Solve the following pair of linear equations by the substitution method.

$\dfrac{3x}{2} - \dfrac{5y}{3} = -2 \text{ and } \dfrac{x}{3} + \dfrac{y}{2} = \dfrac{13}{6}$

Given,

1st equation :

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

2nd equation :

$\Rightarrow \dfrac{x}{3} + \dfrac{y}{2} = \dfrac{13}{6}$ ………(2)

Substituting value of x from equation (1) in equation (2), we get :

$\Rightarrow \dfrac{\dfrac{10y - 12}{9}}{3} + \dfrac{y}{2} = \dfrac{13}{6} \\[1em] \Rightarrow \dfrac{10y - 12}{27} + \dfrac{y}{2} = \dfrac{13}{6} \\[1em] \Rightarrow \dfrac{2(10y - 12) + 27y}{54} = \dfrac{13}{6} \\[1em] \Rightarrow \dfrac{20y - 24 + 27y}{54} = \dfrac{13}{6} \\[1em] \Rightarrow 47y - 24 = \dfrac{13}{6} \times 54 \\[1em] \Rightarrow 47y - 24 = 13 \times 9 \\[1em] \Rightarrow 47y - 24 = 117 \\[1em] \Rightarrow 47y = 117 + 24 \\[1em] \Rightarrow 47y = 141 \\[1em] \Rightarrow y = \dfrac{141}{47} = 3.$

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

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

Hence, x = 2 and y = 3.