Show that: $\left(\dfrac{1}{2} + \dfrac{3}{4}\right) \times \dfrac{8}{3} = \dfrac{1}{2} \times \dfrac{8}{3} + \dfrac{3}{4} \times \dfrac{8}{3}$.

Given,

Equation : $\left(\dfrac{1}{2} + \dfrac{3}{4}\right) \times \dfrac{8}{3} = \dfrac{1}{2} \times \dfrac{8}{3} + \dfrac{3}{4} \times \dfrac{8}{3}$

Solving L.H.S. of the equation :

$\Rightarrow \left(\dfrac{1}{2} + \dfrac{3}{4}\right) \times \dfrac{8}{3} \\[1em] \Rightarrow \left(\dfrac{2}{4} + \dfrac{3}{4}\right) \times \dfrac{8}{3} \quad \text{[L.C.M. of 2 and 4 is 4]} \\[1em] \Rightarrow \dfrac{5}{4} \times \dfrac{8}{3} \\[1em] \Rightarrow \dfrac{40}{12} \\[1em] \Rightarrow \dfrac{10}{3}.$

Solving R.H.S. of the equation :

$\Rightarrow \dfrac{1}{2} \times \dfrac{8}{3} + \dfrac{3}{4} \times \dfrac{8}{3} \\[1em] \Rightarrow \dfrac{8}{6} + \dfrac{24}{12} \\[1em] \Rightarrow \dfrac{4}{3} + 2 \\[1em] \Rightarrow \dfrac{4}{3} + \dfrac{6}{3} \quad \text{[L.C.M. of 3 and 1 is 3]} \\[1em] \Rightarrow \dfrac{10}{3}.$

Since, L.H.S. = R.H.S.

Hence, proved that $\left(\dfrac{1}{2} + \dfrac{3}{4}\right) \times \dfrac{8}{3} = \dfrac{1}{2} \times \dfrac{8}{3} + \dfrac{3}{4} \times \dfrac{8}{3}$.