Answer :
[tex]\frac{1 }{3}(7y-2)+ \frac{1}{9}(6y+10) =\frac{7}{3}y-\frac{2}{3}+\frac{6}{9} y + \frac{10}{9}=\frac{7}{3}y-\frac{6}{9}+\frac{2}{3} y + \frac{10}{9}= \\ \\=\frac{9}{3}y+\frac{4}{9}=3y+\frac{4}{9}[/tex]
Simplify the expression 1/3(7y-2)+1/9(6y+10)