} = - 6 \\ x = - 12x + 6( \frac{2 {x}^{2} - 2x + 1}{x - 1} ) \\ {x}^{2} - x = - 12 {x}^{2} + 12x + 12 {x}^{2} - 12x + 6 \\ {x}^{2} - x - 6 = 0 \\ d = 1 + 24 = 25 \\ x1 = \frac{1 + 5}{2} = 3 \\ x2 = \frac{1 - 5}{2} = - 2$

$y1 = \frac{2 \times {3}^{2 } - 2 \times 3 + 1}{3 \times 3 - 3} = \frac{13}{6} \\ y2 = \frac{2 \times { (- 2)}^{2} - 2 \times ( - 2) + 1 }{3 \times ( - 2) - 3} = - \frac{13}{9}$