To solve the system of equations:

\[
\frac{3}{x-2} + \frac{2}{y+1} = \frac{11}{3}
\]

\[
\frac{2}{x-2} + \frac{1}{y+1} = 3
\]

Let's introduce new variables to simplify the equations. Let:

\[
u = \frac{1}{x-2}
\]

\[
v = \frac{1}{y+1}
\]

Then the system of equations becomes:

\[
3u + 2v = \frac{11}{3}
\]

\[
2u + v = 3
\]

Now we have a system of linear equations in terms of \(u\) and \(v\). We can solve this system using substitution or elimination. Let's use the elimination method.

First, multiply the second equation by 2 to align the coefficients of \(v\):

\[
4u + 2v = 6
\]

Now we have:

\[
3u + 2v = \frac{11}{3}
\]

\[
4u + 2v = 6
\]

Subtract the first equation from the second equation:

\[
(4u + 2v) - (3u + 2v) = 6 - \frac{11}{3}
\]

\[
4u + 2v - 3u - 2v = 6 - \frac{11}{3}
\]

\[
u = 6 - \frac{11}{3}
\]

Convert 6 to a fraction with a common denominator:

\[
6 = \frac{18}{3}
\]

So,

\[
u = \frac{18}{3} - \frac{11}{3} = \frac{7}{3}
\]

Now that we have \(u\), substitute \(u = \frac{7}{3}\) back into the second equation to find \(v\):

\[
2u + v = 3
\]

\[
2 \left(\frac{7}{3}\right) + v = 3
\]

\[
\frac{14}{3} + v = 3
\]

Subtract \(\frac{14}{3}\) from both sides:

\[
v = 3 - \frac{14}{3}
\]

Convert 3 to a fraction with a common denominator:

\[
3 = \frac{9}{3}
\]

So,

\[
v = \frac{9}{3} - \frac{14}{3} = -\frac{5}{3}
\]

Now we have \(u = \frac{7}{3}\) and \(v = -\frac{5}{3}\). Recall that:

\[
u = \frac{1}{x-2}
\]

\[
v = \frac{1}{y+1}
\]

So,

\[
\frac{1}{x-2} = \frac{7}{3}
\]

\[
x - 2 = \frac{3}{7}
\]

\[
x = 2 + \frac{3}{7} = \frac{14}{7} + \frac{3}{7} = \frac{17}{7}
\]

And,

\[
\frac{1}{y+1} = -\frac{5}{3}
\]

\[
y + 1 = -\frac{3}{5}
\]

\[
y = -\frac{3}{5} - 1 = -\frac{3}{5} - \frac{5}{5} = -\frac{8}{5}
\]

So the solution to the system of equations is:

\[
x = \frac{17}{7}, \quad y = -\frac{8}{5}
\]