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!
$$x\equiv 6 \pmod{14}$$
$$x\equiv 24 \pmod{29}$$
 A: Hint: use the Chinese Remainder Theorem (CRT)
We have pairwise prime moduli, so it can be directly applied.
And see similar posts as to how it can be applied to various systems of linear congruences:


*

*Find all solutions to linear congruences 1

*Find all solutions to linear congruences 2

*Solving congruences and the CRT
@Jyriki's comments are very helpful, so I won't duplicate in my post the hints and pointers made in the comments. But feel free to check back/comment with your progress, and/or if/where you might be getting stuck.
A: Applying Easy CRT (below), noting that $\rm\displaystyle\, \frac{-18}{29}\equiv \frac{-4}{1}\ \ (mod\ 14),\ $ quickly yields
$$\begin{array}{ll}\rm x\equiv \  \ 6\ \ (mod\ 14)\\ \rm x\equiv 24\ \ (mod\ 29)\end{array}\rm \!\iff\! x\equiv 24\! +\! 29 \left[\frac{-18}{29}\, mod\ 14\right]\!\equiv 24\!+\!29[\!-4]\equiv -92\equiv 314\,\ (mod\ 406) $$
Theorem (Easy CRT) $\rm\ \ $ If $\rm\ m,n\:$ are coprime integers then $\rm\ n^{-1}\ $ exists $\rm\ (mod\ m)\ \ $ and
$\rm\displaystyle\quad \begin{eqnarray}\rm x&\equiv&\rm\ a\ \ (mod\ m) \\
\rm x&\equiv&\rm\ b\ \ (mod\ n)\end{eqnarray} \! \iff x\ \equiv\ b + n\ \bigg[\frac{a\!-\!b}{n}\ mod\ m\:\bigg]\ \ (mod\ mn)$
Proof $\rm\ (\Leftarrow)\ \ \ mod\ n\!:\,\ x\equiv b + n\ [\cdots]\equiv b,\ $ and $\rm\,\ mod\ m\!:\,\ x\equiv b + (a\!-\!b)\ n/n\: \equiv\: a\:.$
$\rm\ (\Rightarrow)\ \ $ The solution is unique $\rm\ (mod\ mn)\ $ since if $\rm\ x',x\ $ are solutions then $\rm\ x'\equiv x\ $ mod $\rm\:m,n\:$ therefore $\rm\ m,n\ |\ x'-x\ \Rightarrow\ mn\ |\ x'-x\ \ $ since $\rm\ \:m,n\:$ coprime $\rm\:\Rightarrow\ lcm(m,n) = mn\:.\ \ $ QED
Remark $\ $ I chose $\rm\: n,m = 29,14\ $ (vs. $\rm\, 14,29)\:$ since then $\rm\:n \equiv 1\,\ (mod\ m),\:$ making completely trivial the computation of $\rm\,\ n^{-1}\ mod\ m\,\ $ in the bracketed term in the formula.
A: For your problem, note that we want
$$x = 14k_1 + 6 = 29k_2 + 24$$
Hence, we want to find $(k_1,k_2)$ such that
$$29k_2 - 14k_1 = -18 \,\,\,\,\, (\clubsuit)$$
First note that if $(k_1^0,k_2^0)$ are solutions, then $(k_1^0 + 29 n, k_2^0 + 14n)$ are solutions for all $n \in \mathbb{Z}$. Further, these are all the only possible solutions since $\gcd(14,29) = 1$.
The Chinese remainder theorem essentially generalizes the above two lines and you can read more about it from the links @amWhy has posted.
Hence, the goal is to find/guess at-least one solution $(k_1,k_2)$ satisfying $(\clubsuit)$. In your case, $(22,10)$ is one candidate. Hence, all the solutions are of the form
$$(k_1,k_2) = (22+29n,10+14n)$$
This means $$x = 14(22+29n) + 6 = 314+406n$$
A: 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
$$
314\pmod{406}\tag{4}
$$
