Hint $\,\bmod 9\,$ invertibles have form $\,2^{\large n}$ so $\,2^{\large n} x\equiv 2 \iff x \equiv 2^{\large\:\! 7-n},\,\ n = 0,1,2,\ldots,5 $
Example $\ $ For $\,n = 2\,$ the above says that $\, 2^{\large 2} x\equiv 2\iff x\equiv 2^{\large 5}\equiv 5.\,$ Indeed $\,2^{\large 2} 5\equiv 2\,\color{#c00}\checkmark$
Remark $\ $ Since $\,-7\equiv 2\,$ is invertible $\bmod 9\,$ so too is its factor $\,a := 2n\!+\!1.\,$ Or, more explicitly, $\,ax\equiv 2\,\iff ax\,2^{-1}\equiv 1\iff a^{-1}\equiv 2^{-1}x\equiv 5x$
That every invertible has form $\,2^{\large n}$ follows because $\,2\,$ is a primitive root
$\bmod 3^{\large 2}\,$ (generally a p.r. $\,g \bmod p\,$ persists as a p.r. $\bmod p^k\,$ except if $\, g^{\large p-1}\!\equiv 1\pmod{\!p^2};\,$ where instead $\,g\! +\! p\,$ works).
Directly: $\,a\,$ invertible $\!\bmod 9\iff a\,$ invertible $\!\bmod 3\iff a = \pm1 + 3j,\,$ so
$\!\bmod \color{#c00}9\!:\,\ ax = (\pm1 + 3j)x \equiv 2\iff x\equiv \dfrac{2}{\pm1 + 3j} \equiv \dfrac{2(1-\color{#c00}9j^2)}{\pm1 + 3j\ \ } \equiv 2(\pm1 -3j)$
Example $\ \ \ \ \ \ a = 1+3\iff x \equiv 2(1-3)\equiv 5,\,$ same as above
Starting from your result of
$$219y \equiv -422 \pmod{41} \tag{1}\label{eq1A}$$
the first thing I recommend doing is reducing the coefficients to smaller values since these are generally easier to deal with, plus sometimes you may be able to find simplifications you can use, such as common factors relatively prime to $41$ so they can be "removed" using their multiplicative inverses. Doing this, and using that $\gcd(2, 41) = \gcd(7, 41) = 1$, gives
$$\begin{equation}\begin{aligned}
14y & \equiv -12 \pmod{41} \\
7y & \equiv -6 \pmod{41} \\
7y & \equiv 35 \pmod{41} \\
y & \equiv 5 \pmod{41}
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
Using this result in your first congruence equation, along with $\gcd(28, 41) = 1$, gives
$$\begin{equation}\begin{aligned}
28x + 85 & \equiv 18 \pmod{41} \\
28x + 3 & \equiv 18 \pmod{41} \\
28x & \equiv 15 \pmod{41} \\
28x & \equiv 56 \pmod{41} \\
x & \equiv 2 \pmod{41}
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
In general, though, if you can't fairly easily simplify the results as I did above, you can instead use something like either the constructive proof or direct proof methods described in Wikipedia's "Chinese remainder theorem" article.
Best Answer
$\begin{align} {\bf Hint}\ \bmod 9(n\!-\!1)\!:\ \, &3n\!-\!2\:\! =\:\! \overbrace{ 1\,+\,\color{#0a0}\epsilon}^{\large 1\ +\ \color{#0a0}{3(n-1)}\!\!\!\!\!\!\!\!\!\!}\ \ \ \& \ \ \color{#c00}{\varepsilon^2 \equiv 0} \ \ \ \rm so\\[.3em] &\!\!\dfrac{1}{3n\!-\!2}\equiv\dfrac{1-\color{#c00}{\epsilon^2}}{1+\epsilon\ \ }\equiv 1-\color{}\epsilon \equiv 4\!-\!3n\end{align}$
$\ $ which implies that: $\ \ \ \,(3n\!-\!2)\,x\:\!\equiv\:\! a\, \iff\, x\equiv (4\!-\!3n)\,a$
Remark $ $ The same works for higher powers $\color{#c00}{\epsilon^k\equiv 0}\,\Rightarrow\, 1+\epsilon\,\mid\, 1 = 1-\color{#c00}{\epsilon^k}.\,$ This simple inversion method is one of the prototypical methods discussed in this answer on "simpler multiple" methods. More generally this is a special case of successive approximations schemes such as Hensel lifting of solutions, or general Newton's methods.
Or we can invert using the augmented-matrix form of the extended Euclidean algorithm.
$\begin{eqnarray} [\![1]\!]&& && &\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! f := 9n\!-\!9\ =&\, \left<\,\color{#c00}1,\,\color{#0a0}0\,\right>\quad\ \ {\rm i.e.}\ \ \ \ \ \ f\, =\ \color{#c00}1\cdot f\, +\, \color{#0a0}0\cdot g\\ [\![2]\!]&& &&\qquad\ \ \, g :=3n\!-\!2 &\!\!=&\, \left<\,\color{#c00}0,\,\color{#0a0}1\,\right>\quad\ \ {\rm i.e.}\ \ \ \ \ \ g\, =\ \color{#c00}0\cdot f\, +\, \color{#0a0}1\cdot g\\ [\![3]\!]&=&[\![1]\!]-3[\![2]\!]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! &&\qquad\qquad\ \ \ \ \ \ \ \ \,{-}3 \,&\!\!=&\, \left<\,\color{#c00}1,\,\color{#0a0}{-3}\,\right>\ \ \ {\rm i.e.}\ \ \ {-}3\, =\, \color{#c00}1\cdot f\,\color{#0c0}{-\,3}\cdot g\\ [\![4]\!]&=&[\![2]\!]+n[\![3]\!]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! &&\qquad\qquad\qquad\ {-}2 \,&\!\!=&\, \left<\,\color{#c00}{n},\,\color{#0a0}{1\!-\!3n}\,\right>\\ [\![5]\!]&=&[\![4]\!]\ -\ [\![3]\!]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! &&\qquad\qquad\qquad\ \,&\!\!\!\!\!\! 1\ =&\, \left<\,\color{#c00}{n\!-\!1},\,\color{#0a0}{4\!-\!3n}\,\right>\\ \end{eqnarray}$
Hence the prior line implies: $ \ \ \underbrace{1\:\! =\, (\color{#c00}{n\!-\!1})f + (\color{#0a0}{4\!-\!3n})g}_{\large\text{Bezout equation for }\!\gcd(f,g)}\:\!\Rightarrow\, \smash{\bbox[7px,border:1px solid #c00]{ 1\equiv (4\!-\!3n)g\,\pmod{\!f}}} $