`(3)/(2)x-(1)/(2)y=7`
`kx-6y=4`
In the system of linear equation above, is a constant. If the system has no solution, what is the value of k?

Answer

To solve the problem, we need to determine the value of \( k \) such that the system of linear equations has no solution. This occurs when the two equations represent parallel lines, which means they have the same slope but different intercepts.

### Step-by-Step Solution:

1. **Write the equations**:
The given equations are:
\[ \frac{3}{2}x - \frac{1}{2}y = 7 \quad \text{(Equation 1)} \]
\[ kx - 6y = 4 \quad \text{(Equation 2)} \]

2. **Eliminate the fractions in Equation 1**:
To eliminate the fractions, multiply the entire Equation 1 by 2:
\[ 2 \left(\frac{3}{2}x - \frac{1}{2}y\right) = 2 \cdot 7 \]
This simplifies to:
\[ 3x - y = 14 \quad \text{(Equation 1 simplified)} \]

3. **Rearrange Equation 1**:
Rearranging gives:
\[ y = 3x - 14 \]

4. **Find the slope of Equation 1**:
The slope (m) of Equation 1 is 3.

5. **Rearrange Equation 2**:
Rearranging Equation 2 gives:
\[ 6y = kx - 4 \]
\[ y = \frac{k}{6}x - \frac{2}{3} \]

6. **Find the slope of Equation 2**:
The slope of Equation 2 is \( \frac{k}{6} \).

7. **Set the slopes equal for parallel lines**:
For the lines to be parallel (and thus have no solution), their slopes must be equal:
\[ 3 = \frac{k}{6} \]

8. **Solve for \( k \)**:
To find \( k \), multiply both sides by 6:
\[ k = 3 \cdot 6 = 18 \]

### Conclusion:
The value of \( k \) for which the system of equations has no solution is:
\[ \boxed{18} \]