Задание 292 Вариант 3 Самостоятельная работа ГДЗ Дидактические материалы Чесноков 6 класс (Математика)

а)

$$\frac{1}{12}y-\frac{1}{9}y-\frac{1}{4}y=1\frac{2}{3}$$
$$\left(\frac{1}{12}-\frac{1}{9}-\frac{1}{4}\right)y=\frac{5}{3}$$
$$\left(\frac{3}{36}-\frac{4}{36}-\frac{9}{36}\right)y=\frac{5}{3}$$
$$-\frac{10}{36}y=\frac{5}{3}$$
$$-\frac{5}{18}y=\frac{5}{3}$$
$$y=\frac{5}{3}:\left(-\frac{5}{18}\right)$$
$$y=\frac{5}{3}\cdot\left(-\frac{18}{5}\right)=-6$$

б)

$$0{,}6z-\frac{2}{9}z=-3{,}4$$
$$\left(0{,}6-\frac{2}{9}\right)z=-3{,}4$$
$$\left(\frac{6}{10}-\frac{2}{9}\right)z=-3{,}4$$
$$\left(\frac{3}{5}-\frac{2}{9}\right)z=-3{,}4$$
$$\left(\frac{27}{45}-\frac{10}{45}\right)z=-\frac{34}{10}$$
$$\frac{17}{45}z=-\frac{17}{5}$$
$$z=-\frac{17}{5}:\frac{17}{45}$$
$$z=-\frac{17}{5}\cdot\frac{45}{17}=-9$$

Ответ: а) $$y=-6$$; б) $$z=-9$$.