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 step by step, we will break down each part and find the values of \( x \) and \( y \).

### Part (i)
Given:
\[ (3x - 7) + 2iy = -5y + (5 + x)i \]
We can separate the real and imaginary parts:
- Real part: \( 3x - 7 = -5y \)
- Imaginary part: \( 2y = 5 + x \)

**Step 1:** Solve the first equation for \( y \):
\[ 3x - 7 = -5y \implies 5y = -3x + 7 \implies y = \frac{-3x + 7}{5} \]

**Step 2:** Substitute \( y \) into the second equation:
\[ 2\left(\frac{-3x + 7}{5}\right) = 5 + x \]
Multiply through by 5 to eliminate the fraction:
\[ 2(-3x + 7) = 5(5 + x) \implies -6x + 14 = 25 + 5x \]
Combine like terms:
\[ -6x - 5x = 25 - 14 \implies -11x = 11 \implies x = -1 \]

**Step 3:** Substitute \( x = -1 \) back into the equation for \( y \):
\[ y = \frac{-3(-1) + 7}{5} = \frac{3 + 7}{5} = \frac{10}{5} = 2 \]

**Values found:** \( x = -1, y = 2 \)

### Part (ii)
Given:
\[ 2xi + 12 = 3y - 6i \]
Separate real and imaginary parts:
- Real part: \( 12 = 3y \)
- Imaginary part: \( 2x = -6 \)

**Step 1:** Solve for \( y \):
\[ 3y = 12 \implies y = 4 \]

**Step 2:** Solve for \( x \):
\[ 2x = -6 \implies x = -3 \]

**Values found:** \( x = -3, y = 4 \)

### Part (iii)
Given:
\[ z = x + iy \quad \text{and} \quad i(z + 2) + 1 = 0 \]
Substituting for \( z \):
\[ i(x + iy + 2) + 1 = 0 \implies i(x + 2 + iy) + 1 = 0 \]
Distributing \( i \):
\[ ix + 2i - y + 1 = 0 \]
Separate real and imaginary parts:
- Real part: \( 1 - y = 0 \implies y = 1 \)
- Imaginary part: \( x + 2 = 0 \implies x = -2 \)

**Values found:** \( x = -2, y = 1 \)

### Part (iv)
Given:
\[ \frac{(1+i)x - 2i}{3+i} + \frac{(2-3i)y + i}{3-i} = i \]
Multiply both sides by the common denominator \( (3+i)(3-i) = 10 \):
\[ 10\left(\frac{(1+i)x - 2i}{3+i} + \frac{(2-3i)y + i}{3-i}\right) = 10i \]
This simplifies to:
\[ (1+i)x - 2i + (2-3i)y + i = 10i \]
Combine like terms:
\[ (1+i)x + (2-3i)y - i = 10i \]
Separate real and imaginary parts:
- Real part: \( x + 2y = 10 \)
- Imaginary part: \( x - 3y - 1 = 10 \)

**Step 1:** Solve the first equation for \( y \):
\[ y = \frac{10 - x}{2} \]

**Step 2:** Substitute into the second equation:
\[ x - 3\left(\frac{10 - x}{2}\right) - 1 = 10 \]
Multiply through by 2:
\[ 2x - 3(10 - x) - 2 = 20 \implies 2x - 30 + 3x - 2 = 20 \]
Combine:
\[ 5x - 32 = 20 \implies 5x = 52 \implies x = \frac{52}{5} \]

**Step 3:** Substitute \( x \) back to find \( y \):
\[ y = \frac{10 - \frac{52}{5}}{2} = \frac{\frac{50 - 52}{5}}{2} = \frac{-2/5}{2} = -\frac{1}{5} \]

**Values found:** \( x = \frac{52}{5}, y = -\frac{1}{5} \)

### Part (v)
Given:
\[ (3x - 2iy)(2+i)^2 = 10(1+i) \]
Calculate \( (2+i)^2 \):
\[ (2+i)^2 = 4 + 4i - 1 = 3 + 4i \]
Substituting:
\[ (3x - 2iy)(3 + 4i) = 10 + 10i \]
Distributing:
\[ (3x)(3 + 4i) - (2iy)(3 + 4i) = 10 + 10i \]
Separate real and imaginary parts:
- Real part: \( 9x + 8y = 10 \)
- Imaginary part: \( 12x - 6y = 10 \)

**Step 1:** Solve the first equation for \( y \):
\[ 8y = 10 - 9x \implies y = \frac{10 - 9x}{8} \]

**Step 2:** Substitute into the second equation:
\[ 12x - 6\left(\frac{10 - 9x}{8}\right) = 10 \]
Multiply through by 8:
\[ 96x - 6(10 - 9x) = 80 \implies 96x - 60 + 54x = 80 \]
Combine:
\[ 150x = 140 \implies x = \frac{14}{15} \]

**Step 3:** Substitute \( x \) back to find \( y \):
\[ y = \frac{10 - 9\left(\frac{14}{15}\right)}{8} = \frac{10 - \frac{126}{15}}{8} = \frac{\frac{150 - 126}{15}}{8} = \frac{\frac{24}{15}}{8} = \frac{24}{120} = \frac{2}{10} = \frac{1}{5} \]

**Values found:** \( x = \frac{14}{15}, y = \frac{1}{5} \)

### Summary of Values
1. Part (i): \( x = -1, y = 2 \)
2. Part (ii): \( x = -3, y = 4 \)
3. Part (iii): \( x = -2, y = 1 \)
4. Part (iv): \( x = \frac{52}{5}, y = -\frac{1}{5} \)
5. Part (v): \( x = \frac{14}{15}, y = \frac{1}{5} \)