Solve the following pairs of linear equations:

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

$\dfrac{7}{2x + 3y} + \dfrac{4}{3x - 2y} = 2$

Substituting $\dfrac{1}{2x + 3y} = a \text{ and } \dfrac{1}{3x - 2y} = b$ in above eq. we get,

$\dfrac{1}{2}a + \dfrac{12}{7}b = \dfrac{1}{2}$ …..(i)

7a + 4b = 2 …….(ii)

Multiplying eq. (i) by 14 we get,

7a + 24b = 7 ……(iii)

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

⇒ 7a + 24b - (7a + 4b) = 7 - 2

⇒ 7a + 24b - 7a - 4b = 7 - 2

⇒ 20b = 5

⇒ b = $\dfrac{5}{20} = \dfrac{1}{4}$.

$\therefore \dfrac{1}{3x - 2y} = \dfrac{1}{4}$

⇒ 3x - 2y = 4 ……..(iv)

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

⇒ 7a + $24 \times \dfrac{1}{4}$ = 7

⇒ 7a + 6 = 7

⇒ 7a = 1

⇒ a = $\dfrac{1}{7}$.

$\therefore \dfrac{1}{2x + 3y} = \dfrac{1}{7}$

⇒ 2x + 3y = 7 …….(v)

Multiplying eq. (iv) by 2 and eq. (v) by 3 we get,

⇒ 6x - 4y = 8 …….(vi)

⇒ 6x + 9y = 21 …….(vii)

Subtracting eq. (vi) from (vii) we get,

⇒ (6x + 9y) - (6x - 4y) = 21 - 8

⇒ 13y = 13

⇒ y = 1.

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

⇒ 6x - 4(1) = 8

⇒ 6x = 8 + 4

⇒ 6x = 12

⇒ x = 2.

Hence, x = 2 and y = 1.