Solve the following simultaneous linear equations:

x + y = 7xy

2x - 3y + xy = 0

Given,

x + y = 7xy

2x - 3y + xy = 0

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

Now when x ≠ 0 and y ≠ 0.

Dividing x + y = 7xy by xy,

$\Rightarrow \dfrac{x}{xy} + \dfrac{y}{xy} = \dfrac{7xy}{xy} \\[1em] \Rightarrow \dfrac{1}{y} + \dfrac{1}{x} = 7 ……(i)$

Dividing 2x - 3y + xy = 0 by xy,

$\Rightarrow \dfrac{2x}{xy} - \dfrac{3y}{xy} + \dfrac{xy}{xy} = 0 \\[1em] \Rightarrow \dfrac{2}{y} - \dfrac{3}{x} + 1 = 0 \\[1em] \Rightarrow \dfrac{2}{y} - \dfrac{3}{x} = -1 ……(ii)$

Substituting $\dfrac{1}{x}$ as p and $\dfrac{1}{y}$ as q we get,

q + p = 7 …….(iii)

2q - 3p = -1 ……(iv)

Multiplying (iii) by 3 we get,

3q + 3p = 21 …….(v)

Adding (iv) and (v) we get,

⇒ 2q - 3p + (3q + 3p) = -1 + 21

⇒ 2q + 3q = 20

⇒ 5q = 20

⇒ q = $\dfrac{20}{5}$

⇒ q = 4.

$\therefore \dfrac{1}{y} = 4 \text{ or } y = \dfrac{1}{4}$.

Substituting value of q in (iii) we get,

⇒ 4 + p = 7

⇒ p = 3.

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

Hence, x = 0, y = 0 and x = $\dfrac{1}{3}, y = \dfrac{1}{4}$.