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

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 $$x = 14(22+29n) + 6 = 314+406n$$
