` 4x - 3y + 4 = 0, 4x + 3y - 20 = 0 `.

Answer

To solve the given linear equations \(4x - 3y + 4 = 0\) and \(4x + 3y - 20 = 0\), we will follow these steps:

### Step 1: Write the equations in standard form
We have the equations:
1. \(4x - 3y + 4 = 0\) (Equation 1)
2. \(4x + 3y - 20 = 0\) (Equation 2)

### Step 2: Add the two equations
We will add Equation 1 and Equation 2 to eliminate \(y\):
\[ (4x - 3y + 4) + (4x + 3y - 20) = 0 \]
This simplifies to:
\[ 4x + 4x - 3y + 3y + 4 - 20 = 0 \]
\[ 8x - 16 = 0 \]

### 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
We will substitute \(x = 2\) back into Equation 1 to find \(y\):
\[ 4(2) - 3y + 4 = 0 \]
This simplifies to:
\[ 8 - 3y + 4 = 0 \]
\[ 12 - 3y = 0 \]

### 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:
\[ x = 2, \quad y = 4 \]