Determine real values of x and y for which each statement is true
`(x-yi)(2+ 3i) = (x-2i)/(1-i)`

Answer

To solve the equation \((x - yi)(2 + 3i) = \frac{x - 2i}{1 - i}\), we will follow these steps:

### Step 1: Expand the Left Side
We start by expanding the left-hand side of the equation:

\[ (x - yi)(2 + 3i) = x \cdot 2 + x \cdot 3i - yi \cdot 2 - yi \cdot 3i \]

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

\[ = 2x + 3xi - 2yi - 3y(-1) = 2x + 3xi - 2yi + 3y \]

Combining the real and imaginary parts gives us:

\[ = (2x + 3y) + (3x - 2y)i \]

### Step 2: Simplify the Right Side
Next, we simplify the right-hand side, \(\frac{x - 2i}{1 - i}\). To do this, we multiply the numerator and denominator by the conjugate of the denominator:

\[ \frac{x - 2i}{1 - i} \cdot \frac{1 + i}{1 + i} = \frac{(x - 2i)(1 + i)}{(1 - i)(1 + i)} \]

Calculating the denominator:

\[ (1 - i)(1 + i) = 1^2 - i^2 = 1 - (-1) = 2 \]

Now for the numerator:

\[ (x - 2i)(1 + i) = x \cdot 1 + x \cdot i - 2i \cdot 1 - 2i \cdot i = x + xi - 2i + 2 = (x + 2) + (x - 2)i \]

Thus, we have:

\[ \frac{(x + 2) + (x - 2)i}{2} = \frac{x + 2}{2} + \frac{x - 2}{2}i \]

### Step 3: Set the Real and Imaginary Parts Equal
Now we equate the real and imaginary parts from both sides:

1. Real part:
\[ 2x + 3y = \frac{x + 2}{2} \]

2. Imaginary part:
\[ 3x - 2y = \frac{x - 2}{2} \]

### Step 4: Clear the Fractions
To eliminate the fractions, we can multiply both equations by 2:

1. \(4x + 6y = x + 2\)
2. \(6x - 4y = x - 2\)

### Step 5: Rearrange the Equations
Rearranging both equations gives us:

1. \(4x - x + 6y = 2 \Rightarrow 3x + 6y = 2\) (Equation 1)
2. \(6x - x - 4y = -2 \Rightarrow 5x - 4y = -2\) (Equation 2)

### Step 6: Solve the System of Equations
Now we solve the system of equations:

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

2. From Equation 2:
\[ 5x - 4y = -2 \quad \text{(2)} \]

We can multiply Equation (1) by 5 and Equation (2) by 3 to eliminate \(x\):

\[ 15x + 30y = 10 \quad \text{(3)} \]
\[ 15x - 12y = -6 \quad \text{(4)} \]

### Step 7: Subtract the Equations
Subtract Equation (4) from Equation (3):

\[ (15x + 30y) - (15x - 12y) = 10 - (-6) \]
\[ 42y = 16 \Rightarrow y = \frac{16}{42} = \frac{8}{21} \]

### Step 8: Substitute \(y\) Back to Find \(x\)
Now substitute \(y = \frac{8}{21}\) back into Equation (1):

\[ 3x + 6\left(\frac{8}{21}\right) = 2 \]
\[ 3x + \frac{48}{21} = 2 \]
\[ 3x = 2 - \frac{48}{21} \]
Convert 2 to a fraction:
\[ 3x = \frac{42}{21} - \frac{48}{21} = \frac{-6}{21} = -\frac{2}{7} \]
\[ x = -\frac{2}{21} \]

### Final Solution
The real values of \(x\) and \(y\) are:

\[ x = -\frac{2}{21}, \quad y = \frac{8}{21} \]