9. \[\begin{cases} 0{,}4+\frac{y}{3}<\frac{2}{3}x-1{,}2,\\ \frac{2x+9}{7}>\frac{5x-4}{4}. \end{cases}\]

\[ \begin{cases} 0{,}4+\frac{y}{3}<\frac{2}{3}x-1{,}2 \;\Rightarrow\; y<2x-4{,}8,\\[2pt] \frac{2x+9}{7}>\frac{5x-4}{4} \;\Rightarrow\; 4(2x+9)>7(5x-4) \;\Rightarrow\; x<\frac{64}{27}. \end{cases} \] Ответ: \(\displaystyle x<\frac{64}{27},\quad y<2x-4{,}8\).