Solve the following pairs of linear equations:

99x + 101y = 499xy

101x + 99y = 501xy

Given,

99x + 101y = 499xy

101x + 99y = 501xy

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

Now when x ≠ 0 and y ≠ 0.

Dividing the above equations by xy we get,

$\dfrac{99}{y} + \dfrac{101}{x} = 499$ …….(i)

$\dfrac{101}{y} + \dfrac{99}{x} = 501$ …….(ii)

Substituting $\dfrac{1}{x} = p \text{ and } \dfrac{1}{y} = q$ in both equations and multiplying eq. (i) by 101 and (ii) by 99 we get,

9999q + 10201p = 50399 ……(iii)

9999q + 9801p = 49599 ……(iv)

Subtracting (iv) from (iii) we get,

⇒ 9999q + 10201p - (9999q + 9801p) = 50399 - 49599

⇒ 9999q - 9999q + 10201p - 9801p = 800

⇒ 400p = 800

⇒ p = 2.

$\therefore \dfrac{1}{x} = 2 \text{ or } x = \dfrac{1}{2}.$

Substituting value of p in (iii) we get,

⇒ 9999q + 10201(2) = 50399

⇒ 9999q + 20402 = 50399

⇒ 9999q = 29997

⇒ q = 3.

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

Hence, x = 0, y = 0 and x = $\dfrac{1}{2} \text{ and } y = \dfrac{1}{3}.$