Solve the following pairs of linear (simultaneously) equations using method of elimination by substitution:

$\dfrac{2x + 1}{7} + \dfrac{5y - 3}{3} = 12$

$\dfrac{3x + 2}{2} - \dfrac{4y + 3}{9} = 13$

Simplifying first equation :

$\Rightarrow \dfrac{2x + 1}{7} + \dfrac{5y - 3}{3} = 12 \\[1em] \Rightarrow \dfrac{3(2x + 1) + 7(5y - 3)}{21} = 12 \\[1em] \Rightarrow 6x + 3 + 35y - 21 = 12 \times 21 \\[1em] \Rightarrow 6x + 35y - 18 = 252 \\[1em] \Rightarrow 6x + 35y = 252 + 18 \\[1em] \Rightarrow 6x + 35y = 270 \\[1em] \Rightarrow 6x = 270 - 35y \\[1em] \Rightarrow x = \dfrac{270 - 35y}{6} \text{ ……..(1)}$

Simplifying second equation :

$\Rightarrow \dfrac{3x + 2}{2} - \dfrac{4y + 3}{9} = 13 \\[1em] \Rightarrow \dfrac{9(3x + 2) - 2(4y + 3)}{18} = 13 \\[1em] \Rightarrow 27x + 18 - 8y - 6 = 18 \times 13 \\[1em] \Rightarrow 27x - 8y + 12 = 234 \\[1em] \Rightarrow 27x - 8y = 234 -12 \\[1em] \Rightarrow 27x - 8y = 222 \text{ …….(2)}$

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

$\Rightarrow 27 \times \Big(\dfrac{270 - 35y}{6}\Big) - 8y = 222 \\[1em] \Rightarrow \dfrac{9(270 - 35y)}{2} - 8y = 222 \\[1em] \Rightarrow \dfrac{2430 - 315y - 16y}{2} = 222 \\[1em] \Rightarrow 2430 - 331y = 444 \\[1em] \Rightarrow 331y = 2430 - 444 \\[1em] \Rightarrow 331y = 1986 \\[1em] \Rightarrow y = \dfrac{1986}{331} = 6.$

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

$\Rightarrow x = \dfrac{270 - 35 \times 6}{6} \\[1em] = \dfrac{270 - 210}{6} \\[1em] = \dfrac{60}{6} \\[1em] = 10.$

Hence, x = 10 and y = 6.