For which value of k will the following pair of linear equations have no solutions?
`3x+ y=1`
`(2k-1) x + (k-1)y= 2k+1`

Answer

To determine the value of \( k \) for which the given pair of linear equations has no solutions, we need to analyze the conditions under which a system of linear equations is inconsistent.

The given equations are:
1. \( 3x + y = 1 \) (Equation 1)
2. \( (2k-1)x + (k-1)y = 2k + 1 \) (Equation 2)

For the system to have no solutions, the following condition must hold:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \quad \text{and} \quad \frac{c_1}{c_2} \neq \frac{a_1}{a_2} \]

Where:
- \( a_1 = 3 \), \( b_1 = 1 \), \( c_1 = 1 \) (from Equation 1)
- \( a_2 = 2k - 1 \), \( b_2 = k - 1 \), \( c_2 = 2k + 1 \) (from Equation 2)

### Step 1: Set up the ratios
We will first set up the ratios:
\[ \frac{3}{2k - 1} = \frac{1}{k - 1} \]

### Step 2: Cross-multiply the ratios
Cross-multiplying gives us:
\[ 3(k - 1) = 1(2k - 1) \]

### Step 3: Expand both sides
Expanding both sides results in:
\[ 3k - 3 = 2k - 1 \]

### Step 4: Rearrange the equation
Now, we will rearrange the equation to isolate \( k \):
\[ 3k - 2k = -1 + 3 \]
\[ k = 2 \]

### Step 5: Verify the condition for no solutions
Now we need to ensure that this value of \( k \) satisfies the second part of the condition:
\[ \frac{c_1}{c_2} \neq \frac{a_1}{a_2} \]

Substituting \( k = 2 \):
- \( c_2 = 2(2) + 1 = 5 \)
- \( a_2 = 2(2) - 1 = 3 \)

Now check:
\[ \frac{c_1}{c_2} = \frac{1}{5} \quad \text{and} \quad \frac{a_1}{a_2} = \frac{3}{3} = 1 \]

Since \( \frac{1}{5} \neq 1 \), the condition is satisfied.

### Final Answer
Thus, the value of \( k \) for which the pair of linear equations has no solutions is:
\[ \boxed{2} \]