You understand, do you not, that you can write the system of equations as
\(\displaystyle \begin{bmatrix}\frac{dx}{dt} \\ \frac{dy}{dt}\end{bmatrix}= \begin{bmatrix}1 & 3 \\ 2 & 2 \end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}\)
There are several different ways to solve such an equation and since you have not shown any attempt of your own, we don't know which methods you have been taught. This is what I would do:
If P is the matrix with columns being the Eigenvectors of matrix A, then \(\displaystyle P^{-1}AP\) is the diagonal matrix, D, having the eigenvalues of A on its diagonal
So if we have the matrix differential equation \(\displaystyle \frac{dX}{dt}= AX\), we can multiply both sides by \(\displaystyle P^{-1}\) to get \(\displaystyle \frac{d(P^{-1}X)}{dt}= P^{-1}Ax\) and we can write that as \(\displaystyle \frac{d(P^{-1}X)}{dt}= P^{-1}A(PP^{-1})X= (P^{-1}AP)(P^{-1}X)\)
Let \(\displaystyle Y= P^{-1}X\) and we can write the equation as \(\displaystyle \frac{dY}{dt}= DY\) which, because D is diagonal will give two separate, uncoupled, equations for the components of Y. Then, of course, \(\displaystyle X= PY\).