<s>[INST] Exponential Simultaneous Equations Solve the following simultaneous equations:
$$2^x + 2^y = 10$$
$$x + y = 4$$
Looking at it, it is obvious that the answers are $(3,1)$ and $(1,3)$, however, I was wondering if they could be solved algebraically. Here's my approach:
$$2^x + 2^{4-x} = 10$$
$$2^x + \frac{(2^4)}{(2^x)} = 10$$
$$2^x + \frac {16}{2^x} = 10$$
And this is where I get stuck. Any help will be greatly appreciated, thanks in advance.
 [/INST] Notice that:
\begin{align*}
2^x+\frac{16}{2^x}={}&10 \\
(2^x)^2+16={}&10\cdot 2^x \\
(2^x)^2-10\cdot 2^x+16={}&0 \\
(2^x)^2-8\cdot 2^x-2\cdot 2^x+16={}&0
\end{align*}
By factorizing we get
\begin{align*}
&\phantom{2^x-8}(2^x-8)(2^x-2)=0 \iff{} \\
&{}\iff 2^x-8=0\vee2^x-2=0 \iff{} \\
&{}\iff 2^x=8\vee2^x=2\iff {} \\
&{}\iff2^x=2^3\vee2^x=2^1\iff{} \\
&{}\iff x=3\vee x=1.
\end{align*}
Hence, we get that the solutions are:
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{x=1,\ 3}}$$
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