To solve $\,26x\equiv 6\pmod{\!110},\,$ note $\,\gcd(26,110)=2\,$ so there are $\,2\,$ solutions.
Next, $\ 104x\equiv 24\pmod{\!110}\ $ follows by scaling the above by $\,4.$
i.e. $\,\ \ \ \ \ {-}6x\equiv 24\pmod{\!110}$
Hence $\ \ \ \ \,x\equiv -4\pmod{\!55}\ $ by dividing prior by $\,6$.
What is the final answer?
Note $\,x\equiv -4\pmod{\!55}\iff x\equiv 51, 106\pmod{\!110},\,$ so those are the sought $2$ solutions.
However you got lucky. The method you used will generally introduce extraneous solutions because you scaled by a number $(4)$ this is noninvertible $\bmod{110}$. To remedy that in general you need to test which of the derived possible solutions are actual solutions. This is not needed when we scale by invertibles since that yields equivalent congruences, i.e. the congruences are connected by $(\!\!\iff\!\!)$ arrows, vs. $(\Rightarrow)$.
For example if we do the same problem mod $220$ and scale by $8$ we deduce that any solution again satisfies $\,x\equiv -4\pmod{\!55},\,$ but only two of those $4$ possible solutions are actually solutions, viz. $\,x\equiv 51\pmod{\!110}\iff x\equiv 51,\,161\pmod{\!220}$.
You can avoid such problems by cancelling $\,2 =\gcd(26,110)\,$ at the start (vs. end), as in the general solution method explained in this answer.
Another handy way to solve such congruences is to use the fractional extended Euclidean algorithm, which is closely related to the method that you (implicitly) employed.
You could first solve $13x\equiv 3\mod 110$ by the extended Euclidean algorithm which is possible since $gcd(13,110)=1$ and then multiply the congruence by $2$ giving a solution of the original congruence.
$26x \equiv 6 \pmod{110} \iff 110 | (26x - 6) \iff $
$55 | (13x - 3) \iff 13x \equiv 3 \pmod{55}.$
$55 = (4)13 + 3.$
$13 = 4(3) + 1.$
Therefore,
$1 = 13 - 4[55 - 4(13)] = 17(13) - 4(55)$
Therefore, $17 \times 13 \equiv 1 \pmod{55}.$
Therefore, $3 \times (17 \times 13) \equiv 3 \equiv 13 x \pmod{55}.$
Therefore, $x \equiv 51 \pmod{55}.$
Therefore $55 | \{(13[51]) - 3\}$.
Therefore $110 | \{(26[51]) - 6\}$.
Therefore $\{26[51]\} \equiv 6 \pmod{110}.$
Therefore $x \equiv 51 \pmod{110}.$
Avoid scaling by a factor that is not comprime to the modulo; you may create/miss solutions.
From $26x \equiv 6 \pmod {110}$, we can divide everything by $2$ to reach $13 x \equiv 3 \pmod {55}$.
Now scaling by $4$ is valid, since $4$ and $55$ are coprime. We have:
$$12 \equiv 52 x \equiv -3 x \pmod {55}$$
Dividing both sides by $-3$, which is also coprime with $55$:
$$x \equiv -4 \pmod {55}$$
which is the solution you somehow managed to arrive at, despite the questionable method.
It is very simple. After rescaling it to $$13x \equiv 3 \pmod{55}$$, you can simply find the modular inverse of $13$ mod $55$, which in this case is $17$(Denoted by $13^{-1}$, and also note that it exists as 13 and 55 are coprime). Multiplying both sides by $17$ you get $$13 \times 13^{-1} x \equiv 3 \times 13^{-1} \pmod {55} $$ $$13 \times 17x \equiv x \equiv 3 \times 17 \pmod {55} $$ $$x \equiv 51 \pmod {55} $$
