Solve the following pairs of linear equations:

$\dfrac{1}{2(x + 2y)} + \dfrac{5}{3(3x - 2y)} = -\dfrac{3}{2}$

$\dfrac{5}{4(x + 2y)} - \dfrac{3}{5(3x - 2y)} = \dfrac{61}{60}$

Substitute $\dfrac{1}{x + 2y} = p \text{ and } \dfrac{1}{3x - 2y} = q$ in above equations,

$\dfrac{1}{2}p + \dfrac{5}{3}q = -\dfrac{3}{2}$ ……..(i)

$\dfrac{5}{4}p - \dfrac{3}{5}q = \dfrac{61}{60}$ ……..(ii)

Multiplying (i) by $\dfrac{3}{5}$ and (ii) by $\dfrac{5}{3}$ we get,

$\dfrac{3}{10}p + q = -\dfrac{9}{10}$ …….(iii)

$\dfrac{25}{12}p - q = \dfrac{61}{36}$ …….(iv)

Adding (iii) and (iv) we get,

$\Rightarrow \dfrac{3}{10}p + q + \dfrac{25}{12}p - q = -\dfrac{9}{10} + \dfrac{61}{36} \\[1em] \Rightarrow \dfrac{18p + 125p}{60} = \dfrac{-162 + 305}{180} \\[1em] \Rightarrow \dfrac{143}{60}p = \dfrac{143}{180} \\[1em] \Rightarrow p = \dfrac{143 \times 60}{180 \times 143} \\[1em] \Rightarrow p = \dfrac{1}{3} \\[1em] \therefore \dfrac{1}{x + 2y} = \dfrac{1}{3} \\[1em] \Rightarrow x + 2y = 3 …….(v)$

Substituting value of p in (i) we get,

$\Rightarrow \dfrac{1}{2} \times \dfrac{1}{3} + \dfrac{5}{3}q = -\dfrac{3}{2} \\[1em] \Rightarrow \dfrac{1}{6} + \dfrac{5}{3}q = -\dfrac{3}{2} \\[1em] \Rightarrow \dfrac{5}{3}q = -\dfrac{3}{2} - \dfrac{1}{6} \\[1em] \Rightarrow \dfrac{5}{3}q = \dfrac{-9 - 1}{6} \\[1em] \Rightarrow \dfrac{5}{3}q = -\dfrac{10}{6} \\[1em] \Rightarrow q = -\dfrac{10 \times 3}{6 \times 5} \\[1em] \Rightarrow q = -1 \\[1em] \therefore \dfrac{1}{3x - 2y} = -1 \\[1em] \Rightarrow 3x - 2y = -1 \\[1em] \Rightarrow 2y - 3x = 1 ……(vi)$

Subtracting (vi) from (v) we get,

⇒ x + 2y - (2y - 3x) = 3 - 1

⇒ x + 3x = 2

⇒ 4x = 2

⇒ x = $\dfrac{1}{2}$.

Substituting value of x from (v) we get,

$\Rightarrow \dfrac{1}{2} + 2y = 3 \\[1em] \Rightarrow 2y = 3 - \dfrac{1}{2} \\[1em] \Rightarrow 2y = \dfrac{6 - 1}{2} \\[1em] \Rightarrow 2y = \dfrac{5}{2} \\[1em] \Rightarrow y = \dfrac{5}{4}.$

Hence, x = $\dfrac{1}{2} \text{ and } y = \dfrac{5}{4}.$