The system of equation `|x-1|+3y=4,x-|y-1|=2` has

Options

A. no solution
B. A unique solution
C. Two solutions
D. More than two solutions

Answer

To solve the system of equations given by \( |x-1| + 3y = 4 \) and \( x - |y-1| = 2 \), we will analyze each equation by considering the cases for the absolute values.

### Step 1: Analyze the first equation \( |x-1| + 3y = 4 \)

1. **Case 1**: \( x - 1 \geq 0 \) (i.e., \( x \geq 1 \))
- Here, \( |x-1| = x - 1 \).
- Substitute into the equation:
\[ (x - 1) + 3y = 4 \implies x + 3y = 5 \quad \text{(Equation 1)} \]

2. **Case 2**: \( x - 1 < 0 \) (i.e., \( x < 1 \))
- Here, \( |x-1| = -(x - 1) = -x + 1 \).
- Substitute into the equation:
\[ (-x + 1) + 3y = 4 \implies -x + 3y = 3 \implies x - 3y = -3 \quad \text{(Equation 2)} \]

### Step 2: Analyze the second equation \( x - |y-1| = 2 \)

1. **Case 1**: \( y - 1 \geq 0 \) (i.e., \( y \geq 1 \))
- Here, \( |y-1| = y - 1 \).
- Substitute into the equation:
\[ x - (y - 1) = 2 \implies x - y + 1 = 2 \implies x - y = 1 \quad \text{(Equation 3)} \]

2. **Case 2**: \( y - 1 < 0 \) (i.e., \( y < 1 \))
- Here, \( |y-1| = -(y - 1) = -y + 1 \).
- Substitute into the equation:
\[ x - (-y + 1) = 2 \implies x + y - 1 = 2 \implies x + y = 3 \quad \text{(Equation 4)} \]

### Step 3: Solve the system of equations

Now we have four equations to consider based on the cases:

1. **From Equation 1 and Equation 3**:
\[ x + 3y = 5 \quad \text{(1)} \]
\[ x - y = 1 \quad \text{(3)} \]

Substitute \( x = y + 1 \) from Equation (3) into Equation (1):
\[ (y + 1) + 3y = 5 \implies 4y + 1 = 5 \implies 4y = 4 \implies y = 1 \]
Substitute \( y = 1 \) back into \( x - y = 1 \):
\[ x - 1 = 1 \implies x = 2 \]

**Solution**: \( (x, y) = (2, 1) \)

2. **From Equation 1 and Equation 4**:
\[ x + 3y = 5 \quad \text{(1)} \]
\[ x + y = 3 \quad \text{(4)} \]

Substitute \( x = 3 - y \) from Equation (4) into Equation (1):
\[ (3 - y) + 3y = 5 \implies 3 + 2y = 5 \implies 2y = 2 \implies y = 1 \]
Substitute \( y = 1 \) back into \( x + y = 3 \):
\[ x + 1 = 3 \implies x = 2 \]

**Solution**: \( (x, y) = (2, 1) \)

3. **From Equation 2 and Equation 3**:
\[ x - 3y = -3 \quad \text{(2)} \]
\[ x - y = 1 \quad \text{(3)} \]

Substitute \( x = y + 1 \) from Equation (3) into Equation (2):
\[ (y + 1) - 3y = -3 \implies -2y + 1 = -3 \implies -2y = -4 \implies y = 2 \]
Substitute \( y = 2 \) back into \( x - y = 1 \):
\[ x - 2 = 1 \implies x = 3 \]

**Solution**: \( (x, y) = (3, 2) \)

4. **From Equation 2 and Equation 4**:
\[ x - 3y = -3 \quad \text{(2)} \]
\[ x + y = 3 \quad \text{(4)} \]

Substitute \( x = 3 - y \) from Equation (4) into Equation (2):
\[ (3 - y) - 3y = -3 \implies 3 - 4y = -3 \implies -4y = -6 \implies y = \frac{3}{2} \]
Substitute \( y = \frac{3}{2} \) back into \( x + y = 3 \):
\[ x + \frac{3}{2} = 3 \implies x = \frac{3}{2} \]

**Solution**: \( (x, y) = \left(\frac{3}{2}, \frac{3}{2}\right) \)

### Conclusion

The system of equations has multiple solutions, specifically:
- \( (2, 1) \)
- \( (3, 2) \)
- \( \left(\frac{3}{2}, \frac{3}{2}\right) \)

However, the unique solution that satisfies both equations is \( (2, 1) \).