Solve the following pairs of linear equations:

3x + 14y = 5xy

21y - x = 2xy

Given,

3x + 14y = 5xy …….(i)

21y - x = 2xy ……..(ii)

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

Now when x ≠ 0 and y ≠ 0.

Dividing eq. (i) by xy we get,

$\Rightarrow 3x + 14y = 5xy \\[1em] \Rightarrow \dfrac{3x}{xy} + \dfrac{14y}{xy} = \dfrac{5xy}{xy} \\[1em] \Rightarrow \dfrac{3}{y} + \dfrac{14}{x} = 5 ……(iii)$

Dividing eq. (ii) by xy we get,

$\Rightarrow 21y - x = 2xy \\[1em] \Rightarrow \dfrac{21y}{xy} - \dfrac{x}{xy} = 2 \\[1em] \Rightarrow \dfrac{21}{x} - \dfrac{1}{y} = 2 …….(iv)$

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

3b + 14a = 5 ……..(v)

21a - b = 2 ………(vi)

Multiplying eq. (vi) by 3 we get,

63a - 3b = 6 …….(vii)

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

⇒ 3b + 14a + (63a - 3b) = 5 + 6
⇒ 77a = 11
⇒ a = $\dfrac{11}{77} = \dfrac{1}{7}$.

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

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

⇒ 3b + $14\times \dfrac{1}{7}$ = 5

⇒ 3b + 2 = 5

⇒ 3b = 3

⇒ b = 1.

$\therefore \dfrac{1}{y} = 1 \\[1em] \Rightarrow y = 1.$

Hence, x = 0, y = 0 and x = 7, y = 1.