Let's solve the systems of equations one by one.

### System 40:
\[
\begin{cases}
3x + y = 3 \\
2x - y = 7
\end{cases}
\]

1. Add the two equations to eliminate \( y \):
\[
(3x + y) + (2x - y) = 3 + 7 \\
5x = 10 \\
x = 2
\]

2. Substitute \( x = 2 \) back into the first equation to solve for \( y \):
\[
3(2) + y = 3 \\
6 + y = 3 \\
y = 3 - 6 \\
y = -3
\]

So, the solution for system 40 is \( x = 2 \) and \( y = -3 \).

### System 44:
\[
\begin{cases}
x + y = 4 \\
x - y = 6
\end{cases}
\]

1. Add the two equations to eliminate \( y \):
\[
(x + y) + (x - y) = 4 + 6 \\
2x = 10 \\
x = 5
\]

2. Substitute \( x = 5 \) back into the first equation to solve for \( y \):
\[
5 + y = 4 \\
y = 4 - 5 \\
y = -1
\]

So, the solution for system 44 is \( x = 5 \) and \( y = -1 \).