Solve the following system of equations: $\;$ $\dfrac{1}{y - 1} - \dfrac{1}{y + 1} = \dfrac{1}{x}, \;\; y^2 - x - 5 = 0$
Given system of equations:
$\dfrac{1}{y - 1} - \dfrac{1}{y + 1} = \dfrac{1}{x}$ $\;\;\; \cdots \; (1)$
and $\;$ $y^2 - x - 5 = 0$ $\;$ i.e. $\;$ $x = y^2 - 5$ $\;\;\; \cdots \; (2)$
Substituting the value of $x$ from equation $(2)$ in equation $(1)$ gives
$\dfrac{1}{y - 1} - \dfrac{1}{y + 1} = \dfrac{1}{y^2 - 5}$
i.e. $\;$ $\dfrac{y + 1 - y + 1}{y^2 - 1} = \dfrac{1}{y^2 - 5}$
i.e. $\;$ $2 \left(y^2 - 5\right) = y^2 - 1$
i.e. $\;$ $y^2 = 9$ $\implies$ $y = \pm 3$
Substituting $y^2 = 9$ in equation $(2)$ gives
$x = 9 - 5 = 4$
$\therefore \;$ The solution to the given system of equations is $\;$ $\left(x, y\right) = \left\{\left(4, 3\right), \; \left(4, -3\right) \right\}$