 Solution

Let $F=\mathbb{Z}{e}_1\oplus \mathbb{Z}{e}_2$ and $N=\mathbb{Z}(6e_1+9e_2)$. The precise presented group is $A=F/N$, with $x=e_1+N$ and $y=e_2+N$. It is abelian and has exactly the imposed additive relation and its integer multiples.

Change the integral basis to $u=2e_1+3e_2$, $v=e_1+2e_2$. The change-of-basis determinant is $2\cdot 2-3\cdot 1=1$, so these are a basis of the free abelian group. Explicitly, $e_1=2u-3v$ and $e_2=-u+2v$. The relation generator becomes $6e_1+9e_2=3u$. ThereforeThe image of $u$ is nonzero, since $(2,3)$ is not an integer multiple of $(6,9)$, but three times it vanishes. Thus $A$ has an element of exact order three. A free abelian group has no nonzero finite-order element, so $A$ is not free abelian.