Solve the following system of equations: $\;$ $2xy - \dfrac{3x}{y} = 15, \;\; xy + \dfrac{x}{y} = 15$
Given system of equations:
$2xy - \dfrac{3x}{y} = 15$ $\;\;\; \cdots \; (1)$
and $\;$ $xy + \dfrac{x}{y} = 15$ $\;\;\; \cdots \; (2)$
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
$2u - 3v = 15$ $\;\;\; \cdots \; (5)$
and $\;$ $u + v = 15$ $\;\;\; \cdots \; (6)$
We have from equations $(5)$ and $(6)$,
$2u - 3v = u + v$
i.e. $\;$ $u = 4v$ $\;\;\; \cdots \; (7)$
Substituting the value of $u$ in equation $(6)$ gives
$4v + v = 15$ $\;$ i.e. $\;$ $5v = 15$ $\;$ i.e. $\;$ $v = 3$
When $v = 3$, we have from equation $(7)$, $\;$ $u = 4 \times 3 = 12$
Substituting the values of $u$ and $v$ in equations $(4)$ and $(3)$ gives
$\dfrac{x}{y} = 3$ $\;$ i.e. $\;$ $x = 3y$ $\;\;\; \cdots \; (8)$
and $\;\;$ $x \; y = 12$ $\;\;\; \cdots \; (9)$
In view of equation $(8)$, equation $(9)$ can be written as
$3y \cdot y = 12$
i.e. $\;$ $y^2 = 4$ $\implies$ $y = \pm 2$
Substituting the value of $y$ in equation $(8)$ gives
when $\;$ $y = + 2$, $\;$ $x = 3 \times 2 = 6$
and when $\;$ $y = -2$, $\;$ $x = 3 \times \left(-2\right) = -6$
$\therefore \;$ The solution to the given system of equations is $\;$ $\left(x, y\right) = \left\{\left(6, 2\right), \; \left(-6, -2\right) \right\}$