Let's solve the system of equations you've provided step by step. The equations seem to have typos or missing connections, but I will interpret this based on common structures and simplify the last two equations you’ve outlined.

Assuming the equations you meant are:

1. \( \frac{3}{x-2} + \frac{2}{y+1} = \frac{11}{3} \)
2. \( \frac{2}{x-2} + \frac{3}{y+1} = 3 \)

Let's denote \( a = \frac{1}{x-2} \) and \( b = \frac{1}{y+1} \). Then the equations become:

1. \( 3a + 2b = \frac{11}{3} \)
2. \( 2a + 3b = 3 \)

Now let's solve these equations.

### Step 1: Solve the first equation for one of the variables
From the first equation:
\[
3a + 2b = \frac{11}{3}
\]
Rearranging gives:
\[
2b = \frac{11}{3} - 3a
\]
\[
b = \frac{11/3 - 3a}{2} = \frac{11 - 9a}{6}
\]

### Step 2: Substitute \( b \) in the second equation
Substituting this expression for \( b \) into the second equation:
\[
2a + 3\left( \frac{11 - 9a}{6} \right) = 3
\]
Multiply everything by 6 to eliminate the fraction:
\[
12a + 3(11 - 9a) = 18
\]
\[
12a + 33 - 27a = 18
\]
Combine like terms:
\[
-15a + 33 = 18
\]
\[
-15a = 18 - 33
\]
\[
-15a = -15
\]
\[
a = 1
\]

### Step 3: Finding \( b \)
Now substitute \( a = 1 \) back into the expression for \( b \):
\[
b = \frac{11 - 9(1)}{6} = \frac{11 - 9}{6} = \frac{2}{6} = \frac{1}{3}
\]

### Step 4: Find \( x \) and \( y \)
We have:
\[
a = \frac{1}{x - 2} = 1 \implies x - 2 = 1 \implies x = 3
\]
\[
b = \frac{1}{y + 1} = \frac{1}{3} \implies y + 1 = 3 \implies y = 2
\]

### Step 5: Final Solution
The solutions are:
\[
x = 3, \quad y = 2
\]

If you have further equations or details, please provide so I can assist accordingly!