I want to know how to check for which natural numbers fraction $\frac{7n+19}{7n+2}$ can be simplified. The correct answer is supposed to be $n = 2k-1$.
I do understand that the denominator and the numerator have to have a common divisor in order to simplify the fraction, but fail to see how this particular problem should be approached.
$$\frac{7n+19}{7n+2}=1+\frac{17}{7n+2}$$ Thus, in order for $\dfrac{7n+19}{7n+2}$ to be simplified, we have that $\dfrac{17}{7n+2}$ can be simplified as well.
This implies that $17$ and $7n+2$ must have a common divisor that is not $1$. Since $17$ is prime, this implies that $$7n+2 \equiv 0 \pmod {17}$$ This gives us that $$n \equiv 7 \pmod {17}$$ Thus, the answer is that $n=17k+7$ where $k \in \mathbb{Z}$, not $2k-1$ as stated, so your answer is incorrect.