Problems from Extracting common factor

Development:

1. $$\dfrac{1}{5}\cdot\dfrac{2}{3}+\dfrac{1}{3}\cdot\dfrac{1}{5}-\dfrac{1}{3}\cdot\dfrac{1}{5}=\dfrac{1}{5}\cdot\Big(\dfrac{2}{3}+\dfrac{1}{3}-\dfrac{1}{3}\Big)=$$ $$=\dfrac{1}{5}\cdot\dfrac{2}{3}=\dfrac{2}{15}$$

2. $$\dfrac{3}{5}\cdot\dfrac{a}{b}-\dfrac{2}{b}\cdot\dfrac{a}{3}+\dfrac{a}{b}\cdot\dfrac{1}{3}=\dfrac{3}{5}\cdot\dfrac{a}{b}-\dfrac{a}{b}\cdot\dfrac{2}{3}+\dfrac{a}{b}\cdot\dfrac{1}{3}=$$ $$=\dfrac{3}{5}\cdot\dfrac{a}{b}+\dfrac{a}{b}\cdot\Big(-\dfrac{2}{3}\Big)+\dfrac{a}{b}\cdot\dfrac{1}{3}= \dfrac{a}{b}\cdot \Big(\dfrac{3}{5}-\dfrac{2}{3}+\dfrac{1}{3}\Big)=$$ $$\dfrac{a}{b}\cdot \Big(\dfrac{9}{15}-\dfrac{10}{15}+\dfrac{5}{15}\Big)= \dfrac{a}{b}\cdot \Big(\dfrac{9-10+5}{15}\Big)=\dfrac{a}{b}\cdot\dfrac{4}{15}=\dfrac{4a}{15b}$$

Solution:

1. $$\dfrac{2}{15}$$
2. $$\dfrac{4a}{15b}$$

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