Solve the following pairs of linear equations:

$\dfrac{3}{2x} + \dfrac{2}{3y} = 5$

$\dfrac{5}{x} - \dfrac{3}{y} = 1$

Substituting $\dfrac{1}{x} = a \text{ and } \dfrac{1}{y} = b$ in above equations we get,

$\dfrac{3}{2}a + \dfrac{2}{3}b = 5$ ……(i)

5a - 3b = 1 …….(ii)

Solving eq (i) we get,

$\Rightarrow \dfrac{3}{2}a + \dfrac{2}{3}b = 5 \\[1em] \Rightarrow \dfrac{9a + 4b}{6} = 5 \\[1em] \Rightarrow 9a + 4b = 30 …….(iii)$

Multiplying eq. (ii) by 4 we get,

20a - 12b = 4 …….(iv)

Multiplying eq. (iii) by 3 we get,

27a + 12b = 90 …….(v)

Adding eq. (iv) and (v) we get,

⇒ 20a - 12b + 27a + 12b = 4 + 90

⇒ 47a = 94

⇒ a = 2.

$\therefore \dfrac{1}{x} = 2 \\[1em] \Rightarrow x = \dfrac{1}{2}.$

Substituting value of a in eq. (ii) we get,

⇒ 5(2) - 3b = 1

⇒ 10 - 3b = 1

⇒ 3b = 10 - 1

⇒ 3b = 9

⇒ b = 3.

$\therefore \dfrac{1}{y} = 3 \\[1em] \Rightarrow y = \dfrac{1}{3}.$

Hence, $x = \dfrac{1}{2} \text{ and } y = \dfrac{1}{3}.$