Solve the following system of equations by using the method of cross multiplication:

$\dfrac{5}{x + 1} - \dfrac{2}{y - 1} = \dfrac{1}{2},\dfrac{10}{x + 1} + \dfrac{2}{y - 1} = \dfrac{5}{2}$, where x ≠ -1 and x ≠ 1.

$\dfrac{5}{x + 1} - \dfrac{2}{y - 1} = \dfrac{1}{2},\dfrac{10}{x + 1} + \dfrac{2}{y - 1} = \dfrac{5}{2}$

Substituting $\dfrac{1}{x + 1}= u, \dfrac{1}{y - 1} = v$ in $\dfrac{5}{x + 1} - \dfrac{2}{y - 1} = \dfrac{1}{2}$, we get:

⇒ 5u - 2v = $\dfrac{1}{2}$

⇒ 5u - 2v - $\dfrac{1}{2}$ = 0     ….(1)

Substituting $\dfrac{1}{x + 1}= u, \dfrac{1}{y - 1} = v$ in $\dfrac{10}{x + 1} + \dfrac{2}{y - 1} = \dfrac{5}{2}$, we get:

⇒ 10u + 2v = $\dfrac{5}{2}$

⇒ 10u + 2v - $\dfrac{5}{2}$ = 0     ….(2)

Multiply equation (1) and (2) by 2, we get,

⇒ $2\Big(5u - 2v - \dfrac{1}{2}\Big)$= 0

⇒ 10u - 4v - 1 = 0     …….(3)

⇒ $2\Big(10u + 2v - \dfrac{5}{2}\Big)$ = 0

⇒ 20u + 4v - 5 = 0     ………(4)

Applying cross-multiplication method for solving equations (3) and (4), we get :

$\Rightarrow \dfrac{u}{(-4) \times (-5) - (4) \times (-1)} = \dfrac{v}{(-1) \times (20) - (-5) \times (10)} = \dfrac{1}{(10) \times (4) - (20) \times (-4)} \\[1em] \Rightarrow \dfrac{u}{(20) + 4} = \dfrac{v}{-20 + 50} = \dfrac{1}{40 + 80} \\[1em] \Rightarrow \dfrac{u}{24} = \dfrac{v}{30} = \dfrac{1}{120} \\[1em] \Rightarrow \dfrac{u}{24} = \dfrac{1}{120} \text{ and } \dfrac{v}{30} = \dfrac{1}{120} \\[1em] \Rightarrow u = \dfrac{24}{120} \text{ and } v = \dfrac{30}{120} \\[1em] \Rightarrow u = \dfrac{1}{5} \text{ and } v = \dfrac{1}{4}.$

Now we have u = $\dfrac{1}{5} \text{ and } v = \dfrac{1}{4}$,

$\Rightarrow \dfrac{1}{x + 1} = u \\[1em] \Rightarrow \dfrac{1}{x + 1} = \dfrac{1}{5} \\[1em] \Rightarrow x + 1 = 5 \\[1em] \Rightarrow x = 5 - 1 = 4 \\[1em] \Rightarrow \dfrac{1}{y - 1} = v \\[1em] \Rightarrow \dfrac{1}{y - 1} = \dfrac{1}{4} \\[1em] \Rightarrow y - 1 = 4 \\[1em] \Rightarrow y = 4 + 1 = 5.$

Hence, x = 4, y = 5.