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 mod 3, then letting 3x be n+3 gives a solution as can be easily checked.  For n=1 mod 3 (and so n is 1 mod 12, 3x needs to be an integer which is 8 mod 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 mod 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.