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-by-Step Solution:

1. **Identify the Conjugate**:
The conjugate of a complex number \( a + bi \) is given by \( a - bi \). Therefore, the conjugate of \( -6 - 24i \) is:
\[ -6 + 24i \]

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

3. **Multiply the Left Side**:
Now, we will multiply \( (x - iy)(3 + 5i) \):
\[ (x - iy)(3 + 5i) = x \cdot 3 + x \cdot 5i - iy \cdot 3 - iy \cdot 5i \]
Simplifying this, we have:
\[ = 3x + 5xi - 3yi - 5y(-1) = 3x + 5xi + 5y - 3yi \]
Rearranging gives:
\[ = (3x + 5y) + (5x - 3y)i \]

4. **Equate Real and Imaginary Parts**:
Now we equate the real and imaginary parts from both sides:
- Real part: \( 3x + 5y = -6 \) (Equation 1)
- Imaginary part: \( 5x - 3y = 24 \) (Equation 2)

5. **Solve the System of Equations**:
We have the following system of equations:
\[ \begin{align*} 3x + 5y &= -6 \quad \text{(1)} \\ 5x - 3y &= 24 \quad \text{(2)} \end{align*} \]

We can use the elimination method. First, we can multiply Equation (1) by 3 and Equation (2) by 5 to eliminate \( y \):
\[ \begin{align*} 9x + 15y &= -18 \quad \text{(3)} \\ 25x - 15y &= 120 \quad \text{(4)} \end{align*} \]

6. **Add the Two Equations**:
Adding Equations (3) and (4):
\[ 9x + 15y + 25x - 15y = -18 + 120 \]
This simplifies to:
\[ 34x = 102 \]
Therefore:
\[ x = \frac{102}{34} = 3 \]

7. **Substitute \( x \) Back to Find \( y \)**:
Now substitute \( x = 3 \) back into Equation (1):
\[ 3(3) + 5y = -6 \]
This simplifies to:
\[ 9 + 5y = -6 \implies 5y = -6 - 9 \implies 5y = -15 \implies y = -3 \]

8. **Final Answer**:
The values of \( x \) and \( y \) are:
\[ x = 3, \quad y = -3 \]