Solve the following system of equations: $\begin{cases} 3 \times 2^x + 2 \times 3^y = 2.75 \\ 2^x - 3^y = -0.75 \end{cases}$
Given system of equations:
$3 \times 2^x + 2 \times 3^y = 2.75$ $\;\;\; \cdots \; (1)$
$2^x - 3^y = -0.75$ $\;\;\; \cdots \; (2)$
Multiply equation $(2)$ with $2$ and adding to equation $(1)$ gives
$5 \times 2^x = 1.25$
i.e. $\;$ $2^x = 0.25 = \dfrac{1}{4} = \left(\dfrac{1}{2}\right)^2 = 2^{-2}$
i.e. $\;$ $x = \log_2 2^{-2} = -2 \log_2 2 = -2$
Substituting the value of $x$ in equation $(2)$ gives
$2^{-2} - 3^y = -0.75$
i.e. $\;$ $3^y = 0.25 + 0.75 = 1$
i.e. $\;$ $y = \log_3 1 = 0$
$\therefore \;$ The solution to the given system of equations is $\;\;$ $x = \left\{\left(-2, 0\right) \right\}$