Solve :

3(2x + y) = 7xy

3(x + 3y) = 11xy; x ≠ 0, y ≠ 0

6x + 3y = 7xy

3x + 9y = 11xy

Multiply 1 in the first equation and 2 in the second equation, then subtract both equations.

(6x + 3y = 7xy) x 1

(3x + 9y = 11xy) x 2

$\begin{matrix} & 6x & + & 3y & = & 7xy \ & 6x & + & 18y & = & 22xy \ & - & - & & & - \ \hline & & - & 15y & = & 7xy - 22xy \ \Rightarrow & & - & 15y & = & - 15xy \end{matrix}$

⇒ x = $\dfrac{-15y}{-15y}$

⇒ x = 1

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

⇒ 6x + 3y = 7xy

⇒ 6 $\times$ 1 + 3y = 7 $\times$ 1 $\times$ y

⇒ 6 + 3y = 7y

⇒ 6 = 7y - 3y

⇒ 4y = 6

⇒ y = $\dfrac{6}{4}$

⇒ y = $\dfrac{3}{2}$

Hence, the value of x = 1 and y = $\dfrac{3}{2}$.