$\frac{1}{3}x+\frac{4}{3}y=\frac{1}{3}\ast x+\frac{1}{3}\ast 4y=\frac{1}{3}(x+4y)$

$\frac{8}{9}a-\frac{16}{27}b=\frac{24}{27}a-\frac{16}{27}b=\frac{8}{9}\ast a-\frac{8}{9}\ast \frac{2}{3}b=\frac{8}{9}(a-\frac{2}{3}b)$

$\frac{18}{25}a+\frac{12}{35}y=\frac{6}{5}\ast \frac{3}{5}a+\frac{6}{5}\ast \frac{2}{7}y=\frac{6}{5}(\frac{3}{5}a+\frac{2}{7}y)$

$\frac{12}{49}x-\frac{3}{28}y=\frac{3}{7}\ast \frac{4}{7}x-\frac{3}{7}\ast \frac{1}{4}y=\frac{3}{7}(\frac{4}{7}x-\frac{1}{4}y)$