[tex]\frac{1}{5}(y-3)- \frac{2}{3}(y- \frac{3}{5})+\frac{1}{3} = \frac{2}{5} - \frac{1}{3}(2y-3)\ \ |*15 \\\\3(y-3)- 10(y- \frac{3}{5})+5=6- 5(2y-3)\\\\3y-9-10y+6+5=6-10y+15\\\\-7y+2=21-10y\\\\-7y+10y=21-2\\\\3y=19\ \ |:3\\\\y=\frac{19}{3} =6\frac{1}{3}[/tex]
