System of congruences with moduli not coprime and chinese remainder theorem application Find all solutions, if any, to the system of congruences
$$\Bigg\{
 \begin{array}{c}
     x \equiv 1 (\mod 6)\\
     x \equiv 7 (\mod 9)\\
     x \equiv 4 (\mod 15)\\
    \end{array}$$
we can see that
$ x \equiv 1 \pmod 6 \implies 
    \left\{
    \begin{array}{l}
        x\equiv 1 \pmod 2 \\
        x \equiv 1 \pmod 3
    \end{array}
    \right.$
as well as
$ x \equiv 7 \pmod 9\\
    \qquad \Rightarrow \quad x=7+9k\\
   \qquad \Rightarrow \quad x=1+3(2+3k)\\
   \qquad \Rightarrow \quad x\equiv 1 \pmod 3\\
$
and
$  x \equiv 4 \pmod{15} \implies 
    \left\{
    \begin{array}{l}
        x\equiv 4 \pmod 5 \\
        x \equiv 1 \pmod 3
    \end{array}
    \right.
$
After removal of redundant congruences and picking the one with the highest power of 3 we are left with:
$
\Bigg\{
 \begin{array}{c}
     x \equiv 1 \pmod 2\\
     x \equiv 7 \pmod {3^2}\\
     x \equiv 4 \pmod 5\\
    \end{array}
$
Now we reduced it to standard CRT problem.
$ x=2k+1 \\
2k+1 \equiv 7\pmod{9}\\
2k \equiv 6\pmod{9}\quad \gcd{(2,9)}=1\\
k\equiv 3\pmod{9}\\
k=9\cdot l+3$
$x=2\cdot(9\cdot l+3)+1=18l+7$
$
18l+7\equiv 4\pmod{5}\\
18l\equiv -3\pmod{5}\\
18l\equiv 2\pmod{5}\\
3l\equiv 2\pmod{5}\\
\quad \Rightarrow \quad 7\pmod{5}\\
\quad \Rightarrow \quad 12\pmod{5} \quad \gcd{(3,5)}=1\\
l\equiv 4\pmod{5}\\
l=5\cdot m +4\\
$
The solution to the original system:
$
x=2\cdot(9\cdot (5\cdot m +4)+3)+1=90\cdot m+79\\
x\equiv 79\pmod{90}
$
with the smallest x = 79
Is it a correct approach?
It is correct, but you can make the solution a bit shorter using the explicit inverse isomorphism in the C.R.T. for two coprime  moduli $a$ and $b$, given a Bézout's relation $ua+vb=1$:

$$x\equiv \begin{cases}\alpha\mod a,\\\beta\mod b,\end{cases}\iff x\equiv \beta ua+\alpha vb\mod ab$$

As $4\cdot 3^2-7\cdot 5=1$, the last two final congruences yield as a first step $$x\equiv 4\cdot 4\cdot 9-7\cdot 7\cdot 5=144-245=-101\mod 45.$$
Next, from $45-22\cdot 2=1$, you deduce from this congruence and the first above that
$$x\equiv 45+101\cdot 22\cdot 2=4489\equiv 79\mod 90.$$
It is correct, but you can make the solution a bit shorter using the explicit inverse isomorphism in the C.R.T. for two coprime  moduli $a$ and $b$, given a Bézout's relation $ua+vb=1$:

$$x\equiv \begin{cases}\alpha\mod a,\\\beta\mod b,\end{cases}\iff x\equiv \beta ua+\alpha vb\mod ab$$

As $4\cdot 3^2-7\cdot 5=1$, the last two final congruences yield as a first step $$x\equiv 4\cdot 4\cdot 9-7\cdot 7\cdot 5=144-245=-101\mod 45.$$
Next, from $45-22\cdot 2=1$, you deduce from this congruence and the first above that
$$x\equiv 45+101\cdot 22\cdot 2=4489\equiv 79\mod 90.$$