Solve the following pairs of linear equations:

3x + 5y = 4xy

2y - x = xy.

Given,

3x + 5y = 4xy ……..(i)

2y - x = xy …….(ii)

First we note that x = 0, y = 0 is a solution of equations.

Now when x ≠ 0 and y ≠ 0.

Dividing above equations by xy we get,

$\dfrac{3}{y} + \dfrac{5}{x} = 4$ …….(iii)

$\dfrac{2}{x} - \dfrac{1}{y} = 1$ …….(iv)

Substituting $\dfrac{1}{x} = a \text{ and } \dfrac{1}{y} = b$ in eq. (iii) and (iv) we get,

3b + 5a = 4 …….(v)

2a - b = 1 …….(vi)

Multiplying eq. (vi) by 3 we get,

6a - 3b = 3 ……..(vii)

Adding eq. (v) and (vii) we get,

⇒ 3b + 5a + 6a - 3b = 4 + 3

⇒ 11a = 7

⇒ a = $\dfrac{7}{11}$.

$\therefore \dfrac{1}{x} = \dfrac{7}{11} \\[1em] \Rightarrow x = \dfrac{11}{7}.$

Substituting value of a in eq. (vi) we get,

$\Rightarrow 2 \times \dfrac{7}{11} - b = 1 \\[1em] \Rightarrow \dfrac{14}{11} - b = 1 \\[1em] \Rightarrow b = \dfrac{14}{11} - 1 \\[1em] \Rightarrow b = \dfrac{3}{11} \\[1em] \therefore \dfrac{1}{y} = \dfrac{3}{11} \\[1em] \Rightarrow y = \dfrac{11}{3}.$

Hence, x = 0, y = 0 and x = $\dfrac{11}{7}, y = \dfrac{11}{3}.$