To solve the system of linear equations:

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

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

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

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

Simplifying this, we get:

\[
2x - y + x + y = 6
\]

\[
3x = 6
\]

Solving for \( x \):

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

Now that we have \( x = 2 \), we can substitute this value back into one of the original equations to find \( y \). Let's use the second equation:

\[
x + y = 1
\]

Substitute \( x = 2 \):

\[
2 + y = 1
\]

Solving for \( y \):

\[
y = 1 - 2 = -1
\]

So, the solution to the system of equations is:

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

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

1. For \( 2x - y = 5 \):

\[
2(2) - (-1) = 4 + 1 = 5 \quad \text{(True)}
\]

2. For \( x + y = 1 \):

\[
2 + (-1) = 1 \quad \text{(True)}
\]

Both equations are satisfied, so the solution is correct. The solution to the system is:

\[
\boxed{(2, -1)}
\]