Find the complete solution to the simultaneous congruence. I'm having trouble understanding the steps involved to do this question so any step by step reasoning in solving the solution would help me study for my exam. 
Thanks so much!
$$x\equiv 6 \pmod{14}$$
$$x\equiv 24 \pmod{29}$$

I usually solve the two sets of equations
$$
\begin{align}
x_{14}&\equiv1\pmod{14}\\
x_{14}&\equiv0\pmod{29}
\end{align}\tag{1}
$$
and
$$
\begin{align}
x_{29}&\equiv0\pmod{14}\\
x_{29}&\equiv1\pmod{29}
\end{align}\tag{2}
$$
and get a solution to the given equations with $6x_{14}+24x_{29}$.
The Euclid-Wallis Algorithm yields
$$
\begin{array}{r}
&&2&14\\
\hline
1&0&1&-14\\
0&1&-2&29\\
29&14&1&0\\
\end{array}\tag{3}
$$
whose last two columns say $29(1-14k)+14(-2+29k)=1$.
From $(3)$, we get one solution $x_{14}=29(1)$ and $x_{29}=14(-2)$. Thus, $6x_{14}+24x_{29}=-498$.
Solutions are unique $\bmod$ $\mathrm{lcm}(14,29)=406$, so the general solution is
$$
314\pmod{406}\tag{4}
$$
