Solve the following system of equations: $\;$ $\dfrac{x + y}{x - y} + \dfrac{x - y}{x + y} = \dfrac{5}{2}, \;\; x^2 + y^2 = 20$
Given system of equations:
$\dfrac{x + y}{x - y} + \dfrac{x - y}{x + y} = \dfrac{5}{2}$ $\;\;\; \cdots \; (1)$
and $\;$ $x^2 + y^2 = 20$ $\;\;\; \cdots \; (2)$
Equation $(1)$ can be written as
$\left(x + y\right)^2 + \left(x - y\right)^2 = \dfrac{5 \left(x^2 - y^2\right)}{2}$
i.e. $\;$ $x^2 + y^2 + 2xy + x^2 + y^2 - 2xy = \dfrac{5 \left(x^2 - y^2\right)}{2}$
i.e. $\;$ $2 x^2 + 2 y^2 = \dfrac{5 \left(x^2 - y^2\right)}{2}$
i.e. $\;$ $4x^2 + 4y^2 = 5x^2 - 5y^2$
i.e. $\;$ $x^2 = 9y^2$ $\;\;\; \cdots \; (1a)$
In view of equation $(1a)$, equation $(2)$ becomes
$9 y^2 + y^2 = 20$
i.e. $\;$ $10 y^2 = 20$
i.e. $\;$ $y^2 = 2$ $\implies$ $y = \pm \sqrt{2}$
Substituting the value of $y$ in equation $(1a)$ gives
when $\;$ $y = + \sqrt{2}$, $\;$ $x^2 = 9 \times 2 = 18$ $\implies$ $x = \pm 3 \sqrt{2}$
when $\;$ $y = - \sqrt{2}$, $\;$ $x^2 = 9 \times 2 = 18$ $\implies$ $x = \pm 3 \sqrt{2}$
$\therefore \;$ The solution to the given system of equations is $\;$ $\left(x, y\right) = \left\{\left(3 \sqrt{2}, \sqrt{2}\right), \; \left(-3 \sqrt{2}, \sqrt{2}\right), \left(3 \sqrt{2}, - \sqrt{2}\right), \left(-3 \sqrt{2}, - \sqrt{2}\right) \right\}$