Find the value of k for which the linear equations x + 2y = 3 and 5x + ky = 7, does not has a unique solution.

Options

A. 5
B. 7
C. 2
D. 10

Answer

To find the value of \( k \) for which the linear equations \( x + 2y = 3 \) and \( 5x + ky = 7 \) do not have a unique solution, we need to analyze the conditions under which a system of linear equations has either no solution or infinitely many solutions.

### Step-by-step Solution:

1. **Identify the coefficients**:
The first equation is \( x + 2y = 3 \). We can rewrite it in the standard form \( Ax + By = C \):
- \( A_1 = 1 \), \( B_1 = 2 \), \( C_1 = 3 \)

The second equation is \( 5x + ky = 7 \):
- \( A_2 = 5 \), \( B_2 = k \), \( C_2 = 7 \)

2. **Set up the condition for no unique solution**:
For the two equations to not have a unique solution, the ratios of the coefficients must be equal:
\[ \frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2} \]
This means we need to check:
\[ \frac{1}{5} = \frac{2}{k} \quad \text{and} \quad \frac{1}{5} = \frac{3}{7} \]

3. **Equate the first two ratios**:
From \( \frac{1}{5} = \frac{2}{k} \), we can cross-multiply:
\[ 1 \cdot k = 2 \cdot 5 \]
\[ k = 10 \]

4. **Check the second ratio**:
Now, we check if \( \frac{3}{7} \) is equal to \( \frac{1}{5} \):
- Cross-multiplying gives us \( 3 \cdot 5 = 15 \) and \( 1 \cdot 7 = 7 \).
- Since \( 15 \neq 7 \), the condition for no unique solution is satisfied when \( k = 10 \).

5. **Conclusion**:
Therefore, the value of \( k \) for which the system of equations does not have a unique solution is:
\[ \boxed{10} \]