Banuiesc ca[tex]\frac{a}{b}=\frac{3}{5}, 5a=3b(1)\\
^{5)}\frac{a}{2a+b}=}\frac{5a}{2\cdot 5a+5\cdot b}=\frac{3b}{6b+5b}=\frac{3b}{11b}=\frac{3}{11}.\\
\frac{a+b}{b}=^{5)}\frac{a+b}{b}=\frac{5a+5b}{5b}=\frac{3b+5b}{5b}=\frac{8b}{5b}=\frac{8}{5}.\\
\frac{2a+b}{a+3b}=^{5)}\frac{2a+b}{a+3b}=\frac{2\cdot5a+5b}{5a+3\cdot5b}=\frac{2\cdot 3b+5b}{3b+15b}=\frac{11b}{18b}=\frac{11}{18}[/tex]
