`{:(4x - 3y + 4 = 0),(4x + 3y - 20 = 0):}`

Answer

To solve the system of equations given by:

1. \( 4x - 3y + 4 = 0 \)
2. \( 4x + 3y - 20 = 0 \)

we will follow these steps:

### Step 1: Rewrite the equations
We can rewrite the equations for clarity:
- Equation 1: \( 4x - 3y + 4 = 0 \)
- Equation 2: \( 4x + 3y - 20 = 0 \)

### Step 2: Add both equations
Next, we will add both equations together to eliminate \( y \):
\[ (4x - 3y + 4) + (4x + 3y - 20) = 0 \]
This simplifies to:
\[ 4x + 4x - 3y + 3y + 4 - 20 = 0 \]
Combining like terms, we get:
\[ 8x - 16 = 0 \]

### Step 3: Solve for \( x \)
Now, we can solve for \( x \):
\[ 8x = 16 \]
\[ x = \frac{16}{8} = 2 \]

### Step 4: Substitute \( x \) back into one of the original equations
Now that we have \( x = 2 \), we will substitute this value back into one of the original equations to find \( y \). We can use Equation 1:
\[ 4(2) - 3y + 4 = 0 \]
This simplifies to:
\[ 8 - 3y + 4 = 0 \]
Combining the constants:
\[ 12 - 3y = 0 \]

### Step 5: Solve for \( y \)
Now, we can solve for \( y \):
\[ -3y = -12 \]
\[ y = \frac{-12}{-3} = 4 \]

### Final Solution
Thus, the solution to the system of equations is:
\[ x = 2, \quad y = 4 \]