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

$\dfrac{x + 1}{2} + \dfrac{y - 1}{3} = 8$

$\dfrac{x - 1}{3} + \dfrac{y + 1}{2} = 9$

Solving 1st equation,

$\Rightarrow \dfrac{x + 1}{2} + \dfrac{y - 1}{3} = 8 \\[1em] \Rightarrow \dfrac{3(x + 1) + 2(y - 1)}{6} = 8 \\[1em] \Rightarrow \dfrac{3x + 3 + 2y - 2}{6} = 8 \\[1em] \Rightarrow \dfrac{3x + 2y + 1}{6} = 8 \\[1em] \Rightarrow 3x + 2y + 1 = 48 \\[1em] \Rightarrow 3x + 2y = 47 ……(i)$

Solving 2nd equation,

$\Rightarrow \dfrac{x - 1}{3} + \dfrac{y + 1}{2} = 9 \\[1em] \Rightarrow \dfrac{2(x - 1) + 3(y + 1)}{6} = 9 \\[1em] \Rightarrow \dfrac{2x - 2 + 3y + 3}{6} = 9 \\[1em] \Rightarrow \dfrac{2x + 3y + 1}{6} = 9 \\[1em] \Rightarrow 2x + 3y + 1 = 54 \\[1em] \Rightarrow 2x + 3y = 53 …….(ii)$

Multiplying eq. (i) by 2 and eq. (ii) by 3 we get,

6x + 4y = 94 ……(iii)

6x + 9y = 159 …….(iv)

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

⇒ 6x + 9y - (6x + 4y) = 159 - 94

⇒ 5y = 65

⇒ y = 13.

Substituting value of y in eq. (iii) we get,

⇒ 6x + 4(13) = 94

⇒ 6x + 52 = 94

⇒ 6x = 42

⇒ x = 7.

Hence, x = 7 and y = 13.