Solve the following pairs of linear equations:

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

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

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

3a + 2b = 3 …….(i)

2a + 3b = $\dfrac{11}{3}$ …….(ii)

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

6a + 4b = 6 …….(iii)

6a + 9b = 11 …….(iv)

Subtracting eq. (iii) from iv we get,

⇒ 6a + 9b - (6a + 4b) = 11 - 6

⇒ 6a + 9b - 6a - 4b = 5

⇒ 5b = 5

⇒ b = 1.

$\therefore \dfrac{1}{x - y} = 1 \\[1em] \Rightarrow x - y = 1 …….(v)$

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

⇒ 6a + 4(1) = 6

⇒ 6a + 4 = 6

⇒ 6a = 2

⇒ a = $\dfrac{2}{6} = \dfrac{1}{3}.$

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

⇒ x + y = 3

⇒ x = 3 - y.

Putting above value of x in eq. (v) we get,

⇒ (3 - y) - y = 1

⇒ 3 - 2y = 1

⇒ 2y = 3 - 1

⇒ 2y = 2

⇒ y = 1.

⇒ x = 3 - y = 3 - 1 = 2.

Hence, x = 2 and y = 1.