Since the GCD of the moduli is $(15,21)=3$, it is necessary that $x$ be the same thing in both equations mod $3$. That is,
$$
x\equiv5\pmod{15}\implies x\equiv2\pmod{3}
$$
and
$$
x\equiv8\pmod{21}\implies x\equiv2\pmod{3}
$$
If we didn't get that $x\equiv2\pmod{3}$ from both equations, a solution would not be possible.
This prompts us to look at $\frac{x-2}3\pmod{\frac{15}3}$ and $\frac{x-2}3\pmod{\frac{21}3}$. That is,
$$
\frac{x-2}3\equiv1\pmod{5}\tag{1}
$$
and
$$
\frac{x-2}3\equiv2\pmod{7}\tag{2}
$$
Using the Extended Euclidean Algorithm as implemented in this answer,
$$
\begin{array}{r}
&&1&2&2\\\hline
1&0&1&-2&5\\
0&1&-1&3&-7\\
7&5&2&1&0
\end{array}\tag{3}
$$
we get that
$$
\underbrace{5(3)}_{\large\color{#C00000}{15}}+\underbrace{\!7(-2)\!}_{\large\color{#00A000}{-14}}=1\tag{4}
$$
We can use $(4)$ to see that
$$
\begin{align}
\color{#00A000}{-14}&\equiv\color{#0000F0}{1}\pmod{5}\\
\color{#00A000}{-14}&\equiv\color{#0000F0}{0}\pmod{7}
\end{align}\tag{5}
$$
and that
$$
\begin{align}
\color{#C00000}{15}&\equiv\color{#0000F0}{0}\pmod{5}\\
\color{#C00000}{15}&\equiv\color{#0000F0}{1}\pmod{7}
\end{align}\tag{6}
$$
To solve $(1)$ and $(2)$ we can add $1$ times $(5)$ to $2$ times $(6)$ to get
$$
\begin{align}
16&\equiv\color{#0000F0}{1}\pmod{5}\\
16&\equiv\color{#0000F0}{2}\pmod{7}
\end{align}\tag{7}
$$
Equations $(7)$ tell us that $\frac{x-2}3\equiv16\pmod{35}$ or that
$$
\bbox[5px,border:2px solid #C0A000]{x\equiv50\pmod{105}}\tag{8}
$$