Solve the equation `((1+i) x-2i)/(3+i)+ ((2-3i )y + i)/(3-i)=i, x y in R, i= sqrt(-1)`

Answer

To solve the equation

\[ \frac{(1+i)(x-2i)}{3+i} + \frac{(2-3i)(y+i)}{3-i} = i \]

where \(x\) and \(y\) are real numbers and \(i = \sqrt{-1}\), we will follow these steps:

### Step 1: Rewrite the equation
Start by rewriting the equation clearly:

\[ \frac{(1+i)(x-2i)}{3+i} + \frac{(2-3i)(y+i)}{3-i} = i \]

### Step 2: Find a common denominator
The common denominator for the fractions is \((3+i)(3-i)\). Therefore, we can write:

\[ \frac{(1+i)(x-2i)(3-i) + (2-3i)(y+i)(3+i)}{(3+i)(3-i)} = i \]

### Step 3: Simplify the denominator
Calculate the denominator:

\[ (3+i)(3-i) = 9 + 1 = 10 \]

### Step 4: Multiply both sides by the denominator
Multiply both sides by \(10\):

\[ (1+i)(x-2i)(3-i) + (2-3i)(y+i)(3+i) = 10i \]

### Step 5: Expand the first term
Expand \((1+i)(x-2i)(3-i)\):

1. First, calculate \((1+i)(x-2i)\):

\[ (1+i)(x-2i) = x - 2i + ix + 2 = (x + 2) + (i - 2i)x = (x + 2) + (1 - 2)x \]

2. Now multiply by \((3-i)\):

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

### Step 6: Expand the second term
Expand \((2-3i)(y+i)(3+i)\):

1. First, calculate \((y+i)(3+i)\):

\[ (y+i)(3+i) = 3y + yi + 3i - 1 = (3y - 1) + (y + 3)i \]

2. Now multiply by \((2-3i)\):

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

### Step 7: Combine the terms
Combine all the terms from both expansions and set equal to \(10i\).

### Step 8: Separate real and imaginary parts
Equate the real and imaginary parts to form a system of equations.

### Step 9: Solve the system of equations
Solve the resulting system of equations for \(x\) and \(y\).

### Step 10: Final values
After solving, you will find the values of \(x\) and \(y\).