Question

Помогите решить номер 8 и 9

Answer 1

$8) \frac{2x + 3}{2x - 1} = \frac{x - 5}{x + 3} \\ (2x + 3) \times (x + 3) = (x - 5)(2x - 1) \\ \\ 2x {}^{2} + 6x + 3x + 9 = 2x {}^{2} - x - 10x + 5 \\ \\ 6 + 3x + 9 = - x - 10x + 5 \\ 9x + 9 = - 11x + 5 \\ 9x + 11x = 5 - 9 \\ 20x = - 4 \\ x = - \frac{1}{5}$

$9) \frac{1 + 3x}{1 - 2x} = \frac{5 - 3x}{1 + 2x} \\ ( 1 + 3x)(1 + 2x) = (5 - 3x)(1 - 2x) \\ \\ 1 + 2x + 3x + 6x {}^{2} = 5 - 10x - 3x + 6x {}^{2} \\ \\ 1 + 2x + 3x = 5 - 10x - 3x \\ \\ 1 + 5x = 5 - 13x \\ 5x + 13x = 5 - 1 \\ 18x = 4 \\ x = \frac{2}{9}$