Solve the following pairs of linear equations:

$\dfrac{20}{x + 1} + \dfrac{4}{y - 1} = 5$

$\dfrac{10}{x + 1} - \dfrac{4}{y - 1} = 1$

Substituting $\dfrac{1}{x + 1} = a \text{ and } \dfrac{1}{y - 1} = b$ in above eq. we get,

20a + 4b = 5 …….(i)

10a - 4b = 1 …….(ii)

Multiplying equation (ii) by 2 we get,

20a - 8b = 2 …….(iii)

Subtracting eq. (iii) from (i) we get,

⇒ 20a + 4b - (20a - 8b) = 5 - 2

⇒ 12b = 3

⇒ b = $\dfrac{3}{12} = \dfrac{1}{4}$.

$\therefore \dfrac{1}{y - 1} = \dfrac{1}{4} \\[1em] \Rightarrow y - 1 = 4 \\[1em] \Rightarrow y = 5.$

Substituting value of b in (iii) we get,

⇒ 20a - $8 \times \dfrac{1}{4}$ = 2

⇒ 20a - 2 = 2

⇒ 20a = 2 + 2

⇒ a = $\dfrac{4}{20}$

⇒ a = $\dfrac{1}{5}$

$\therefore \dfrac{1}{x + 1} = \dfrac{1}{5} \\[1em] \Rightarrow x + 1 = 5 \\[1em] \Rightarrow x = 4.$

Hence, x = 4 and y = 5.