Solve :

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

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

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

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

Let $\dfrac{1}{x + 2y}$ = a and $\dfrac{1}{3x - 2y}$ = b.

So, the equations are

$\dfrac{1}{2}a + \dfrac{5}{3}b = -\dfrac{3}{2}$

Multiplying the complete equation with 6,

⇒ 3a + 10b = - 9 ……………………..(1)

$\dfrac{5}{4}a - \dfrac{3}{5}b = \dfrac{61}{60}$

Multiplying the complete equation with 60,

⇒ 75a - 36b = 61 ……………………..(2)

Multiplying 25 in equation (1), we get

⇒ (3a + 10b = -9) x 25

⇒ 75a + 250b = -225 ……………….(3)

Subtracting equation (3) and (2), we get:

$\begin{matrix} & 75a & + & 250b & = & -225 \ & 75a & - & 36b & = & 61 \ & - & &- & & - \ \hline & & & 286b & = & -286 \ \Rightarrow & & & b & = & \dfrac{-286}{286} \ \end{matrix}$

⇒ b = -1

Putting the value of b in equation (2), we get

⇒ 75a - 36 x (-1) = 61

⇒ 75a + 36 = 61

⇒ 75a = 61 - 36

⇒ 75a = 25

⇒ a = $\dfrac{25}{75} = \dfrac{1}{3}$

So, $\dfrac{1}{x + 2y}$ = a = $\dfrac{1}{3}$ and $\dfrac{1}{3x - 2y}$ = b = -1

⇒ x + 2y = 3 and 3x - 2y = -1

Adding both equation, we get:

$\begin{matrix} & x & + & 2y & = & 3 \ & 3x & - & 2y & = & -1 \ & + & &+ & & + \ \hline & 4x & & & = & 2 \ \Rightarrow &x & & & = & \dfrac{2}{4} \ \end{matrix}$

⇒ x = $\dfrac{1}{2}$

And putting the value of x in equation x + 2y = 3

⇒ $\dfrac{1}{2}$ + 2y = 3

⇒ 2y = 3 - $\dfrac{1}{2}$

⇒ 2y = $\dfrac{5}{2}$

⇒ y = $\dfrac{5}{4}$

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