Question

алгебра. способ подстановки систем

Answer 1

а)
$- x = 6 - 4y \\ x = 4y - 6 \\ \\ 8y - 6(4y - 6) = 7 \\ 8y - 24y + 36 = 7 \\ - 16y = - 29 \\ y = \frac{29}{16} = 1.8125\\ x = 4 \times \frac{29}{16} - 6 = \frac{29}{4} - 6 = \frac{29 - 24}{4} = \frac{5}{4}$

б)

$- y = 10 - 3x \\ y = 3x - 10 \\ 6x + 2(3x - 10) = 1 \\ 6x + 6x - 20 = 1 \\ 12x = 21 \\ x = \frac{21}{12} = \frac{7}{4} \\ y = 3 \times \frac{7}{4} - 10 = \frac{21 - 40}{4} = - \frac{19}{4}$

Answer 2

Ответ в приложенном фото