The value of k for which the equations `9x+4y=9` and `7x+ky=5`, have no solution is

Options

A. `9/5`
B. `9/7`
C. `9/28`
D. `28/9`

Answer

To find the value of \( k \) for which the equations \( 9x + 4y = 9 \) and \( 7x + ky = 5 \) have no solution, we will use the condition for no solution in simultaneous linear equations.

### Step-by-Step Solution:

1. **Identify the coefficients from the equations**:
- For the first equation \( 9x + 4y = 9 \):
- \( A_1 = 9 \) (coefficient of \( x \))
- \( B_1 = 4 \) (coefficient of \( y \))
- \( C_1 = 9 \) (constant term)
- For the second equation \( 7x + ky = 5 \):
- \( A_2 = 7 \) (coefficient of \( x \))
- \( B_2 = k \) (coefficient of \( y \))
- \( C_2 = 5 \) (constant term)

2. **Apply the condition for no solution**:
The condition for two linear equations to have no solution is:
\[ \frac{A_1}{A_2} = \frac{B_1}{B_2} \quad \text{and} \quad \frac{A_1}{A_2} \neq \frac{C_1}{C_2} \]

3. **Set up the ratio**:
From the coefficients, we have:
\[ \frac{A_1}{A_2} = \frac{9}{7} \quad \text{and} \quad \frac{B_1}{B_2} = \frac{4}{k} \]
Setting these equal gives:
\[ \frac{9}{7} = \frac{4}{k} \]

4. **Cross-multiply to solve for \( k \)**:
Cross-multiplying gives:
\[ 9k = 4 \times 7 \]
Simplifying the right side:
\[ 9k = 28 \]

5. **Solve for \( k \)**:
Dividing both sides by 9:
\[ k = \frac{28}{9} \]

### Final Answer:
The value of \( k \) for which the equations have no solution is:
\[ \boxed{\frac{28}{9}} \]