The problem here is that the matrix $\begin{pmatrix} 2 & 3 \\ 1 & -1 \\ \end{pmatrix}$ is invertible but not unimodular and hence the elements $2x+3y$ and $x-y$ generate a free abelian group of rank 2 but still a proper subgroup of $\langle x,y\rangle$. But the question was to prove that $2x+3y$ and $x-y$ form a basis.
This question is problem 3 of section-67 in Topology by Munkres.
EDIT: The question verbatim from Munkres book is this: If $G$ is free abelian with basis $\{x,y\}$, show that $\{2x+3y, x-y\}$ is also a basis for $G$ Thanks