[tex]\dfrac{4}{a}+\dfrac{5}{3a} < 3\ \ \ |-3\ \ \ assume\ a\neq0\\\\\dfrac{3\cdot4}{3a}+\dfrac{5}{3a}-\dfrac{3a\cdot3}{3a} < 0\\\\\dfrac{12+5-9a}{3a} < 0\\\\\dfrac{-9a+17}{3a} < 0\iff 3a(-9a+17) < 0\\\\3a=0\to a=0\\\\-9a+17=0\to a=\dfrac{17}{9}[/tex]
Look at the picture.
[tex]Answer:\ a\in\left(0;\ \dfrac{17}{9}\right)[/tex]
