Answer:
[tex]\large\boxed{x<-\dfrac{8}{15}}[/tex]
Step-by-step explanation:
[tex]-\dfrac{1}{2}x+\dfrac{1}{3}>\dfrac{3}{5}\\\\\text{Find common denominator of the fractions}\\\\2,\ 3,\ 5\to2\cdot3\cdot5=30\\\\-\dfrac{1}{2}x+\dfrac{1}{3}>\dfrac{3}{5}\qquad\text{multiply both sides by 30}\\\\30\!\!\!\!\!\diagup^{15}\left(-\dfrac{1}{2\!\!\!\!\diagup_1}x\right)+30\!\!\!\!\!\diagup^{10}\left(\dfrac{1}{3\!\!\!\!\diagup_1}\right)>30\!\!\!\!\!\diagup^{6}\left(\dfrac{3}{5\!\!\!\!\diagup_1}\right)\\\\-15x+10>(6)(3)\\\\-15x+10>18\qquad\text{subtract 10 from both sides}[/tex]
[tex]-15x>8\qquad\text{change the signs}\\\\15x<-8\qquad\text{divide both sides by 15}\\\\x<-\dfrac{8}{15}[/tex]
