To solve the system of linear equations:

\[
\begin{cases}
x - 3y = 5 \\
2x + 3y = 1
\end{cases}
\]

we can use the method of elimination or substitution. Here, we'll use elimination.

First, add the two equations together to eliminate \( y \):

\[
(x - 3y) + (2x + 3y) = 5 + 1
\]

This simplifies to:

\[
3x = 6
\]

Solving for \( x \):

\[
x = \frac{6}{3} = 2
\]

Now that we have \( x = 2 \), substitute this value back into one of the original equations to solve for \( y \). We'll use the first equation:

\[
x - 3y = 5
\]

Substitute \( x = 2 \):

\[
2 - 3y = 5
\]

Solving for \( y \):

\[
-3y = 5 - 2
\]

\[
-3y = 3
\]

\[
y = \frac{3}{-3} = -1
\]

So, the solution to the system of equations is:

\[
x = 2, \quad y = -1
\]

To verify, substitute \( x = 2 \) and \( y = -1 \) back into the original equations:

1. \( 2 - 3(-1) = 5 \)
\[ 2 + 3 = 5 \]
\[ 5 = 5 \] (True)

2. \( 2(2) + 3(-1) = 1 \)
\[ 4 - 3 = 1 \]
\[ 1 = 1 \] (True)

Both equations are satisfied, so the solution \( x = 2 \) and \( y = -1 \) is correct.