`kx-3y=4`
`4x- 5y =7`
In the system of equations above, k is a constant and x and y are variables. For what value of k will the system of equations have no solution?

Options

A. `12/5`
B. `16/7`
C. `-16/7`
D. `-12/5`

Answer

To determine the value of \( k \) for which the system of equations has no solution, we start with the given equations:

1. \( kx - 3y = 4 \) (Equation 1)
2. \( 4x - 5y = 7 \) (Equation 2)

### Step 1: Identify the coefficients
From the equations, we can identify the coefficients:
- For Equation 1:
- \( a_1 = k \)
- \( b_1 = -3 \)
- \( c_1 = 4 \)

- For Equation 2:
- \( a_2 = 4 \)
- \( b_2 = -5 \)
- \( c_2 = 7 \)

### Step 2: Set up the condition for no solution
For a system of linear equations to have no solution, the following condition must hold:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \quad \text{and} \quad \frac{a_1}{a_2} \neq \frac{c_1}{c_2} \]

### Step 3: Apply the condition
Substituting the coefficients into the condition:
1. From \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \):
\[ \frac{k}{4} = \frac{-3}{-5} = \frac{3}{5} \]

2. From \( \frac{c_1}{c_2} \):
\[ \frac{c_1}{c_2} = \frac{4}{7} \]

### Step 4: Solve for \( k \)
Now, we solve the equation \( \frac{k}{4} = \frac{3}{5} \):
\[ k = 4 \cdot \frac{3}{5} = \frac{12}{5} \]

### Step 5: Verify the second condition
We need to ensure that:
\[ \frac{3}{5} \neq \frac{4}{7} \]
Calculating both fractions:
- \( \frac{3}{5} = 0.6 \)
- \( \frac{4}{7} \approx 0.571 \)

Since \( 0.6 \neq 0.571 \), the condition holds true.

### Conclusion
Thus, the value of \( k \) for which the system of equations has no solution is:
\[ \boxed{\frac{12}{5}} \]