Solve the following system of equations: $\begin{cases} 2^x + 2^y = 12 \\ x + y = 5 \end{cases}$
Given system of equations:
$2^x + 2^y = 12$ $\;\;\; \cdots \; (1)$
$x + y = 5$ $\;\;\; \cdots \; (2)$
Let $\;$ $2^x = p$ $\;\;\; \cdots \; (3a)$
$\implies$ $x = \log_2 p$ $\;\;\; \cdots \; (3b)$
and $\;$ $2^y = q$ $\;\;\; \cdots \; (4a)$
$\implies$ $y = \log_2 q$ $\;\;\; \cdots \; (4b)$
In view of equations $(3a)$ and $(4a)$ equation $(1)$ becomes
$p + q = 12$ $\;\;\; \cdots \; (5)$
In view of equations $(3b)$ and $(4b)$ equation $(2)$ becomes
$\log_2 p + \log_ q = 5$
i.e. $\;$ $\log_2 \left(pq\right) = 5$
i.e. $\;$ $pq = 2^5 = 32$ $\;\;\; \cdots \; (6)$
Now, $\;$ $\left(p - q\right)^2 = \left(p + q\right)^2 - 4pq$
i.e. $\;$ $\left(p - q\right)^2 = 144 - 128 = 16$ $\;\;\;$ [by equations $(5)$ and $(6)$]
i.e. $\;$ $p - q = 4$ $\;\;\; \cdots \; (7a)$ $\;$ or $\;$ $p - q = -4$ $\;\;\; \cdots \; (7b)$
Solving equations $(5)$ and $(7a)$ simultaneously gives
$2p = 16$ $\implies$ $p = 8$ $\;\;\; \cdots \; (8a)$
Substituting the value of $p$ from equation $(8a)$ in equation $(5)$ gives
$q = 4$ $\;\;\; \cdots \; (8b)$
Solving equations $(5)$ and $(7b)$ simultaneously gives
$2p = 8$ $\implies$ $p = 4$ $\;\;\; \cdots \; (9a)$
Substituting the value of $p$ from equation $(9a)$ in equation $(5)$ gives
$q = 8$ $\;\;\; \cdots \; (9b)$
In view of equations $(8a)$ and $(8b)$, equations $(3b)$ and $(4b)$ become
$x = \log_2 8 = \log_2 2^3 = 3 \log_2 2 = 3$
$y = \log_2 4 = \log_2 2^2 = 2 \log_2 2 = 2$
In view of equations $(9a)$ and $(9b)$, equations $(3b)$ and $(4b)$ become
$x = \log_2 4 = \log_2 2^2 = 2 \log_2 2 = 2$
$y = \log_2 8 = \log_2 2^3 = 3 \log_2 2 = 3$
$\therefore \;$ The solution to the given system of equations is $\;\;$ $x = \left\{\left(2, 3\right), \; \left(3, 2\right) \right\}$