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

Options

A. ` x = -2, y=2`
B. ` x = -3, y=3`
C. ` x = 3, y=-3`
D. ` x = -4, y=1`

Explanation

To solve the problem, we need to find the real numbers \( x \) and \( y \) such that

\[ (x - iy)(3 + 5i) = \text{conjugate of } (-6 - 24i) \]

### Step 1: Find the conjugate of \(-6 - 24i\)

The conjugate of a complex number \( a + bi \) is \( a - bi \). Therefore, the conjugate of \(-6 - 24i\) is:

\[ -6 + 24i \]

### Step 2: Set up the equation

Now we can set up the equation:

\[ (x - iy)(3 + 5i) = -6 + 24i \]

### Step 3: Expand the left side

We will expand the left-hand side:

\[ (x - iy)(3 + 5i) = x \cdot 3 + x \cdot 5i - iy \cdot 3 - iy \cdot 5i \]

Using \( i^2 = -1 \), we can simplify:

\[ = 3x + 5xi - 3yi - 5y(-1) \]
\[ = 3x + 5xi + 5y - 3yi \]

Now, we can combine the real and imaginary parts:

\[ = (3x + 5y) + (5x - 3y)i \]

### Step 4: Set the real and imaginary parts equal

Now we equate the real and imaginary parts to those of \(-6 + 24i\):

1. Real part:
\[ 3x + 5y = -6 \quad \text{(Equation 1)} \]

2. Imaginary part:
\[ 5x - 3y = 24 \quad \text{(Equation 2)} \]

### Step 5: Solve the system of equations

We have the following system of equations:

1. \( 3x + 5y = -6 \)
2. \( 5x - 3y = 24 \)

To eliminate \( x \), we can multiply Equation 1 by 5 and Equation 2 by 3:

- From Equation 1:
\[ 15x + 25y = -30 \quad \text{(Equation 3)} \]

- From Equation 2:
\[ 15x - 9y = 72 \quad \text{(Equation 4)} \]

### Step 6: Subtract the equations

Now, we subtract Equation 4 from Equation 3:

\[ (15x + 25y) - (15x - 9y) = -30 - 72 \]
\[ 25y + 9y = -102 \]
\[ 34y = -102 \]

### Step 7: Solve for \( y \)

Now we can solve for \( y \):

\[ y = \frac{-102}{34} = -3 \]

### Step 8: Substitute \( y \) back to find \( x \)

Now we substitute \( y = -3 \) back into Equation 1:

\[ 3x + 5(-3) = -6 \]
\[ 3x - 15 = -6 \]
\[ 3x = 9 \]
\[ x = 3 \]

### Final Answer

Thus, the values of \( x \) and \( y \) are:

\[ x = 3, \quad y = -3 \]