If we factor $n_k$ into primes, $n_k = p_{1}^{b_{1}}\cdots p_r^{b_{r}}$, then the Chinese Remainder Theorem tells us that $x\equiv a_k\pmod{n_k}$ is equivalent to the system of congruences
$$\begin{align*}
x&\equiv a_k\pmod{p_1^{b_{1}}}\\
x&\equiv a_k\pmod{p_2^{b_{2}}}\\
&\vdots\\
x&\equiv a_k\pmod{p_r^{b_{r}}}
\end{align*}$$
Thus, we can replace the given system of congruences with one in which every modulus is a prime power, $n_i = p_i^{b_i}$.
Note that the assumption that $a_i\equiv a_j\pmod{\gcd(n_i,n_j)}$ "goes through" this replacement (if they were congruend modulo $\gcd(n_i,n_j)$, then they are congruent modulo the gcds of the prime powers as well).
So, we may assume without loss of generality that every modulus is a prime power.
I claim that we can deal with each prime separately, again by the Chinese Remainder Theorem. If we can solve all congruences involving the prime $p_1$ to obtain a solution $x_1$ (which will be determined modulo the highest power of $p_1$ that occurs); and all congruences involving the prime $p_2$ to obtain a solution $x_2$ (which will be determined modulo the highest power of $p_2$ that occurs); and so on until we obtain a solution $x_n$ for all congruences involving the prime $p_n$ (determined modulo the highest power of $p_n$ that occurs), then we can obtain a simultaneous solution by solving the usual Chinese Remainder Theorem system
$$\begin{align*}
x &\equiv x_1 \pmod{p_1^{m_1}}\\
&\vdots\\
x &\equiv x_n\pmod{p_n^{m_n}}
\end{align*}$$
(where $m_i$ is the highest power of $p_i$ that occurs as a modulus).
So we are reduced to solving figuring out whether we can solve the system
$$\begin{align*}
x &\equiv a_1\pmod{p^{b_1}}\\
x &\equiv a_2\pmod{p^{b_2}}\\
&\vdots\\
x & \equiv a_n\pmod{p^{b_n}}
\end{align*}$$
with, without loss of generality, $b_1\leq b_2\leq\cdots\leq b_n$.
When can this be solved? Clearly, this can be solved if and only if $a_i\equiv a_j\pmod{p^{b_{\min(i,j)}}}$: any solution must satisfy this condition, and if this condition is satisfied, then $a_n$ is a solution.
For example: say the original moduli had been $n_1 = 2^3\times 3\times 7^2$, $n_2= 2^2\times 5\times 7$, $n_3=3^2\times 5^3$. First we replace the system with the system of congruences
$$\begin{align*}
x&\equiv a_1 \pmod{2^3}\\
x&\equiv a_2\pmod{2^2}\\
x&\equiv a_1\pmod{3}\\
x&\equiv a_3\pmod{3^2}\\
x&\equiv a_2\pmod{5}\\
x&\equiv a_3\pmod{5^3}\\
x&\equiv a_1\pmod{7^2}\\
x&\equiv a_2\pmod{7}.
\end{align*}$$
Then we separately solve the systems:
$$\begin{align*}
x_1&\equiv a_1 \pmod{2^3} &x_2&\equiv a_1\pmod{3}\\
x_1&\equiv a_2\pmod{2^2}&x_2&\equiv a_3\pmod{3^2}\\
\strut\\
x_3&\equiv a_2\pmod{5}&x_4&\equiv a_1\pmod{7^2}\\
x_3&\equiv a_3\pmod{5^3}&x_4&\equiv a_2\pmod{7}.
\end{align*}$$
Assuming we can solve these, $x_1$ is determined modulo $2^3$, $x_2$ modulo $3^2$, $x_3$ modulo $5^3$, and $x_4$ modulo $7^2$, so we then solve the system
$$\begin{align*}
x &\equiv x_1\pmod{2^3}\\
x &\equiv x_2\pmod{3^2}\\
x&\equiv x_3 \pmod{5^3}\\
x&\equiv x_4\pmod{7^2}
\end{align*}$$
and obtain a solution to the original system.
Hence, if the condition $a_i\equiv a_j\pmod{\gcd(n_i,n_j)}$ holds in the original system, then we obtain a solution for each prime, and from the solution for each prime we obtain a solution to the original system by applying the usual Chinese Remainder Theorem twice.
To use your example, trying to solve:
$$x\equiv 10\pmod{24}\\x\equiv 18\pmod{64}$$
You take each in terms of its prime factors:
$$x\equiv 10\pmod {3}\\
x\equiv 10\pmod{8}\\x\equiv 18\pmod {64}
$$
Note that $x\equiv{18}\pmod {64}$ implies $x\equiv 10\pmod 8$, so we are left with:
$$x\equiv 10\pmod {3}\\x\equiv 18\pmod {64}$$
which is a pair of equations with coprime moduli.
You can always do this. Prime factorization is hard, however, so if you wanted an algorithm, you'd want to come up with a more general way than just prime-factorization. There are ways, but I'd have to call them up from memory. But the general gist is that you need to always reduce the problem to the co-prime case, and some way to check for contradictions that rule out solutions.
Best Answer
The Chinese remainder theorem works for pairwise co-prime moduli,
and, as you noted, yours are not pairwise co-prime. Now
$x\equiv3\bmod10\iff x\equiv1\bmod2$ and $x\equiv3\bmod5$,
$x\equiv8\bmod15\iff x\equiv2\bmod3$ and $x\equiv3\bmod5$, and
$x\equiv5\bmod84\iff x\equiv2\bmod3, x\equiv1\bmod4$ and $x\equiv5\bmod7$.
Putting this all together, your system is satisfied when
$x\equiv3\bmod5, x\equiv2\bmod3, x\equiv5\bmod7, $ and $x\equiv1\bmod4$,
which you solved (per comments) using CRT to get $x\equiv173\bmod420$.