Solve the following system of equations: $\begin{cases} \dfrac{1}{x} - \dfrac{1}{y} = \dfrac{2}{15} \\ \log_3 x + \log_3 y = 1 + \log_3 5 \end{cases}$
Given system of equations:
$\dfrac{1}{x} - \dfrac{1}{y} = \dfrac{2}{15}$ $\;\;\; \cdots \; (1)$
$\log_3 x + \log_3 y = 1 + \log_3 5$ $\;\;\; \cdots \; (2)$
We have from equation $(2)$,
$\log_3 x + \log_3 y - \log_3 5 = 1$
i.e. $\;$ $\log_3 \left(\dfrac{x \; y}{5}\right) = 1$
i.e. $\;$ $\dfrac{x \; y}{5} = 3^1 = 3$
$\implies$ $x = \dfrac{15}{y}$ $\;\;\; \cdots \; (3)$
In view of equation $(3)$, equation $(1)$ becomes
$\dfrac{y}{15} - \dfrac{1}{y} = \dfrac{2}{15}$
i.e. $\;$ $y^2 - 2y - 15 = 0$
i.e. $\;$ $\left(y - 5\right) \left(y + 3\right) = 0$
i.e. $\;$ $y = 5$ $\;$ or $\;$ $y = -3$
When $y = -3$, the term $\log_3 y$ in equation $(2)$ becomes $\log_3 \left(-3\right)$.
But logarithm of a negative number is not defined.
$\therefore \;$ $y = -3$ is not a valid solution.
When $y = 5$, we have from equation $(3)$, $\;\;$ $x = \dfrac{15}{5} = 3$
$\therefore \;$ The solution to the given system of equations is $\;\;$ $x = \left\{\left(3, \; 5\right) \right\}$