`3x+4y=7`
`kx-2y=1`
For what value of k will the above system of equations have no solution?

Options

A. `-3`
B. `-(3)/(2)`
C. `0`
D. `3/2`

Answer

To determine the value of \( k \) for which the system of equations

1. \( 3x + 4y = 7 \)
2. \( kx - 2y = 1 \)

has no solution, we need to analyze the conditions under which two linear equations are parallel. Two lines are parallel if their slopes are equal but their y-intercepts are different.

### Step-by-Step Solution:

1. **Identify the coefficients**:
- For the first equation \( 3x + 4y = 7 \), we can identify:
- \( a_1 = 3 \)
- \( b_1 = 4 \)
- \( c_1 = 7 \)
- For the second equation \( kx - 2y = 1 \), we identify:
- \( a_2 = k \)
- \( b_2 = -2 \)
- \( c_2 = 1 \)

2. **Set up the condition for no solution**:
- The condition for the system to have no solution is given by:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \quad \text{and} \quad \frac{a_1}{a_2} \neq \frac{c_1}{c_2} \]
- Plugging in the values we have:
\[ \frac{3}{k} = \frac{4}{-2} \]

3. **Simplify the ratio**:
- Calculate \( \frac{4}{-2} \):
\[ \frac{4}{-2} = -2 \]
- Now we have:
\[ \frac{3}{k} = -2 \]

4. **Solve for \( k \)**:
- Cross-multiply to solve for \( k \):
\[ 3 = -2k \]
- Rearranging gives:
\[ k = -\frac{3}{2} \]

5. **Conclusion**:
- The value of \( k \) for which the system of equations has no solution is:
\[ k = -\frac{3}{2} \]