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!
$$x\equiv 6 \pmod{14}$$
$$x\equiv 24 \pmod{29}$$

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.
