<s>[INST] Can either of these two simultaneous equations be solved? I have been struggling with the elimination of the variables.
$\begin{array}{c|c}
 \left(1-2^k\right) a+2^k b=-2^k+3^k & \left(-1+2^k\right) a-2^k b=2^k-3^k\\
 b+\left(-1+2^k\right) c=-1+3^k & a+2^k c=3^k\\
 a+2^k c=3^k & b+\left(-1+2^k\right) c=-1+3^k \\
\end{array}$
Note: each equation has two variables because I haven't figured out how to combine them to make 3 3-variable equations.
Some pointers would be nice.
 [/INST] Note that using Cramer’s rule may be much better. But, let us go with a brute-force method.
From the last equation, we have: $c = \frac{3^k-a}{2^k}$. Substituting this in the second equation, gives us: $$b + (2^k-1)[\frac{3^k-a}{2^k}]=3^k -1$$ $$\implies b -a+\frac{a}{2^k}=\frac{3^k}{2^k} -1 \implies (1-2^k)a+2^kb= 3^k -2^k$$ which is the same as the first equation! 
This means there are infinite such triples $$(a,b,c) \equiv (a, a - \frac{a}{2^k}+\frac{3^k}{2^k}-1, \frac{3^k-a}{2^k})$$ a fact which can be confirmed using Cramer’s rule as the resultant matrix will have only two non-zero rows, implying the existence of infinite such triples.
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