Finid the vlues of x and y if:
(i) `(x+iy)(1+i)=1-i`
(ii) `((1+i)x-2i)/(3+i)+((2-3i)y+i)/(3-i)=i`, where `i=sqrt(-1)`.

Answer

Let's solve the given equations step by step.

### Part (i): Solve \((x + iy)(1 + i) = 1 - i\)

1. **Expand the left-hand side**:
\[ (x + iy)(1 + i) = x(1 + i) + iy(1 + i) = x + xi + iy + i^2y \]
Since \(i^2 = -1\), we have:
\[ = x + xi + iy - y = (x - y) + i(x + y) \]

2. **Set the real and imaginary parts equal**:
\[ (x - y) + i(x + y) = 1 - i \]
This gives us two equations:
- Real part: \(x - y = 1\) (1)
- Imaginary part: \(x + y = -1\) (2)

3. **Solve the system of equations**:
From equation (1):
\[ x = y + 1 \]
Substitute \(x\) in equation (2):
\[ (y + 1) + y = -1 \Rightarrow 2y + 1 = -1 \Rightarrow 2y = -2 \Rightarrow y = -1 \]
Now substitute \(y\) back into equation (1):
\[ x - (-1) = 1 \Rightarrow x + 1 = 1 \Rightarrow x = 0 \]

**Values found**: \(x = 0\), \(y = -1\)

### Part (ii): Solve \(\frac{(1+i)x - 2i}{3+i} + \frac{(2-3i)y + i}{3-i} = i\)

1. **Rationalize the denominators**:
For \(\frac{(1+i)x - 2i}{3+i}\), multiply numerator and denominator by \(3 - i\):
\[ \frac{((1+i)x - 2i)(3-i)}{(3+i)(3-i)} = \frac{(3(1+i)x - 2i(3-i))}{9 + 1} = \frac{(3(1+i)x - 6i + 2)}{10} \]

For \(\frac{(2-3i)y + i}{3-i}\), multiply numerator and denominator by \(3 + i\):
\[ \frac{((2-3i)y + i)(3+i)}{(3-i)(3+i)} = \frac{((2-3i)y(3+i) + i(3+i))}{10} \]

2. **Combine the fractions**:
The equation becomes:
\[ \frac{(3(1+i)x - 6i + 2) + (6y + (2-3i)y)}{10} = i \]
Simplifying the numerator:
\[ 3(1+i)x - 6i + 2 + 6y + (2-3i)y = 3x + 3xi - 6i + 2 + 6y + 2y - 3iy \]
Combine like terms:
\[ = (3x + 6y + 2) + i(3x - 6 - 3y) \]

3. **Set equal to \(10i\)**:
\[ \frac{(3x + 6y + 2) + i(3x - 6 - 3y)}{10} = i \]
Multiply both sides by 10:
\[ (3x + 6y + 2) + i(3x - 6 - 3y) = 10i \]
This gives us two equations:
- Real part: \(3x + 6y + 2 = 0\) (3)
- Imaginary part: \(3x - 6 - 3y = 10\) (4)

4. **Solve the system of equations**:
From equation (3):
\[ 3x + 6y = -2 \Rightarrow x + 2y = -\frac{2}{3} \quad (5) \]
From equation (4):
\[ 3x - 3y = 16 \Rightarrow x - y = \frac{16}{3} \quad (6) \]

5. **Substitute (5) into (6)**:
Substitute \(x = -\frac{2}{3} - 2y\) into equation (6):
\[ -\frac{2}{3} - 2y - y = \frac{16}{3} \]
Combine like terms:
\[ -\frac{2}{3} - 3y = \frac{16}{3} \Rightarrow -3y = \frac{16}{3} + \frac{2}{3} = \frac{18}{3} = 6 \Rightarrow y = -2 \]
Substitute \(y\) back into equation (5):
\[ x + 2(-2) = -\frac{2}{3} \Rightarrow x - 4 = -\frac{2}{3} \Rightarrow x = -\frac{2}{3} + 4 = -\frac{2}{3} + \frac{12}{3} = \frac{10}{3} \]

**Values found**: \(x = \frac{10}{3}\), \(y = -2\)

### Summary of Solutions:
- For part (i): \(x = 0\), \(y = -1\)
- For part (ii): \(x = \frac{10}{3}\), \(y = -2\)