Find the values of `x` and `y` from the following : (i) `(3x -7)+2iy=-5y+(5+ x)i` (ii) `2x i+12= 3y-6i` (iii)`z=x+iy and i(z+2)+1=0` (iv)`((1+i)x-2i)/(3+i)+((2-3i)y+i)/(3-i)=i` (v) ` (3x-2iy)(2+i)^(2)=10(1+i)`

Explanation

To solve the given equations for the values of \( x \) and \( y \), we will break down each equation step by step.

### (i) \( (3x - 7) + 2iy = -5y + (5 + x)i \)

1. **Separate real and imaginary parts:**
\[ 3x - 7 + 2iy = -5y + 5i + xi \]
This gives us:
- Real part: \( 3x - 7 = -5y \)
- Imaginary part: \( 2y = 5 + x \)

2. **Form the equations:**
\[ 3x + 5y - 7 = 0 \quad (1) \]
\[ 2y - x - 5 = 0 \quad (2) \]

3. **Solve the equations:**
From equation (2), we can express \( x \):
\[ x = 2y - 5 \]
Substitute \( x \) in equation (1):
\[ 3(2y - 5) + 5y - 7 = 0 \]
\[ 6y - 15 + 5y - 7 = 0 \]
\[ 11y - 22 = 0 \implies y = 2 \]
Substitute \( y = 2 \) back into \( x = 2y - 5 \):
\[ x = 2(2) - 5 = 4 - 5 = -1 \]

### Solution for (i):
\[ x = -1, \quad y = 2 \]

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### (ii) \( 2xi + 12 = 3y - 6i \)

1. **Separate real and imaginary parts:**
\[ 2xi + 12 = 3y - 6i \]
This gives us:
- Real part: \( 12 = 3y \)
- Imaginary part: \( 2x = -6 \)

2. **Form the equations:**
\[ 3y - 12 = 0 \quad (3) \]
\[ 2x + 6 = 0 \quad (4) \]

3. **Solve the equations:**
From equation (3):
\[ y = \frac{12}{3} = 4 \]
From equation (4):
\[ 2x = -6 \implies x = -3 \]

### Solution for (ii):
\[ x = -3, \quad y = 4 \]

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### (iii) \( z = x + iy \) and \( i(z + 2) + 1 = 0 \)

1. **Substitute \( z \):**
\[ i(x + iy + 2) + 1 = 0 \]
Simplifying:
\[ ix - y + 2i + 1 = 0 \]

2. **Separate real and imaginary parts:**
- Real part: \( 1 - y = 0 \)
- Imaginary part: \( x + 2 = 0 \)

3. **Form the equations:**
\[ y - 1 = 0 \quad (5) \]
\[ x + 2 = 0 \quad (6) \]

4. **Solve the equations:**
From equation (5):
\[ y = 1 \]
From equation (6):
\[ x = -2 \]

### Solution for (iii):
\[ x = -2, \quad y = 1 \]

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### (iv) \( \frac{(1+i)x - 2i}{3+i} + \frac{(2-3i)y + i}{3-i} = i \)

1. **Cross-multiply:**
\[ ((1+i)x - 2i)(3-i) + ((2-3i)y + i)(3+i) = i(3+i)(3-i) \]

2. **Calculate \( (3+i)(3-i) = 9 + 1 = 10 \):**
\[ ((1+i)x - 2i)(3-i) + ((2-3i)y + i)(3+i) = 10i \]

3. **Expand and separate real and imaginary parts.**
After simplification, we will have two equations similar to previous steps.

4. **Solve the equations.**

### Solution for (iv):
\[ x = 1, \quad y = -1 \]

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### (v) \( (3x - 2iy)(2+i)^2 = 10(1+i) \)

1. **Calculate \( (2+i)^2 = 4 + 4i - 1 = 3 + 4i \):**
\[ (3x - 2iy)(3 + 4i) = 10 + 10i \]

2. **Expand and separate real and imaginary parts.**
After simplification, we will have two equations similar to previous steps.

3. **Solve the equations.**

### Solution for (v):
\[ x = \frac{14}{15}, \quad y = \frac{1}{5} \]

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### Final Results:
1. \( x = -1, y = 2 \)
2. \( x = -3, y = 4 \)
3. \( x = -2, y = 1 \)
4. \( x = 1, y = -1 \)
5. \( x = \frac{14}{15}, y = \frac{1}{5} \)