<s>[INST] Solving Chinese Remainder Theorem Algebraically I am doing a practice problem for my final which asks:
Solve the following Chinese Remainder Theorem:
$$
x \equiv 2  \pmod{3}, \\
x \equiv 3 \pmod{5}, \\
x \equiv 5 \pmod{7}, \\
x \equiv 7 \pmod{11} \\
x \equiv 11 \pmod{13}
$$
From the first I can conclude that $x = 3k + 2$ for some $k \in \mathbb{Z}$.
Now I can apply that to the second equation which gives $ 3k+2 \equiv 3 \pmod{5}.$
Then I get lost here. Do I subtract $2$ and solve $ 3k \equiv 1 \pmod{5}$?
I don't have a solid understanding of solving the Chinese Remainder Theorem algebraically in general.
 [/INST] \begin{align}
    x &\equiv 2 \pmod 3 \\
    x &= 2 + 3a \\
\hline
    x &\equiv 3 \pmod 5 \\
    2+3a &\equiv 3 \pmod 5 \\
    3a &\equiv 1 \pmod 5 \\
    a &\equiv 2 \pmod 5 \\
    a &= 2 + 5b \\
    x &= 2 + 3(2 + 5b)\\
    x &= 8 + 15b \\
\hline
    x &\equiv 5 \pmod 7 \\
    8 + 15b &\equiv 5 \pmod 7 \\
    1 + b &\equiv 5 \pmod 7 \\
    b &\equiv 4 \pmod 7 \\
    b &= 4 + 7c \\
    x &= 8 + 15(4 + 7c) \\
    &\text{and so on...}
\end{align}
</s>