Solve the following system of equations: $\;$ $\dfrac{1}{3x} - \dfrac{1}{2y} = \dfrac{1}{3}, \;\; \dfrac{1}{9x^2} - \dfrac{1}{4y^2} = \dfrac{1}{4}$
Given system of equations:
$\dfrac{1}{3x} - \dfrac{1}{2y} = \dfrac{1}{3}$ $\;\;\; \cdots \; (1)$
and $\;$ $\dfrac{1}{9x^2} - \dfrac{1}{4y^2} = \dfrac{1}{4}$ $\;\;\; \cdots \; (2)$
Let $\;$ $\dfrac{1}{3x} = u$ $\;\;\; \cdots \; (3)$ $\;$ and $\;$ $\dfrac{1}{2y} = v$ $\;\;\; \cdots \; (4)$
In view of equations $(3)$ and $(4)$, equations $(1)$ and $(2)$ respectively become
$u - v = \dfrac{1}{3}$ $\;$ i.e. $\;$ $u = v + \dfrac{1}{3}$ $\;\;\; \cdots \; (5)$
and $\;$ $u^2 - v^2 = \dfrac{1}{4}$ $\;\;\; \cdots \; (6)$
Substituting the value of $u$ from equation $(5)$ in equation $(6)$ gives
$\left(v + \dfrac{1}{3}\right)^2 - v^2 = \dfrac{1}{4}$
i.e. $\;$ $\dfrac{2v}{3} + \dfrac{1}{9} = \dfrac{1}{4}$
i.e. $\;$ $\dfrac{2v}{3} = \dfrac{5}{36}$
i.e. $\;$ $v = \dfrac{5}{24}$
Substituting the value of $v$ in equation $(5)$ gives
$u = \dfrac{5}{24} + \dfrac{1}{3} = \dfrac{13}{24}$
For $\;$ $u = \dfrac{13}{24}$, $\;$ we have from equation $(3)$
$\dfrac{1}{3x} = \dfrac{13}{24}$ $\implies$ $x = \dfrac{8}{13}$
For $\;$ $v = \dfrac{5}{24}$, $\;$ we have from equation $(4)$
$\dfrac{1}{2y} = \dfrac{5}{24}$ $\implies$ $y = \dfrac{12}{5}$
$\therefore \;$ The solution to the given system of equations is $\;$ $\left(x, y\right) = \left\{\left(\dfrac{8}{13}, \dfrac{12}{5}\right) \right\}$