Solve the following system of equations: $\;$ $\dfrac{x + y}{x - y} + \dfrac{x - y}{x + y} = \dfrac{13}{6}, \;\; x \; y = 5$
Given system of equations:
$\dfrac{x + y}{x - y} + \dfrac{x - y}{x + y} = \dfrac{13}{6}$
i.e. $\;$ $\dfrac{\left(x + y\right)^2 + \left(x - y\right)^2}{\left(x + y\right) \left(x - y\right)} = \dfrac{13}{6}$
i.e. $\;$ $\dfrac{x^2 + y^2 + 2xy + x^2 + y^2 - 2xy}{x^2 - y^2} = \dfrac{13}{6}$
i.e. $\;$ $2 \left(x^2 + y^2\right) = \dfrac{13}{6} \left(x^2 - y^2\right)$
i.e. $\;$ $12 x^2 + 12 y^2 = 13 x^2 - 13 y^2$
i.e. $\;$ $x^2 = 25 y^2$ $\;\;\; \cdots \; (1)$
and $\;$ $x \; y = 5$ $\;$ i.e. $\;$ $x = \dfrac{5}{y}$ $\;\;\; \cdots \; (2)$
In view of equation $(2)$, equation $(1)$ becomes
$\left(\dfrac{5}{y}\right)^2 = 25 y^2$
i.e. $\;$ $\dfrac{25}{y^2} = 25 y^2$ $\;$ i.e. $\;$ $y^4 = 1$ $\;\;\; \cdots \; (3)$
Let $\;$ $y^2 = p$ $\;\;\; \cdots \; (4)$
Then equation $(3)$ can be written as
$p^2 = 1$ $\implies$ $p = \pm 1$
Substituting the value of $p$ in equation $(4)$ gives
when $\;$ $p = +1$, $\;$ $y^2 = 1$ $\implies$ $y = \pm 1$
and when $\;$ $p = -1$, $\;$ $y^2 = -1$
But $y^2 = -1$ is not a valid solution since square of any number cannot be negative.
Substituting the value of $y$ in equation $(2)$ gives
when $\;$ $y = 1$, $\;$ $x = \dfrac{5}{1} = 5$
and when $\;$ $y = -1$, $\;$ $x = \dfrac{5}{-1} = -5$
$\therefore \;$ The solution to the given system of equations is $\;$ $\left(x, y\right) = \left\{\left(5, 1\right), \left(-5, -1\right) \right\}$