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

2x - 3y - 3 = 0

$\dfrac{2x}{3} + 4y + \dfrac{1}{2} = 0$

Given, equations :

⇒ 2x - 3y - 3 = 0 ………….(1)

⇒ $\dfrac{2x}{3} + 4y + \dfrac{1}{2} = 0$ …….(2)

Simplifying second equation :

$\Rightarrow \dfrac{2x}{3} + 4y + \dfrac{1}{2} = 0 \\[1em] \Rightarrow \dfrac{4x + 24y + 3}{6} = 0 \\[1em] \Rightarrow 4x + 24y + 3 = 0 \text{ …….(3)}$

Multiplying equation (1) by 2, we get :

⇒ 2(2x - 3y - 3) = 2 × 0

⇒ 4x - 6y - 6 = 0 ………(4)

Subtracting equation (4) from (3), we get :

⇒ 4x + 24y + 3 - (4x - 6y - 6) = 0

⇒ 4x - 4x + 24y + 6y + 3 + 6 = 0

⇒ 30y + 9 = 0

⇒ 30y = -9

⇒ y = $-\dfrac{9}{30} = -\dfrac{3}{10}$.

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

$\Rightarrow 2x - 3y - 3 = 0 \\[1em] \Rightarrow 2x - 3 \times -\dfrac{3}{10} - 3 = 0 \\[1em] \Rightarrow 2x + \dfrac{9}{10} - 3 = 0\\[1em] \Rightarrow 2x = 3 - \dfrac{9}{10} \\[1em] \Rightarrow 2x = \dfrac{30 - 9}{10} \\[1em] \Rightarrow x = \dfrac{21}{20}.$

Hence, x = $\dfrac{21}{20} \text{ and } y = -\dfrac{3}{10}$.