For solving each pair of equations, use the method of elimination by equating coefficients :

$\dfrac{5y}{2} - \dfrac{x}{3} = 8$

$\dfrac{y}{2} + \dfrac{5x}{3} = 12$

Given equations :

$\Rightarrow \dfrac{5y}{2} - \dfrac{x}{3} = 8 ……….(1) \\[1em] \Rightarrow \dfrac{y}{2} + \dfrac{5x}{3} = 12 ……(2)$

Multiplying equation (1) by 30, we get :

$\Rightarrow 30\Big(\dfrac{5y}{2} - \dfrac{x}{3}\Big) = 30 \times 8 \\[1em] \Rightarrow 75y - 10x = 240 ……(3)$

Multiplying equation (2) by 6, we get :

$\Rightarrow 6\Big(\dfrac{y}{2} + \dfrac{5x}{3}\Big) = 12 \times 6 \\[1em] \Rightarrow 3y + 10x = 72 ………..(4)$

Adding equations (3) and (4), we get :

⇒ 75y - 10x + 3y + 10x = 240 + 72

⇒ 78y = 312

⇒ y = $\dfrac{312}{78}$ = 4.

Substituting value of y in equation (1), we get :

$\Rightarrow \dfrac{5 \times 4}{2} - \dfrac{x}{3} = 8 \\[1em] \Rightarrow 10 - \dfrac{x}{3} = 8 \\[1em] \Rightarrow \dfrac{x}{3} = 10 - 8 \\[1em] \Rightarrow x = 3 \times 2 = 6.$

Hence, x = 6 and y = 4.