One can use common subexpressions to get a simple answer. Note that $a=n(n+3)$ must be a multiple of $4$. Setting $y=3x$, we look for $a+y$ is a multiple of $12$ and $a(a+y)$ is a multiple of $12y$. If we can pick $y$ to meet $a+y$ is a multiple of $12$, then $y$ is an integer and it suffices to also pick $y$ being a divisor of $a$. This is possible for $n=5$ and in general $y$ being a $2 mod 3$ divisor of $n(n+3)$ should work. I find $x=(n+3)/3$ works for many $n$.
Edit 2018.01.16
The problem is a little more intriguing. If $n$ is $0$ or $2 \equiv1 \pmod 3$, then letting $3x$ be $n+3$ gives a solution as can be easily checked. For $n=1 \equiv1 \pmod 3$ (and so $n$ is $1 \equiv1 \pmod {12}$, $3x$ needs to be an integer which is $8 \equiv1 \pmod {12}$ to satisfy the first relation, and setting $y=3x$ and $a=n(n+3)$ gives $12y$ has to divide $a(a+y)$. This is easy if $y$ divides $a$, while if $y$ does not divide a then with $d$ being the greatest divisor of $a$ and $y$, and $b =y/d$, we get $b$ has to divide $d$. For small $n$ with $n$ being $1 \equiv1 \pmod {12}$, we can find such a divisor $y$ which divides $a$, but it is not clear that we can always do that. So far $y=8$ or $20$ works for small $n$.
End Edit 2018.01.16
Gerhard "Addition More Complex Than Multiplication?" Paseman, 2018.01.16.