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

$\dfrac{x}{6} + \dfrac{y}{15} = 4$

$\dfrac{x}{3} - \dfrac{y}{12} = 4\dfrac{3}{4}$

Given,

Equations : $\dfrac{x}{6} + \dfrac{y}{15} = 4, \dfrac{x}{3} - \dfrac{y}{12} = 4\dfrac{3}{4}$

$\Rightarrow \dfrac{x}{6} + \dfrac{y}{15} = 4 \\[1em] \Rightarrow \dfrac{x}{6} = 4 - \dfrac{y}{15} \\[1em] \Rightarrow \dfrac{x}{6} = \dfrac{60 - y}{15} \\[1em] \Rightarrow x = \dfrac{6(60 - y)}{15} \\[1em] \Rightarrow x = \dfrac{2(60 - y)}{5} \\[1em] \Rightarrow x = \dfrac{120 - 2y}{5} ……(1)$

Substituting value of x from equation (1) in $\dfrac{x}{3} - \dfrac{y}{12} = 4\dfrac{3}{4}$, we get :

$\Rightarrow \dfrac{\dfrac{120 - 2y}{5}}{3} - \dfrac{y}{12} = 4\dfrac{3}{4} \\[1em] \Rightarrow \dfrac{120 - 2y}{15} - \dfrac{y}{12} = \dfrac{19}{4} \\[1em] \Rightarrow \dfrac{4(120 - 2y) - 5y}{60} = \dfrac{19}{4} \\[1em] \Rightarrow 480 - 8y - 5y = \dfrac{19}{4} \times 60 \\[1em] \Rightarrow 480 - 13y = 19 \times 15 \\[1em] \Rightarrow 480 - 13y = 285 \\[1em] \Rightarrow 13y = 480 - 285 \\[1em] \Rightarrow 13y = 195 \\[1em] \Rightarrow y = \dfrac{195}{13} = 15.$

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

$\Rightarrow x = \dfrac{120 - 2 \times 15}{5} \\[1em] = \dfrac{120 - 30}{5} \\[1em] = \dfrac{90}{5} \\[1em] = 18.$

Hence, x = 18 and y = 15.