Solve the following system of equations: $\;$ $\dfrac{1}{x + 1} + \dfrac{1}{y} = \dfrac{1}{3}, \;\; \dfrac{1}{\left(x + 1\right)^2} - \dfrac{1}{y^2} = \dfrac{1}{4}$
Given system of equations:
$\dfrac{1}{x + 1} + \dfrac{1}{y} = \dfrac{1}{3}$ $\;\;\; \cdots \; (1)$
and $\;$ $\dfrac{1}{\left(x + 1\right)^2} - \dfrac{1}{y^2} = \dfrac{1}{4}$
i.e. $\;$ $\left(\dfrac{1}{x + 1} + \dfrac{1}{y}\right) \left(\dfrac{1}{x + 1} - \dfrac{1}{y}\right) = \dfrac{1}{4}$
i.e. $\;$ $\dfrac{1}{3} \left(\dfrac{1}{x + 1} - \dfrac{1}{y}\right) = \dfrac{1}{4}$ $\;\;$ [in view of equation $(1)$]
i.e. $\;$ $\dfrac{1}{x + 1} - \dfrac{1}{y} = \dfrac{3}{4}$ $\;\;\; \cdots \; (2)$
Adding equations $(1)$ and $(2)$ gives
$\dfrac{2}{x + 1} = \dfrac{1}{3} + \dfrac{3}{4} = \dfrac{13}{12}$
i.e. $\;$ $24 = 13x + 13$
i.e. $\;$ $13x = 11$ $\implies$ $x = \dfrac{11}{13}$
Substitute $x = \dfrac{11}{13}$ in equation $(1)$ to get
$\dfrac{1}{\dfrac{11}{13} + 1} + \dfrac{1}{y} = \dfrac{1}{3}$
i.e. $\;$ $\dfrac{13}{24} + \dfrac{1}{y} = \dfrac{1}{3}$
i.e. $\;$ $\dfrac{1}{y} = \dfrac{1}{3} - \dfrac{13}{24} = \dfrac{-5}{24}$ $\implies$ $y = \dfrac{-24}{5}$
$\therefore \;$ The solution to the given system of equations is $\;$ $\left(x, y\right) = \left\{\left(\dfrac{11}{13}, \dfrac{-24}{5}\right) \right\}$