Solve the following system of equations: $\;$ $x \sqrt{y} + y \sqrt{x} = 6$, $\;$ $x^2 y + y^2 x = 20$
Given system of equations:
$x \sqrt{y} + y \sqrt{x} = 6$ $\;\;\; \cdots \; (1)$, $\;\;$ $x^2 y + y^2 x = 20$ $\;\;\; \cdots \; (2)$
Squaring equation $(1)$ gives
$\left(x \sqrt{y} + y \sqrt{x} \right)^2 = 6^2$
i.e. $\;$ $x^2 y + y^2 x + 2 xy \sqrt{xy} = 36$
i.e. $\;$ $20 + 2x^{\frac{3}{2}} y^{\frac{3}{2}} = 36$ $\;\;\;$ [in view of equation $(2)$]
i.e. $\;$ $x^{\frac{3}{2}} y^{\frac{3}{2}} = 8 = 2^3$
i.e. $\;$ $\sqrt{xy} = 2$
i.e. $\;$ $xy = 4$ $\implies$ $x = \dfrac{4}{y}$ $\;\;\; \cdots \; (3)$
Substituting $\;$ $x = \dfrac{4}{y}$ $\;$ in equation $(1)$ gives
$\dfrac{4}{y} \times \sqrt{y} + y \times \dfrac{2}{\sqrt{y}} = 6$
i.e. $\;$ $\dfrac{2}{\sqrt{y}} + \sqrt{y} = 3$
i.e. $\;$ $y - 3 \sqrt{y} + 2 = 0$
i.e. $\;$ $\left(\sqrt{y} - 2\right) \left(\sqrt{y} - 1\right) = 0$
i.e. $\;$ $\sqrt{y} = 2$ $\;\;\;$ or $\;\;\;$ $\sqrt{y} = 1$
i.e. $\;$ $y = 4$ $\;\;\;$ or $\;\;\;$ $y = 1$
Substituting the value of $y$ in equation $(3)$ gives
when $\;$ $y = 4$, $\;$ $x = \dfrac{4}{4} = 1$
when $\;$ $y = 1$, $\;$ $x = \dfrac{4}{1} = 4$
$\therefore \;$ The solution to the given system of equations is $\left(x, y\right) = \left\{\left(1, 4\right), \left(4, 1\right) \right\}$