Solve the following system of equations: $\;$ $xy - \dfrac{x}{y} = \dfrac{16}{3}, \;\; xy - \dfrac{y}{x} = \dfrac{9}{2}$
Given system of equations:
$xy - \dfrac{x}{y} = \dfrac{16}{3}$ $\;\;\; \cdots \; (1)$
and $\;$ $xy - \dfrac{y}{x} = \dfrac{9}{2}$ $\;\;\; \cdots \; (2)$
Let $\;$ $x \; y = u$ $\;\;\; \cdots \; (3)$ $\;$ and $\;$ $\dfrac{x}{y} = v$ $\;\;\; \cdots \; (4)$
In view of equations $(3)$ and $(4)$, equations $(1)$ and $(2)$ become
$u - v = \dfrac{16}{3}$ $\;\;\; \cdots \; (5)$
and $\;$ $u - \dfrac{1}{v} = \dfrac{9}{2}$ $\;\;\; \cdots \; (6)$
Solving equations $(5)$ and $(6)$ simultaneously gives
$v - \dfrac{1}{v} = \dfrac{9}{2} - \dfrac{16}{3} = \dfrac{-5}{6}$
i.e. $\;$ $6v^2 - 6 = -5v$
i.e. $\;$ $6v^2 + 5v - 6 = 0$
i.e. $\;$ $\left(2v + 3\right) \left(3v - 2\right) = 0$
i.e. $\;$ $v = \dfrac{-3}{2}$ $\;$ or $\;$ $v = \dfrac{2}{3}$
We have from equation $(5)$, when $\;$ $v = \dfrac{-3}{2}$, $\;$ $u = \dfrac{16}{3} - \dfrac{3}{2} = \dfrac{23}{6}$
and when $\;$ $v = \dfrac{2}{3}$, $\;$ $u = \dfrac{16}{3} + \dfrac{2}{3} = 6$
Substituting $u$ and $v$ in equations $(3)$ and $(4)$ gives
when $\;$ $u = \dfrac{23}{6}$, $\;$ $v = \dfrac{-3}{2}$
we have $\;$ $\dfrac{x}{y} = \dfrac{-3}{2}$ $\;$ i.e. $\;$ $x = \dfrac{-3 y}{2}$ $\;\;\; \cdots \; (7)$
and $\;$ $xy = \dfrac{23}{6}$
i.e. $\;$ $\dfrac{-3 y^2}{2} = \dfrac{23}{6}$ $\;\;\;$ [in view of equation $(7)$]
i.e. $\;$ $y^2 = \dfrac{-23}{9}$
But square of any number cannot be negative.
$\therefore \;$ $u = \dfrac{23}{6}$, $\;$ $v = \dfrac{-3}{2}$ $\;$ are not acceptable values.
When $\;$ $u = 6$, $\;$ $v = \dfrac{2}{3}$
we have $\;$ $\dfrac{x}{y} = \dfrac{2}{3}$ $\;$ i.e. $\;$ $x = \dfrac{2 y}{3}$ $\;\;\; \cdots \; (8)$
and $\;$ $xy = 6$
i.e. $\;$ $\dfrac{2 y^2}{3} = 6$ $\;\;\;$ [in view of equation $(8)$]
i.e. $\;$ $y^2 = 9$ $\implies$ $y = \pm 3$
Substituting the value of $y$ in equation $(8)$ gives
$x = \pm \dfrac{2}{3} \times 3 = \pm 2$
$\therefore \;$ The solution to the given system of equations is $\;$ $\left(x, y\right) = \left\{\left(2, 3\right), \; \left(-2, -3\right) \right\}$