The solution of the simultaneous equations $\dfrac{x}{2} - \dfrac{y}{3} = 0$ and $\dfrac{3x}{2} + \dfrac{2y}{3} + 10 = 0$ is :

1. x = 4, y = 6

2. x = 4, y = -6

3. x = -4, y = 6

4. x = -4, y = -6

Given,

Equations: $\dfrac{x}{2} - \dfrac{y}{3} = 0, \dfrac{3x}{2} + \dfrac{2y}{3} + 10 = 0$

Solving equation $\dfrac{x}{2} - \dfrac{y}{3} = 0$,

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

$\dfrac{3x}{2} + \dfrac{2y}{3} + 10 = 0 \text{ ….(2)}$

Substituting value of x from equation (1) in (2), we get :

$\Rightarrow \dfrac{3\Big(\dfrac{2y}{3}\Big)}{2} + \dfrac{2y}{3} + 10 = 0 \\[1em] \Rightarrow \dfrac{2y}{2} + \dfrac{2y}{3} + 10 = 0 \\[1em] \Rightarrow \dfrac{6y + 4y + 60}{6} = 0 \\[1em] \Rightarrow 10y + 60 = 0 \\[1em] \Rightarrow 10y = -60 \\[1em] \Rightarrow y = \dfrac{-60}{10} = -6.$

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

$\Rightarrow x = \dfrac{2y}{3} \\[1em] \Rightarrow x = \dfrac{2 \times (-6)}{3} \\[1em] \Rightarrow x = \dfrac{-12}{3} = -4.$

Hence, option 4 is the correct option.