If ` ((1 + i) x- 2 i)/( 3 + i) + (( 2 - 3 i ) y + i)/( 3 - i) = i ` then the real value of x and y are given by

Options

A. x = - 3, y = - 1
B. x= 3, y = - 1
C. x = 3, y = 1
D. x= 1, y = - 3

Answer

To solve the equation

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

we will follow these steps:

### Step 1: Clear the denominators
Multiply both sides by the product of the denominators, \((3 + i)(3 - i)\):

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

### Step 2: Simplify the right-hand side
Calculate the right-hand side:

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

### Step 3: Expand the left-hand side
Now, expand both terms on the left-hand side:

1. For \((1 + i)x - 2i)(3 - i)\):
\[ = (1 + i)x(3 - i) - 2i(3 - i) = (3x + 3ix - ix - x) - (6i - 2i^2) = (3x - x) + (3ix - ix - 6i + 2) = 2x + (2x - 6i + 2). \]

2. For \(((2 - 3i)y + i)(3 + i)\):
\[ = (2 - 3i)y(3 + i) + i(3 + i) = (6y + 2iy - 9iy - 3i^2) + (3i - 1) = (6y + 3) + (-7i + 2iy). \]

### Step 4: Combine the results
Combine both expanded parts:

\[ (2x + 6y + 3) + (2iy - 6i + 2) = 10i. \]

### Step 5: Separate real and imaginary parts
From the equation, we can separate the real and imaginary parts:

1. Real part:
\[ 2x + 6y + 3 = 0. \]

2. Imaginary part:
\[ 2y - 6 = 10 \implies 2y = 16 \implies y = 8. \]

### Step 6: Substitute \(y\) back to find \(x\)
Substituting \(y = 8\) into the real part equation:

\[ 2x + 6(8) + 3 = 0 \implies 2x + 48 + 3 = 0 \implies 2x + 51 = 0 \implies 2x = -51 \implies x = -\frac{51}{2}. \]

### Final Answer
Thus, the values of \(x\) and \(y\) are:

\[ x = -\frac{51}{2}, \quad y = 8. \]