<s>[INST] When is $(12x+5)/(12y+2)$ not in lowest terms? I am struggling to solve this problem and would appreciate any help:

When is $\frac{12x+5}{12y+2}$ NOT in lowest terms? ($x$,$y$ are nonnegative integers)

I have found that it is not in lowest terms for $x=6$ and $y=9$ because numerator and denominator are divisible by $11$, but I'm stuck here.

EDIT: Apparently "lowest terms" isn't in common usage in maths, so I will have to explain what it means. A fraction $p/q$ with $p,q\in \mathbb{Z}$ and $q\ne 0$ is in lowest terms when $\gcd(p,q)=1$. Otherwise, it is not in lowest terms.
For example, $\frac{3}{5}$ and $\frac{9}{2}$ are in lowest terms, but $\frac{15}{3}$ and $\frac{17}{34}$ are not.
 [/INST] When is $\dfrac{12x+5}{12y+2} $ irreducible?
This is really just a partial answer.
So, when does $12x + 5 = A$ and $12y + 2 = B$ where $\gcd(A, B)=1$
Note that
\begin{align}
   12 &\mid 2A - 5B \\
   5B &\equiv 2A \pmod{12} \\
   B &\equiv 10A \pmod{12} \\
   B &= 12n + 10A
\end{align}
So $\gcd(A,B)=1$ becomes $\gcd(12n, A) = 1$.
\begin{align}
   12y + 2 &= B \\
   12y + 2 &= 12n + 10A \\
   12y &= 12n + 10A - 2 \\
\hline
   10A &\equiv 2 \pmod{12} \\
   5A &\equiv 1 \pmod 6 \\
    A &\equiv - 1 \pmod 6 \\
    A &= 6\alpha - 1 \\
\hline 
   12y &= 12n + 10(6\alpha - 1) - 2 \\
   12y &= 12n + 60\alpha - 12 \\
   y &= n + 5\alpha - 1 \\
\hline
   B &= 12n + 60\alpha - 10
\end{align}
Finally, we solve for $\alpha$.
\begin{align}
   12x + 5 &= A \\
   12x &= 6\alpha - 6 \\
    2x &= \alpha - 1 \\
  \alpha &= 2x + 1
\end{align}

Pick any integer value for $x$.
$ \alpha = 2x + 1 $
$ A = 12x + 5 $
Pick any $n$ such that $\gcd(12n, A) = 1$.
$B = 12n + 10A$
$y = n + 10x + 4$ implies
$$ \dfrac{12x+5}{12y+2} = \dfrac{12x+5}{12n + 120x + 50}$$
which is irreducible as long as $\gcd(12n, 12x+5) = 1$.
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