Answer:
[tex]x>\frac{10}{3}[/tex]
Step-by-step explanation:
[tex]3/5x+1/3<4/5x-1/3\\\\\frac{3}{5}x+\frac{1}{3}<\frac{4}{5}x-\frac{1}{3}\\\\\mathrm{Subtract\:}\frac{1}{3}\mathrm{\:from\:both\:sides}\\\\\frac{3}{5}x+\frac{1}{3}-\frac{1}{3}<\frac{4}{5}x-\frac{1}{3}-\frac{1}{3}\\\\Simplify\\\\\frac{3}{5}x<\frac{4}{5}x-\frac{2}{3}\\\\\mathrm{Subtract\:}\frac{4}{5}x\mathrm{\:from\:both\:sides}\\\\\frac{3}{5}x-\frac{4}{5}x<\frac{4}{5}x-\frac{2}{3}-\frac{4}{5}x\\\\Simplify\\\\-\frac{1}{5}x<-\frac{2}{3}\\[/tex]
[tex]\mathrm{Multiply\:both\:sides\:by\:-1\:\left(reverse\:the\:inequality\right)}\\\\\left(-\frac{1}{5}x\right)\left(-1\right)>\left(-\frac{2}{3}\right)\left(-1\right)\\\\Simplify\\\\\frac{1}{5}x>\frac{2}{3}\\\\\mathrm{Multiply\:both\:sides\:by\:}5\\\\5\times\frac{1}{5}x>\frac{2\times\:5}{3}\\\\Simplify\\\\x>\frac{10}{3}[/tex]
