Solve the following systems of simultaneous linear equations by the elimination method

$2x + \dfrac{x - y}{6} = 2$

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

Solving 1st equation,

$\Rightarrow 2x + \dfrac{x - y}{6} = 2 \\[1em] \Rightarrow \dfrac{12x + x - y}{6} = 2 \\[1em] \Rightarrow 13x - y = 12 …….(i)$

Solving 2nd equation,

$\Rightarrow x - \dfrac{2x + y}{3} = 1 \\[1em] \Rightarrow \dfrac{3x - 2x - y}{3} = 1 \\[1em] \Rightarrow x - y = 3 ……..(ii)$

Multiplying eq. (ii) by 13 we get,

⇒ 13x - 13y = 39 ………(iii)

Subtracting eq. (iii) from (i) we get,

⇒ 13x - y - (13x - 13y) = 12 - 39

⇒ 13x - 13x - y + 13y = -27

⇒ 12y = -27

⇒ y = $-\dfrac{9}{4}$.

Substituting value of y in eq (ii) we get,

$\Rightarrow x - \Big(-\dfrac{9}{4}\Big) = 3 \\[1em] \Rightarrow x + \dfrac{9}{4} = 3 \\[1em] \Rightarrow x = 3 - \dfrac{9}{4} \\[1em] \Rightarrow x = \dfrac{12 - 9}{4} \\[1em] \Rightarrow x = \dfrac{3}{4}.$

Hence, $x = \dfrac{3}{4} \text{ and } y = -\dfrac{9}{4}.$