Solve the following system of equations: $\;$ $\sqrt{\dfrac{x}{y}} + \sqrt{\dfrac{y}{x}} = \dfrac{5}{2}$, $\;$ $x + y = 5$
Given system of equations:
$\sqrt{\dfrac{x}{y}} + \sqrt{\dfrac{y}{x}} = \dfrac{5}{2}$ $\;\;\; \cdots \; (1)$, $\;\;$ $x + y = 5$ $\;\;\; \cdots \; (2)$
We have from equation $(1)$,
$\dfrac{x + y}{\sqrt{xy}} = \dfrac{5}{2}$
i.e. $\;$ $\dfrac{5}{\sqrt{xy}} = \dfrac{5}{2}$ $\;\;\;$ [in view of equation $(2)$]
i.e. $\;$ $\dfrac{1}{\sqrt{xy}} = \dfrac{1}{2}$
i.e. $\;$ $\sqrt{xy} = 2$
i.e. $\;$ $xy = 4$ $\implies$ $x = \dfrac{4}{y}$ $\;\;\; \cdots \; (3)$
In view of equation $(3)$, equation $(2)$ becomes
$\dfrac{4}{y} + y = 5$
i.e. $\;$ $y^2 - 5y + 4 = 0$
i.e. $\;$ $\left(y - 1\right) \left(y - 4\right) = 0$
i.e. $\;$ $y = 1$ $\;\;\;$ or $\;\;\;$ $y = 4$
When $\;$ $y = 1$, $\;$ we have from equation $(2)$, $\;$ $x = 5 - 1 = 4$
When $\;$ $y = 4$, $\;$ we have from equation $(2)$, $\;$ $x = 5 - 4 = 1$
$\therefore \;$ The solution to the given system of equations is $\left(x, y\right) = \left\{\left(4, 1\right), \left(1, 4\right) \right\}$