Solve the following system of equations: $\;$ $\dfrac{x^2}{y} + \dfrac{y^2}{x} = 18, \;\; x + y = 12$
Given system of equations:
$\dfrac{x^2}{y} + \dfrac{y^2}{x} = 18$ $\;\;\; \cdots \; (1)$ $\;$ and $\;$ $x + y = 12$ $\;\;\; \cdots \; (2)$
Equation $(1)$ can be written as
$x^3 + y^3 = 18 xy$
i.e. $\;$ $\left(x + y\right)^3 - 3xy \left(x + y\right) = 18xy$
i.e. $\;$ $12^3 - 3 x y \times 12 = 18xy$ $\;\;\;$ [by equation $(2)$]
i.e. $\;$ $1728 - 36xy = 18xy$
i.e. $\;$ $54 xy = 1728$ $\;\;\;$ i.e. $\;$ $xy = 32$ $\implies$ $x = \dfrac{32}{y}$ $\;\;\; \cdots \; (3)$
In view of equation $(3)$, equation $(2)$ becomes
$\dfrac{32}{y} + y = 12$
i.e. $\;$ $y^2 - 12y + 32 = 0$
i.e. $\;$ $\left(y - 8\right) \left(y - 4\right) = 0$
i.e. $\;$ $y = 8$ $\;\;$ or $\;\;$ $y = 4$
Substituting the value of $y$ in equation $(2)$ gives
when $\;$ $y = 8$, $\;$ $x = 12 - 8 = 4$
and when $\;$ $y = 4$, $\;$ $x = 12 - 4 = 8$
$\therefore \;$ The solution to the given system of equations is $\;\;$ $\left(x, y\right) = \left\{\left(4, 8\right), \; \left(8, 4\right) \right\}$