Solve the following simultaneous equations:

$\dfrac{x + y - 8}{2} = \dfrac{x + 2y - 14}{3} = \dfrac{3x + y - 12}{11}$

Given,

$\dfrac{x + y - 8}{2} = \dfrac{x + 2y - 14}{3} = \dfrac{3x + y - 12}{11}$

Solving L.H.S. of the given equation,

$\Rightarrow \dfrac{x + y - 8}{2} = \dfrac{x + 2y - 14}{3} \\[1em] \Rightarrow 3(x + y - 8) = 2(x + 2y - 14) \\[1em] \Rightarrow 3x + 3y - 24 = 2x + 4y - 28 \\[1em] \Rightarrow 3x + 3y - 2x - 4y = -28 + 24 \\[1em] \Rightarrow x - y = -4 \\[1em] \Rightarrow x = -4 + y \text{ ….(1)}$

Solving R.H.S. of the given equation,

$\Rightarrow \dfrac{x + 2y - 14}{3} = \dfrac{3x + y - 12}{11} \\[1em] \Rightarrow 11(x + 2y - 14) = 3(3x + y - 12) \\[1em] \Rightarrow 11x + 22y - 154 = 9x + 3y - 36 \\[1em] \Rightarrow 11x + 22y - 9x - 3y = -36 + 154 \\[1em] \Rightarrow 2x + 19y = 118 \text{ ……..(2)}$

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

⇒ 2(-4 + y) + 19y = 118

⇒ -8 + 2y + 19y = 118

⇒ 21y = 118 + 8

⇒ 21y = 126

⇒ y = $\dfrac{126}{21} = 6$.

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

⇒ x = -4 + y

⇒ x = -4 + 6

⇒ x = 2.

Hence, x = 2, y = 6.