ГДЗ по алгебре 7 класс Макарычев Задание 465

\[\boxed{\text{465\ (465).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[если\ x = - 3 \Longrightarrow\]

\[\Longrightarrow y = - \frac{2}{3} \cdot ( - 3) = 2;\]

\[если\ x = 3 \Longrightarrow \ \]

\[\Longrightarrow y = - \frac{2}{3} \cdot 3 = - 2;\]

\[если\ x = \frac{2}{3} \Longrightarrow \ \]

\[\Longrightarrow y = - \frac{2}{3} \cdot \frac{2}{3} = - \frac{4}{9};\ \]

\[если\ x = - \frac{2}{3} \Longrightarrow\]

\[\Longrightarrow y = - \frac{2}{3} \cdot \left( - \frac{2}{3} \right) = \frac{4}{9};\ \]

\[если\ x = 2,4 \Longrightarrow\]

\[\Longrightarrow y = - \frac{2}{3} \cdot 2,4 = - 1,6.\]

\[если\ y = 1:\]

\[1 = - \frac{2}{3}x\]

\[x = 1\ :\left( - \frac{2}{3} \right) = - 1 \cdot \frac{3}{2}\]

\[x = - \frac{3}{2} = - 1,5.\]

\[если\ y = - 6:\ \ \]

\[- 6 = - \frac{2}{3}x\]

\[x = - 6\ :\left( - \frac{2}{3} \right) = 6 \cdot \frac{3}{2}\]

\[x = 9.\]

\[если\ y = - 10,2:\]

\[- 10,2 = - \frac{2}{3}x\]

\[x = - 10,2\ :\left( - \frac{2}{3} \right) = \frac{102}{10} \cdot \frac{3}{2} =\]

\[= \frac{153}{10}\]

\[x = 15,3.\]