Hint $\quad\begin{eqnarray}\rm\ mod\ m_1\!:\ \ x_0 &=&\:\rm m_1 b_1 a_2 + m_2 b_2 a_1 \\ &\equiv&\rm\ 0\cdot b_1 a_2 + \ \ 1\ \cdot\:\ a_1\: \equiv\: \ldots\ \ \end{eqnarray} $ by $\rm\ m_1\equiv 0,\ \:m_2b_2\equiv 1$.
Key is: $\rm\:mod\ (m_1,m_2)\!:\:\ m_1 b_1 \equiv (0,1),\ \ m_2 b_2\equiv (1,0),\:$ and these vectors span since
$$\rm (a_1,a_2)\ =\ a_1\:(1,0)\: +\: a_2\:(0,1)$$
The innate algebraic structure will be clearer when you study the Peirce direct sum decomposition induced by (orthogonal) idempotents.
It's usually easier to solve the congruences successively, i.e. a pair at a time, replacing a pair of congruences by their solution's congruence. So we take the general solution of the first congruence, substitute it into the second, solve that, then substitute that solution into the third, etc, continually replacing the top $\,2\,$ congruences by an equivalent congruence, as summarized below
$\qquad\quad\begin{align} x\equiv 3\!\!\!\pmod{\! 5}\\ x\equiv 1\!\!\!\pmod{\! 4}\\ x\equiv 2\!\!\!\pmod{\! 3}\end{align}\! $
$\iff \begin{align} \\ x\equiv \color{#0a0}{13}\!\!\!\pmod{\!\color{#0a0}{20}}\\ x\equiv \ \ 2 \!\!\!\pmod{\!3} \ \ \end{align}\!\!$
$\begin{align} \\ \\ \iff x\equiv \color{#c00}{53}\!\!\!\pmod{\!60}\end{align}$
Below is the congruence arithmetic justifying the above reductions.
${\rm mod}\ 5\!:\,\ 3\equiv x\iff x = \color{#90f}{3\! +\! 5\:\!j}.\,$ Substituting this value of $\,x\,$ in the 2nd congruence we get
${\rm mod}\ 4\!:\,\ 1\equiv x = \color{#90f}{3\!+\!5\:\!j}\equiv -1\!+\!j\iff j\equiv 2\iff \color{#0a0}{j=2\!+\!4k}$
Thus the first $2$ congruence are equivalent to $\,x = \color{#90f}{3\!+\!5}\:\!\color{#0a0}j \equiv \color{#0a0}{13\!+\!20k}.\,$ Again, substituting this value of $\,x\,$ into the 3rd congruence yields
${\rm mod}\ 3\!:\,\ 2\equiv x = \color{#0a0}{13+20k}\equiv 1\!-\!k\iff k\equiv 2\iff \color{#c00}{k=2\!+\!3n} $
Therefore the $3$ congruences are equivalent to $\,x = \color{#0a0}{13\!+\!20}\:\!\color{#c00}k = \color{#c00}{53+60n}$
Remark $ $ Below is a summary of the reduction of the top two congruences:
$\qquad \begin{align} x\equiv 3\!\!\!\pmod{\! 5}\\ x\equiv 1\!\!\!\pmod {\!4}\end{align}$
$\iff \begin{align} x = \color{#90f}{3\! +\! 5\color{#0a0}j}\\ \color{#0a0}{j=2\!+\!4k}\end{align}$ $\iff x =\color{#90f}{3\!+\!5}(\color{#0a0}{2\!+\!4k})\overbrace{= \color{#0a0}{13+20k}}^{\large\,\ \equiv\ \ \color{#0a0}{13}\!\pmod{\!\color{#0a0}{\!\!20}}\!}$
Note that all reductions are $\iff$ i.e. "if and only if", i.e. equivalences, because we desire to replace two congruences by an equivalent congruence. This means each reduction step must be reversible. In particular, this means that if we encounter a congruence of the form $\,ac\, x\equiv bc\pmod{\!mc}\,$ then any cancelling of $\,c\neq 0\,$ must also be done to the modulus, i.e.
$$ a\color{#c00}c\,x\equiv b\color{#c00}c\!\!\!\pmod{m\color{#c00}c}\iff mc\mid (ax\!-\!b)c \iff m\mid ax\!-\!b\iff ax\equiv b\!\!\!\pmod m$$
i.e. we must cancel $\,\color{#c00}{c\,\ \rm everywhere}\,$ (compare here for a fractional view of this). Such cancellations occur frequently in the general case when the moduli are not pair-coprime.
Notice we ordered the congruences by largest-moduli-first, so that when the numbers are bigger near the end, the moduli are smaller, so modular arithmetic is easier. This optimization pays off much more when larger moduli are present.
See here for a general formula (Easy CRT) for solving a pair of congruences, including the case of noncoprime moduli.
Best Answer
$\begin{eqnarray}&&x\equiv\ \ 7\equiv \color{#c00}{16}\pmod 9\\ &&x\equiv\ \ 4\equiv \color{#c00}{16}\pmod {12}\\ &&x\equiv 16\equiv \color{#c00}{16}\pmod{21}\end{eqnarray}$ $\iff$ $\,9,12,21\mid x\!-\!\color{#c00}{16}$ $\iff$ ${\rm lcm}(9,12,21)\mid x\!-\!\color{#c00}{16}$
Finally $\ {\rm lcm}(9,12,21) = {\rm lcm}(3^{\large\color{#0a0} 2},\,3\cdot 2^{\large \color{#0a0}2},\,3\cdot 7) = 3^{\large\color{#0a0}2}\cdot 2^{\large\color{#0a0}2}\cdot 7 = 252.$
Alternatively, algorithmically, by the third congruence $\ x = 16+21n\,$ for an integer $\,n.\,$ Hence
${\rm mod}\ 12\!:\ 4\equiv x= 16+21n\equiv 4-3n\iff 3n\equiv 0\iff 12\mid 3n\iff 4\mid n\iff\ n = 4k$
${\rm mod}\,\ \ 9\!:\,\ 7 \equiv x = 16+84k\equiv 7+3k\ \iff 3k\equiv 0\iff\,\ 9\mid3k\,\iff 3\mid k\,\iff\, k = 3j$
We've proved $\ x = 16+84(3j) = 16+252j.$
Remark $\ $ Note, in particular, that there is no need to split into pairwise coprime moduli as in David's answer. Generally, proceeding as above will yield a simpler method - often much so.