Solve the following pair of (simultaneous) equations using method of elimination by substitution :

$\dfrac{3x}{2} - \dfrac{5y}{3} + 2 = 0$

$\dfrac{x}{3} + \dfrac{y}{2} = 2\dfrac{1}{6}$

Given,

Equations : $\dfrac{3x}{2} - \dfrac{5y}{3} + 2 = 0, \dfrac{x}{3} + \dfrac{y}{2} = 2\dfrac{1}{6}$

$\Rightarrow \dfrac{3x}{2} - \dfrac{5y}{3} + 2 = 0 \\[1em] \Rightarrow \dfrac{9x - 10y + 12}{6} = 0 \\[1em] \Rightarrow 9x - 10y + 12 = 0 \\[1em] \Rightarrow 10y = 9x + 12 \\[1em] \Rightarrow y = \dfrac{9x + 12}{10} …….(1)$

Substituting value of y from equation (1) in $\dfrac{x}{3} + \dfrac{y}{2} = 2\dfrac{1}{6}$, we get :

$\Rightarrow \dfrac{x}{3} + \dfrac{\dfrac{9x + 12}{10}}{2} = 2\dfrac{1}{6} \\[1em] \Rightarrow \dfrac{x}{3} + \dfrac{9x + 12}{20} = \dfrac{13}{6} \\[1em] \Rightarrow \dfrac{20x + 3(9x + 12)}{60} = \dfrac{13}{6} \\[1em] \Rightarrow \dfrac{20x + 27x + 36}{60} = \dfrac{13}{6} \\[1em] \Rightarrow \dfrac{47x + 36}{60} = \dfrac{13}{6} \\[1em] \Rightarrow 47x + 36 = \dfrac{13}{6} \times 60 \\[1em] \Rightarrow 47x + 36 = 130 \\[1em] \Rightarrow 47x = 94 \\[1em] \Rightarrow x = \dfrac{94}{47} = 2.$

Substituting value of x in equation (1), we get :

$\Rightarrow y = \dfrac{9x + 12}{10} \\[1em] \Rightarrow y = \dfrac{9 \times 2 + 12}{10} \\[1em] \Rightarrow y = \dfrac{18 + 12}{10} \\[1em] \Rightarrow y = \dfrac{30}{10} = 3.$

Hence, x = 2 and y = 3.