Solve the following simultaneous linear equations:

3x + 2y = 2xy

$\dfrac{1}{x} + \dfrac{2}{y} = 1\dfrac{1}{6}$

Given,

3x + 2y = 2xy …….(i)

$\dfrac{1}{x} + \dfrac{2}{y} = 1\dfrac{1}{6}$ ……(ii)

Dividing eq. (i) by xy we get,

$\dfrac{3}{y} + \dfrac{2}{x} = 2$ …….(iii)

Substituting $\dfrac{1}{x}$ as p and $\dfrac{1}{y}$ as q in (ii) and (iii) we get,

p + 2q = $\dfrac{7}{6}$ ……(iv)

3q + 2p = 2 ……..(v)

Multiplying (iv) by 6 and (v) by 3 we get,

6p + 12q = 7 ……..(vi)

9q + 6p = 6 …….(vii)

Subtracting (vii) from (vi) we get,

6p + 12q - (9q + 6p) = 7 - 6

3q = 1

q = $\dfrac{1}{3}$.

$\therefore \dfrac{1}{y} = \dfrac{1}{3} \text{ or } y = 3.$

Substituting value of q in (iv) we get,

$\Rightarrow p + 2 \times \dfrac{1}{3} = \dfrac{7}{6} \\[1em] \Rightarrow p + \dfrac{2}{3} = \dfrac{7}{6} \\[1em] \Rightarrow p = \dfrac{7}{6} - \dfrac{2}{3} \\[1em] \Rightarrow p = \dfrac{7 - 4}{6} \\[1em] \Rightarrow p = \dfrac{3}{6} \\[1em] \therefore \dfrac{1}{x} = \dfrac{3}{6} \\[1em] \Rightarrow x = \dfrac{6}{3} = 2.$

Hence, x = 2 and y = 3.