<s>[INST] Is there a positive integer n such that the fraction $(9n+5)/(10n+3)$ is not in the lowest term? Is there a positive integer n such that the fraction $(9n+5)/(10n+3)$ is not in the lowest term? Please explain. 
I found $n=2$ be a solution. Is it correct?
 [/INST] Your example $n=2$ is correct. We will find all possible examples.
Suppose that $d$ divides $9n+5$ and $10n+3$. Then $d$ divides 
$$10(9n+5)-9(10n+3),$$
so $d$ divides $23$. 
Since $23$ is prime, the only conceivable (positive) common divisors of $9n+5$ and $10n+3$ are $1$ and $23$. We will find the values of $n$ for which $23$ is a common divisor.
To make sure that $23$ divides both, all we need is to make sure that $23$ divides $9n+5$. This is because from $10(9n+5)-9(10n+3)=23$ we can conclude that if $23$ divides $9n+5$, it must divide $10n+3$.
You noticed that $23$ divides $9n+5$ when $n=2$. It follows that $23$ divides $9n+5$ if and only if $n=2+23k$ for some integer $k$.  For if $23$  divides $9n+5$, then $23$ divides $9n+5-23$, so $23$ divides $9(n-2)$, and therefore $n-2$ is a multiple of $23$. Conversely, if $n-2$ is a multiple of $23$, say $n-2=23k$, then a calculation shows that $23$ divides $9n+5$.
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