I have to prove that $2^x + 6(9^x) \equiv 0 \pmod{7}$ is correct for any integer x.
I know that $2 \equiv 9 \pmod{7}$ and so $2^x \equiv 9^x \pmod{7}$. I replace the $9^x$ for the $2^x$ in the equation, so it gives me $2^x + 6(2^x) \equiv 0 \pmod{7}$ in the end the result is $7(2^x)\equiv 0 \pmod{7}$ which proves it. But I don't know it this is correct, as far as I know there is not any property like that.
Any way, if you are interested in elementary solution, you can prove it without congruence. You have to prove $$7\mid 6\cdot 9^x+2^x$$ Write \begin{eqnarray}6\cdot 9^x+2^x &=& 7\cdot 9^x -9^x+2^x \\ &=&7\cdot 9^x -(9^x-2^x) \\&=& 7\cdot 9^x -(9-2)(\underbrace{9^{x-1}+9^{x-2}\cdot 2+...+9\cdot2^{x-2}+2^{x-1}}_{k\in\mathbb{Z}})\\ &=& 7 \cdot 9^x+7k \end{eqnarray} and we are done.