Question

$(2\frac{1}{2}+3\frac{1}{3})+\frac{2}{9}(3\frac{1}{8}-2\frac{1}{12})$

Answer 1

$\left(2\right(\frac{1}{2})+3(\frac{1}{3}\left)\right)+\left(\frac{2}{9}\right)\left(3\right(\frac{1}{8})-2(\frac{1}{12}\left)\right)=\left(\right(\frac{5}{2})+(\frac{10}{3}\left)\right)+\left(\frac{2}{9}\right)\left(\right(\frac{25}{8})-(\frac{25}{12}\left)\right)=\left(\frac{15+20}{6}\right)+\left(\frac{2}{9}\right)\left(\right(\frac{75-50}{24}\left)\right)=\left(\frac{35}{6}\right)+\left(\frac{2}{9}\right)\times \left(\frac{25}{24}\right)$

$=\left(\frac{35}{6}\right)+\left(\frac{25}{108}\right)$

$=\frac{630+25}{108}$

$=\frac{655}{108}$

$=6\frac{7}{108}$