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Let's solve the given complex equations step by step to find the solutions for \( x \) and \( y \).

### Equation 23
\[ (x + i y)(2 - i) = 5 \]

Expanding and equating the real and imaginary parts, we get:
\[ (x + iy)(2 - i) = 2x + 2iy - ix - y = (2x + y) + i(2y - x) \]
\[ 2x + y = 5 \]
\[ 2y - x = 0 \]

From \( 2y = x \), we substitute into the first equation:
\[ 2(2y) + y = 5 \]
\[ 4y + y = 5 \]
\[ 5y = 5 \]
\[ y = 1 \]

Substituting \( y \) back into \( x = 2y \):
\[ x = 2 \cdot 1 = 2 \]

So, the solution is \( x = 2 \), \( y = 1 \).

### Equation 24
\[ (x - i y)(3 + 2i) = 12 - 5i \]

Expanding and equating the real and imaginary parts, we get:
\[ (x - iy)(3 + 2i) = 3x + 2ix - 3iy - 2y = (3x - 2y) + i(2x - 3y) \]
\[ 3x - 2y = 12 \]
\[ 2x - 3y = -5 \]

Solving these two equations simultaneously:
1. \( 3x - 2y = 12 \)
2. \( 2x - 3y = -5 \)

Multiplying the first equation by 2 and the second by 3:
\[ 6x - 4y = 24 \]
\[ 6x - 9y = -15 \]

Subtracting the second from the first:
\[ -4y + 9y = 24 + 15 \]
\[ 5y = 39 \]
\[ y = \frac{39}{5} = 7.8 \]

Substituting \( y \) back into the first equation:
\[ 3x - 2(7.8) = 12 \]
\[ 3x - 15.6 = 12 \]
\[ 3x = 27.6 \]
\[ x = \frac{27.6}{3} = 9.2 \]

So, the solution is \( x = 9.2 \), \( y = 7.8 \).

### Equation 25
\[ (x + i y)(2 + 3i) = 1 - 5i \]

Expanding and equating the real and imaginary parts, we get:
\[ (x + iy)(2 + 3i) = 2x + 3ix + 2iy - 3y = (2x - 3y) + i(3x + 2y) \]
\[ 2x - 3y = 1 \]
\[ 3x + 2y = -5 \]

Solving these two equations simultaneously:
1. \( 2x - 3y = 1 \)
2. \( 3x + 2y = -5 \)

Multiplying the first equation by 3 and the second by 2:
\[ 6x - 9y = 3 \]
\[ 6x + 4y = -10 \]

Subtracting the second from the first:
\[ -9y - 4y = 3 + 10 \]
\[ -13y = 13 \]
\[ y = \frac{13}{-13} = -1 \]

Substituting \( y \) back into the first equation:
\[ 2x - 3(-1) = 1 \]
\[ 2x + 3 = 1 \]
\[ 2x = -2 \]
\[ x = \frac{-2}{2} = -1 \]

So, the solution is \( x = -1 \), \( y = -1 \).

### Summary of solutions:
1. \( x = 2 \), \( y = 1 \)
2. \( x = 9.2 \), \( y = 7.8 \)
3. \( x = -1 \), \( y = -1 \)

Would you like further details on any of these steps or equations? Here are some related questions you might find interesting:

1. How do you multiply complex numbers?
2. What is the significance of equating real and imaginary parts in complex equations?
3. Can these methods be used for solving equations with more variables?
4. How does the imaginary unit \( i \) affect the solution of equations?
5. Are there other methods to solve complex equations?

**Tip:** When solving complex equations, always separate the real and imaginary parts to simplify the problem.

Learn how to solve complex equations 23, 24, and 25 step-by-step with detailed solutions. This problem covers complex number operations, including multiplication and solving equations with real and imaginary components. Suitable for advanced mathematics enthusiasts and students.

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