Answer:
[tex]\large\boxed{x>9}[/tex]
Step-by-step explanation:
[tex]6-\dfrac{2}{3}x<x-9\qquad\text{multiply both sides by 3}\\\\3\cdot6-3\!\!\!\!\diagup^1\cdot\dfrac{2}{3\!\!\!\!\diagup_1}x<3x-3\cdot9\\\\18-2x<3x-27\qquad\text{subtract 18 from both sides}\\\\-2x<3x-45\qquad\text{subtract 3x from both sides}\\\\-5x<-45\qquad\text{change the signs}\\\\5x>45\qquad\text{divide both sides by 5}\\\\x>9[/tex]
