It seems like there are two equations presented in your input. I'll clarify them and solve the system of equations.

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

Let's simplify the second equation first:

-2 - 3y = 5

Adding 2 to both sides gives:
\[
-3y = 7
\]

Dividing both sides by -3 results in:
\[
y = -\frac{7}{3}
\]

Now that we have \(y\), we can substitute it back into the first equation to find \(x\):

Substituting \(y = -\frac{7}{3}\) into the first equation:
\[
3x + \left(-\frac{7}{3}\right) = 3
\]

Multiplying through by 3 to eliminate the fraction:
\[
9x - 7 = 9
\]

Adding 7 to both sides yields:
\[
9x = 16
\]

Dividing by 9 gives:
\[
x = \frac{16}{9}
\]

Thus, the solution to the system of equations is:
\[
x = \frac{16}{9}, \quad y = -\frac{7}{3}
\]