Find the real numbers `x` and `y` if `(x-i y)(3+5i)` is the conjugate of `-6-24 i`.

Explanation

To solve the problem, we need to find the real numbers \( x \) and \( y \) such that \( (x - iy)(3 + 5i) \) is the conjugate of \( -6 - 24i \).

### Step 1: Understand the conjugate
The conjugate of a complex number \( z = a + bi \) is given by \( z^* = a - bi \). For our case, the conjugate of \( -6 - 24i \) is \( -6 + 24i \).

### Step 2: Set up the equation
We need to set up the equation:
\[ (x - iy)(3 + 5i) = -6 + 24i \]

### Step 3: Expand the left-hand side
Now, we will expand the left-hand side:
\[ (x - iy)(3 + 5i) = x \cdot 3 + x \cdot 5i - iy \cdot 3 - iy \cdot 5i \]
This simplifies to:
\[ 3x + 5xi - 3iy - 5y(-1) = 3x + 5xi + 5y - 3iy \]
Combining the real and imaginary parts, we get:
\[ (3x + 5y) + (5x - 3y)i \]

### Step 4: Set up the equations
Now we equate the real and imaginary parts to the conjugate \( -6 + 24i \):
1. Real part: \( 3x + 5y = -6 \) (Equation 1)
2. Imaginary part: \( 5x - 3y = 24 \) (Equation 2)

### Step 5: Solve the equations
We will solve these two equations simultaneously.

From Equation 1:
\[ 3x + 5y = -6 \quad \text{(1)} \]
From Equation 2:
\[ 5x - 3y = 24 \quad \text{(2)} \]

### Step 6: Multiply and eliminate \( y \)
To eliminate \( y \), we can multiply Equation 1 by 3 and Equation 2 by 5:
\[ 9x + 15y = -18 \quad \text{(3)} \]
\[ 25x - 15y = 120 \quad \text{(4)} \]

### Step 7: Add the equations
Now, we add Equations (3) and (4):
\[ (9x + 15y) + (25x - 15y) = -18 + 120 \]
This simplifies to:
\[ 34x = 102 \]
Dividing both sides by 34 gives:
\[ x = 3 \]

### Step 8: Substitute \( x \) back to find \( y \)
Now we substitute \( x = 3 \) back into Equation 1:
\[ 3(3) + 5y = -6 \]
This simplifies to:
\[ 9 + 5y = -6 \]
Subtracting 9 from both sides gives:
\[ 5y = -15 \]
Dividing by 5 gives:
\[ y = -3 \]

### Final Solution
Thus, the values of \( x \) and \( y \) are:
\[ x = 3, \quad y = -3 \]