Find the value of `k` for which the following system of equation has no solution : `k x-5y=2` , `6x+2y=7`

Options

A. -10
B. 10
C. -15
D. 15

Explanation

To find the value of \( k \) for which the system of equations has no solution, we need to analyze the given equations:

1. **Equations Given:**
\[ kx - 5y = 2 \quad \text{(1)} \]
\[ 6x + 2y = 7 \quad \text{(2)} \]

2. **Rearranging to Standard Form:**
We can rewrite both equations in the standard form \( Ax + By + C = 0 \).

For equation (1):
\[ kx - 5y - 2 = 0 \quad \Rightarrow \quad A_1 = k, \quad B_1 = -5, \quad C_1 = -2 \]

For equation (2):
\[ 6x + 2y - 7 = 0 \quad \Rightarrow \quad A_2 = 6, \quad B_2 = 2, \quad C_2 = -7 \]

3. **Condition for No Solution:**
A system of linear equations has no solution if the lines represented by the equations are parallel. The condition for parallel lines is:
\[ \frac{A_1}{A_2} = \frac{B_1}{B_2} \quad \text{and} \quad \frac{A_1}{A_2} \neq \frac{C_1}{C_2} \]

Substituting the values:
\[ \frac{k}{6} = \frac{-5}{2} \]

4. **Cross-Multiplying:**
Cross-multiply to solve for \( k \):
\[ 2k = 6 \cdot (-5) \]
\[ 2k = -30 \]
\[ k = -15 \]

5. **Checking the Condition for No Solution:**
Now we need to ensure that this value of \( k \) does not satisfy the second part of the condition:
\[ \frac{k}{6} \neq \frac{-2}{-7} \]
Substituting \( k = -15 \):
\[ \frac{-15}{6} \neq \frac{2}{7} \]
Simplifying:
\[ -\frac{5}{2} \neq \frac{2}{7} \]
This condition holds true.

Thus, the value of \( k \) for which the system of equations has no solution is:
\[ \boxed{-15} \]