Solve the following system of equations: $\begin{cases} 4^{\left(x-y\right)^2 - 1} = 1 \\ 5^{x + y} = 125 \end{cases}$
Given system of equations:
$4^{\left(x-y\right)^2 - 1} = 1$ $\;\;\; \cdots \; (1)$
$5^{x + y} = 125$ $\;\;\; \cdots \; (2)$
Equation $(1)$ $\implies$ $\left(x - y\right)^2 - 1 = \log_4 1$
i.e. $\;$ $\left(x - y\right)^2 - 1 = 0$
i.e. $\;$ $\left(x - y\right)^2 = 1$
i.e. $\;$ $x - y = 1$ $\;\;\; \cdots \; (3a)$ $\;$ or $\;$ $x - y = -1$ $\;\;\; \cdots \; (3b)$
Equation $(2)$ $\implies$ $x + y = \log_5 125$
i.e. $\;$ $x + y = \log_5 5^3$
i.e. $\;$ $x + y = 3 \log_5 5$
i.e. $\;$ $x + y = 3$ $\;\;\; \cdots \; (4)$
Solving equations $(3a)$ and $(4)$ simultaneously [adding equations $(3a)$ and $(4)$] gives
$2x = 4$ $\implies$ $x = 2$
Substituting the value of $x$ in equation $(4)$ gives
$y = 3 - 2 = 1$
Solving equations $(3b)$ and $(4)$ simultaneously [adding equations $(3b)$ and $(4)$] gives
$2x = 2$ $\implies$ $x = 1$
Substituting the value of $x$ in equation $(4)$ gives
$y = 3 - 1 = 2$
$\therefore \;$ The solution to the given system of equations is $\;\;$ $x = \left\{\left(1, 2\right), \; \left(2, 1\right) \right\}$