`{:(2x + 3y + 5 = 0),(3x - 2y - 12 = 0):}`

Answer

To solve the system of equations given by:

1. \( 2x + 3y + 5 = 0 \) (Equation 1)
2. \( 3x - 2y - 12 = 0 \) (Equation 2)

we will follow these steps:

### Step 1: Rewrite the equations in standard form
We can rewrite the equations as follows:
- From Equation 1: \( 2x + 3y = -5 \)
- From Equation 2: \( 3x - 2y = 12 \)

### Step 2: Multiply the equations to eliminate one variable
To eliminate one variable, we can multiply Equation 1 by 2 and Equation 2 by 3. This gives us:

- \( 2(2x + 3y) = 2(-5) \) → \( 4x + 6y = -10 \) (Equation 3)
- \( 3(3x - 2y) = 3(12) \) → \( 9x - 6y = 36 \) (Equation 4)

### Step 3: Add the two new equations
Now, we add Equation 3 and Equation 4:

\[ (4x + 6y) + (9x - 6y) = -10 + 36 \]

This simplifies to:

\[ 13x + 0y = 26 \]

### Step 4: Solve for x
From the equation \( 13x = 26 \), we can solve for \( x \):

\[ x = \frac{26}{13} = 2 \]

### Step 5: Substitute x back into one of the original equations
Now that we have \( x = 2 \), we can substitute this value back into Equation 1 to find \( y \):

\[ 2(2) + 3y + 5 = 0 \]

This simplifies to:

\[ 4 + 3y + 5 = 0 \implies 3y + 9 = 0 \]

### Step 6: Solve for y
Now, we can solve for \( y \):

\[ 3y = -9 \implies y = \frac{-9}{3} = -3 \]

### Final Solution
Thus, the solution to the system of equations is:

\[ x = 2, \quad y = -3 \]