`2(a x-b y)+(a+4b)=0 , 2(b x+a y)+(b-4a)=0`

Answer

To solve the system of equations given by:

1) \( 2(a x - b y) + (a + 4b) = 0 \)
2) \( 2(b x + a y) + (b - 4a) = 0 \)

We will follow these steps:

### Step 1: Rewrite the equations in standard form
Let's rewrite the equations clearly.

From the first equation:
\[ 2ax - 2by + a + 4b = 0 \]
From the second equation:
\[ 2bx + 2ay + b - 4a = 0 \]

### Step 2: Label the equations
Let’s label these equations for reference:
- Equation (1): \( 2ax - 2by + a + 4b = 0 \)
- Equation (2): \( 2bx + 2ay + b - 4a = 0 \)

### Step 3: Multiply the equations
We will multiply Equation (1) by \( b \) and Equation (2) by \( a \).

Multiplying Equation (1) by \( b \):
\[ b(2ax - 2by + a + 4b) = 0 \implies 2abx - 2b^2y + ab + 4b^2 = 0 \]

Multiplying Equation (2) by \( a \):
\[ a(2bx + 2ay + b - 4a) = 0 \implies 2abx + 2a^2y + ab - 4a^2 = 0 \]

### Step 4: Subtract the equations
Now, we will subtract the modified Equation (2) from modified Equation (1):
\[ (2abx - 2b^2y + ab + 4b^2) - (2abx + 2a^2y + ab - 4a^2) = 0 \]

This simplifies to:
\[ -2b^2y - 2a^2y + 4b^2 + 4a^2 = 0 \]
\[ (-2b^2 - 2a^2)y + 4(b^2 + a^2) = 0 \]

### Step 5: Solve for \( y \)
Rearranging gives:
\[ (-2(b^2 + a^2))y + 4(b^2 + a^2) = 0 \]
If \( b^2 + a^2 \neq 0 \), we can divide through by \( b^2 + a^2 \):
\[ -2y + 4 = 0 \implies 2y = 4 \implies y = 2 \]

### Step 6: Substitute \( y \) back to find \( x \)
Now we will substitute \( y = 2 \) back into one of the original equations. Let’s use Equation (1):
\[ 2ax - 2b(2) + a + 4b = 0 \]
This simplifies to:
\[ 2ax - 4b + a + 4b = 0 \implies 2ax + a = 0 \]
Factoring out \( a \):
\[ a(2x + 1) = 0 \]

### Step 7: Solve for \( x \)
If \( a \neq 0 \):
\[ 2x + 1 = 0 \implies 2x = -1 \implies x = -\frac{1}{2} \]

### Final Solution
Thus, the solution to the system of equations is:
\[ x = -\frac{1}{2}, \quad y = 2 \]