If 8x + 9y = 42xy and 2x + 3y = 12xy, then the value of $\dfrac{1}{xy}$ is :

1. 1

2. 4

3. 6

4. $\dfrac{1}{6}$

Given,

Equations:

⇒ 8x + 9y = 42xy

⇒ 2x + 3y = 12xy

Dividing both the sides of first equation by xy, we get :

$\Rightarrow \dfrac{8x + 9y}{xy} = \dfrac{42xy}{xy} \\[1em] \Rightarrow \dfrac{8x}{xy} + \dfrac{9y}{xy} = 42 \\[1em] \Rightarrow \dfrac{8}{y} + \dfrac{9}{x} = 42 \text{ …….(1)}$

Dividing both the sides of second equation by xy, we get :

$\Rightarrow \dfrac{2x + 3y}{xy} = \dfrac{12xy}{xy} \\[1em] \Rightarrow \dfrac{2x}{xy} + \dfrac{3y}{xy} = 12 \\[1em] \Rightarrow \dfrac{2}{y} + \dfrac{3}{x} = 12.$

Multiplying both sides of the above equation by 4, we get :

$\Rightarrow 4\Big(\dfrac{2}{y} + \dfrac{3}{x}\Big) = 12 \times 4 \\[1em] \Rightarrow \dfrac{8}{y} + \dfrac{12}{x} = 48 \text{ …….(2)}$

Subtracting equation (1) from (2), we get:

$\Rightarrow \dfrac{8}{y} + \dfrac{12}{x} - \Big(\dfrac{8}{y} + \dfrac{9}{x}\Big) = 48 - 42 \\[1em] \Rightarrow \dfrac{8}{y} + \dfrac{12}{x} - \dfrac{8}{y} - \dfrac{9}{x} = 6 \\[1em] \Rightarrow \dfrac{12}{x} - \dfrac{9}{x} = 6 \\[1em] \Rightarrow \dfrac{12 - 9}{x} = 6 \\[1em] \Rightarrow \dfrac{3}{x} = 6 \\[1em] \Rightarrow x = \dfrac{3}{6} = \dfrac{1}{2}.$

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

$\Rightarrow \dfrac{8}{y} + \dfrac{9}{x} = 42 \\[1em] \Rightarrow \dfrac{8}{y} + \dfrac{9}{\dfrac{1}{2}} = 42 \\[1em] \Rightarrow \dfrac{8}{y} + 18 = 42 \\[1em] \Rightarrow \dfrac{8}{y} = 42 - 18 \\[1em] \Rightarrow \dfrac{8}{y} = 24 \\[1em] \Rightarrow y = \dfrac{8}{24} = \dfrac{1}{3}.$

Calculating xy,

xy = $\dfrac{1}{2} \times \dfrac{1}{3} = \dfrac{1}{6}$.

$\therefore \dfrac{1}{xy} = 6$

Hence, option 3 is the correct option.