Find all the numbers $n$ such that $\frac{6n-8}{2n-5}$ can't be reduced.
Attempt:
It can't be reduced when $\gcd(6n-8,2n-5)=\color{red}1$
$$1 = \gcd(6n-8,2n-5)=\gcd(4n-3,2n-5)=\gcd(2n+2,2n-5)=\gcd(7,2n-5)$$
that's equale to one when $2n-5\not\equiv 0 \pmod 7$ i.e $n\not\equiv 6 \pmod 7$
Is my attempt correct?
Yes, that is correct. By the way you can get $1 = \gcd(6n - 8, 2n - 5) = \gcd(7, 2n - 5)$ directly by applying on step of the Euclidean algorithm.
Your attempt is completely correct. It might be fun to check what happens when $n \equiv 6 \pmod{7}.$ In this case you can write $n = 6 + 7k$ for some $k$. Then the fraction can be written as $$ \frac{6(6+7k)-8}{2(6+7k)-5}=\frac{28+42k}{7+14k}=\frac{4+6k}{1+2k}. $$ By the same argument as yours we find that $$\gcd(4+6k,1+2k)=\gcd(1,1+2k)=1$$ for all $k$. We already knew from your findings that this has to be the case of course.
