Solve the following system of equations: $\;$ $\sqrt{\dfrac{20 y}{x}} = \sqrt{x + y} + \sqrt{x - y}$, $\;$ $\sqrt{\dfrac{16 x}{5 y}} = \sqrt{x + y} - \sqrt{x - y}$
Given system of equations:
$\sqrt{\dfrac{20 y}{x}} = \sqrt{x + y} + \sqrt{x - y}$ $\;\;\; \cdots \; (1)$, $\;\;$ $\sqrt{\dfrac{16x}{5y}} = \sqrt{x + y} - \sqrt{x - y}$ $\;\;\; \cdots \; (2)$
Multiplying equations $(1)$ and $(2)$ gives
$\sqrt{\dfrac{20 y}{x} \times \dfrac{16 x}{5 y}} = \left(\sqrt{x + y} + \sqrt{x - y}\right) \left(\sqrt{x + y} - \sqrt{x - y}\right)$
i.e. $\;$ $2 \times 4 = x + y - \sqrt{x^2 - y^2} + \sqrt{x^2 - y^2} - x + y$
i.e. $\;$ $2y = 8$ $\implies$ $y = 4$
Substituting $y = 4$ in equation $(1)$ gives
$\sqrt{\dfrac{20 \times 4}{x}} = \sqrt{x + 4} + \sqrt{x - 4}$
i.e. $\;$ $\dfrac{4 \sqrt{5}}{\sqrt{x}} = \sqrt{x + 4} + \sqrt{x - 4}$ $\;\;\; \cdots \; (3)$
Squaring equation $(3)$ gives
$\dfrac{80}{x} = x + 4 + x - 4 + 2 \sqrt{x^2 - 16}$
i.e. $\;$ $\dfrac{40}{x} - x = \sqrt{x^2 + 16}$ $\;\;\; \cdots \; (4)$
Squaring equation $(4)$ gives
$\dfrac{1600}{x^2} + x^2 - 80 = x^2 - 16$
i.e. $\;$ $\dfrac{1600}{x^2} = 64$
i.e. $\;$ $x^2 = \dfrac{1600}{64} = 25$ $\implies$ $x = \pm 5$
When $\;$ $x = -5, \; y = 4$, $\;$ $\sqrt{x + y} = -1$ $\;$ i.e. an imaginary number.
$\therefore \;$ The solution $\;$ $\left(x, y\right) = \left(-5, 4\right)$ is discarded.
$\therefore \;$ The solution to the given system of equations is $\left(x, y\right) = \left\{\left(5, 4\right) \right\}$