Solve the following system of equations: $\begin{cases} 3^x \times 2^y = 972 \\ \log_{\sqrt{3}} \left(x - y\right) = 2 \end{cases}$
Given system of equations:
$3^x \times 2^y = 972$ $\;\;\; \cdots \; (1)$
$\log_{\sqrt{3}} \left(x - y\right) = 2$ $\;\;\; \cdots \; (2)$
Equation $(2)$ $\implies$ $x - y = \sqrt{3}^2 = 3$
i.e. $\;$ $x = y + 3$ $\;\;\; \cdots \; (3)$
Substituting the value of $x$ from equation $(3)$ in equation $(1)$ gives
$3^{y + 3} \times 2^y = 972$
i.e. $\;$ $3^y \times 3^3 \times 2^y = 972$
i.e. $\;$ $6^y = \dfrac{972}{27} = 36 = 6^2$
$\implies$ $y = 2$
Substituting the value of $y$ in equation $(3)$ gives $\;\;$ $x = 2 + 3 = 5$
$\therefore \;$ The solution to the given system of equations is $\;\;$ $x = \left\{\left(5, \; 2\right) \right\}$