Solve : $\dfrac{2}{x} + \dfrac{2}{3y} = \dfrac{1}{6}$ and $\dfrac{3}{x} + \dfrac{2}{y} = 0$.

Hence, find 'm' for which y = mx - 4.

Lets take $\dfrac{1}{x}$ = u and $\dfrac{1}{y}$ = v. The given equation becomes,

2u + $\dfrac{2}{3}$ v = $\dfrac{1}{6}$ ⇒ 12u + 4v = 1

And, 3u + 2v = 0

Multiply second equation by 4 and subtract from first equation, we get

(3u + 2v = 0) x 4

$\begin{matrix} & 12u & + & 4v & = & 1 \ & 12u & + & 8v & = & 0 \ & - & - & & & - \ \hline & & - & 4v & = & 1 - 0 \ \Rightarrow & & - & 4v & = & 1 \end{matrix}$

⇒ v = -$\dfrac{1}{4}$

Substituting the value of v in equation (2), we get:

⇒ 3u + 2 $\times (-\dfrac{1}{4})$ = 0

⇒ 3u - $\dfrac{1}{2}$ = 0

⇒ 3u = $\dfrac{1}{2}$

⇒ u = $\dfrac{1}{6}$

So, x = $\dfrac{1}{u}$ = 6 and y = $\dfrac{1}{v}$ = -4

Substituting the value of x and y in y = mx - 4, we get:

⇒ -4 = m $\times$ 6 - 4

⇒ -4 + 4 = 6m

⇒ 0 = 6m

⇒ m = $\dfrac{0}{6}$

⇒ m = 0

Hence, the value of x = 6 , y = -4 and m = 0.