Solve for x and y:
`2x + 3y + 1 = 0, 3x + 2y - 11 = 0`

Answer

To solve the system of equations given by:

1. \( 2x + 3y + 1 = 0 \) (Equation 1)
2. \( 3x + 2y - 11 = 0 \) (Equation 2)

we can follow these steps:

### Step 1: Rearrange the equations
First, we can rearrange both equations to express them in a more standard form.

From Equation 1:
\[ 2x + 3y = -1 \quad (1) \]

From Equation 2:
\[ 3x + 2y = 11 \quad (2) \]

### Step 2: Add the equations
Now, we will add both equations (1) and (2):

\[ (2x + 3y) + (3x + 2y) = -1 + 11 \]

This simplifies to:
\[ 5x + 5y = 10 \]

Dividing the entire equation by 5:
\[ x + y = 2 \quad (3) \]

### Step 3: Subtract the equations
Next, we will subtract Equation 1 from Equation 2:

\[ (3x + 2y) - (2x + 3y) = 11 - (-1) \]

This simplifies to:
\[ x - y = 12 \quad (4) \]

### Step 4: Solve the system of equations
Now we have a new system of equations:
1. \( x + y = 2 \) (Equation 3)
2. \( x - y = 12 \) (Equation 4)

We can add these two equations together:

\[ (x + y) + (x - y) = 2 + 12 \]

This simplifies to:
\[ 2x = 14 \]

Dividing by 2:
\[ x = 7 \]

### Step 5: Substitute to find y
Now that we have \( x = 7 \), we can substitute this value back into Equation 3 to find \( y \):

\[ 7 + y = 2 \]

Subtracting 7 from both sides:
\[ y = 2 - 7 \]
\[ y = -5 \]

### Final Solution
Thus, the solution to the system of equations is:
\[ x = 7, \quad y = -5 \]