Solve for `x: (1-i)x + (1+i)y=1 - 3i`.

Explanation

To solve the equation \((1-i)x + (1+i)y = 1 - 3i\) for \(x\), we will follow these steps:

### Step 1: Expand the equation
We start with the equation:
\[ (1-i)x + (1+i)y = 1 - 3i \]
Expanding both sides gives:
\[ x - ix + y + iy = 1 - 3i \]

### Step 2: Rearrange the equation
Now, we can group the real and imaginary parts:
\[ (x + y) + (-x + y)i = 1 - 3i \]
This means:
\[ x + y = 1 \quad \text{(Real part)} \]
\[ -y + x = -3 \quad \text{(Imaginary part)} \]

### Step 3: Set up the system of equations
From the equations we derived, we have:
1. \(x + y = 1\) (Equation 1)
2. \(y - x = 3\) (Equation 2)

### Step 4: Solve the system of equations
We can express \(y\) from Equation 1:
\[ y = 1 - x \]
Now, substitute \(y\) in Equation 2:
\[ (1 - x) - x = 3 \]
Simplifying this gives:
\[ 1 - 2x = 3 \]
\[ -2x = 3 - 1 \]
\[ -2x = 2 \]
\[ x = -1 \]

### Step 5: Find \(y\)
Now, substitute \(x = -1\) back into Equation 1:
\[ -1 + y = 1 \]
\[ y = 1 + 1 = 2 \]

### Final Answer
Thus, the solution for \(x\) is:
\[ \boxed{-1} \]