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.
$\begin{eqnarray}&&x\equiv\ \ 7\equiv \color{#c00}{16}\pmod 9\\ &&x\equiv\ \ 4\equiv \color{#c00}{16}\pmod {12}\\ &&x\equiv 16\equiv \color{#c00}{16}\pmod{21}\end{eqnarray}$ $\iff$ $\,9,12,21\mid x\!-\!\color{#c00}{16}$ $\iff$ ${\rm lcm}(9,12,21)\mid x\!-\!\color{#c00}{16}$
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.
Best Answer
Since these moduli aren't pairwise relatively prime, we need to double-check to be certain there is a solution.
$x\equiv5\pmod 6$ leads to $x\equiv 1\pmod2$ and $x\equiv 2\pmod3$.
$x\equiv3\pmod {10}$ leads to $x\equiv 1\pmod2$ and $x\equiv 3\pmod5$.
$x\equiv8\pmod {15}$ leads to $x\equiv 2\pmod3$ and $x\equiv 3\pmod5$.
Since these are all compatible, there is a solution, which is also the solution to $x\equiv 1\pmod2$ and $x\equiv 2\pmod3$ and $x\equiv 3\pmod5$. You should have better luck solving that system since the moduli are pairwise relatively prime.