Show that

$\left(\frac{1}{2} + \frac{3}{4}\right) \times \frac{8}{3} = \left(\frac{1}{2}\right) \times \left(\frac{8}{3}\right) + \left(\frac{3}{4}\right) \times \left(\frac{8}{3}\right). \text{ [Page No. 49]}$

$$\quad\text{LHS}=\frac{1}{2}+\frac{3}{4}=\frac{2}{4}+\frac{3}{4}=\frac{5}{4}$$

$$\frac{5}{4} \times \frac{8}{3} = \frac{40}{12} = \frac{10}{3}$$

$$\text{RHS} = \left(\frac{1}{2}\right)\left(\frac{8}{3}\right) + \left(\frac{3}{4}\right)\left(\frac{8}{3}\right)$$

$$= \frac{8}{6} + \frac{24}{12} = \frac{4}{3} + \frac{6}{3} = \frac{10}{3}$$

$$\text{LHS} = \text{RHS}$$

Hence, verified.