For what value of k, the following equations have no solutions?
`9x+4y=9`
`7x+ky=5`

Options

A. 3
B. 4.7
C. `28//9`
D. `9//28`

Answer

To determine the value of \( k \) for which the equations \( 9x + 4y = 9 \) and \( 7x + ky = 5 \) have no solutions, we can follow these steps:

### Step 1: Rewrite the equations in standard form
We can rewrite both equations in the standard form \( Ax + By + C = 0 \).

1. For the first equation \( 9x + 4y = 9 \):
\[ 9x + 4y - 9 = 0 \]
Here, \( A_1 = 9 \), \( B_1 = 4 \), and \( C_1 = -9 \).

2. For the second equation \( 7x + ky = 5 \):
\[ 7x + ky - 5 = 0 \]
Here, \( A_2 = 7 \), \( B_2 = k \), and \( C_2 = -5 \).

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

### Step 3: Set up the equations
1. Calculate \( \frac{A_1}{A_2} \):
\[ \frac{A_1}{A_2} = \frac{9}{7} \]

2. Set up the equation for \( \frac{B_1}{B_2} \):
\[ \frac{B_1}{B_2} = \frac{4}{k} \]
Therefore, we have:
\[ \frac{9}{7} = \frac{4}{k} \]

### Step 4: Cross-multiply to solve for \( k \)
Cross-multiplying gives:
\[ 9k = 7 \cdot 4 \]
\[ 9k = 28 \]
Now, solve for \( k \):
\[ k = \frac{28}{9} \]

### Step 5: Verify the second condition
Now we need to check the second condition:
\[ \frac{C_1}{C_2} = \frac{-9}{-5} = \frac{9}{5} \]
We need to ensure that:
\[ \frac{9}{7} \neq \frac{9}{5} \]
This is true since \( \frac{9}{7} \) is not equal to \( \frac{9}{5} \).

### Conclusion
Thus, the value of \( k \) for which the equations have no solutions is:
\[ \boxed{\frac{28}{9}} \]