The solution of the simultaneous equations $2x + \dfrac{1}{y} = 0$ and $3x + \dfrac{1}{2y} = -2$ is :

1. x = 1, y = $\dfrac{1}{2}$

2. x = 1, y = $-\dfrac{1}{2}$

3. x = -1, y = $\dfrac{1}{2}$

4. x = -1, y = $-\dfrac12$

Given,

Equations: $2x + \dfrac{1}{y} = 0$ and $3x + \dfrac{1}{2y} = -2$

Solving equation 1,

⇒ $2x + \dfrac{1}{y} = 0$

⇒ 2x = $-\dfrac{1}{y}$

⇒ x = $\dfrac{-1}{2y}$     …….(1)

Substituting value of x from equation (1) in $3x + \dfrac{1}{2y} = -2$, we get :

$\Rightarrow 3\Big(\dfrac{-1}{2y}\Big) + \dfrac{1}{2y} = -2 \\[1em] \Rightarrow \dfrac{-3}{2y} + \dfrac{1}{2y} = -2 \\[1em] \Rightarrow \dfrac{-2}{2y} = -2 \\[1em] \Rightarrow y = \dfrac{-2}{-2 \times 2} \\[1em] \Rightarrow y = \dfrac{1}{2}.$

Substituting y = $\dfrac{1}{2}$ in equation (1), we get :

$\Rightarrow x = \dfrac{-1}{2y} \\[1em] \Rightarrow x = \dfrac{-1}{2\Big(\dfrac{1}{2}\Big)} \\[1em] \Rightarrow x = -1.$

Hence, option 3 is the correct option.