The pair of linear equations 3x + 5y = 3 and 6x + ky = 8 do not have a solution if:

Options

A. `k = 5`
B. `k= 10`
C. `k ne 10`
D. ` k ne 5`

Answer

To determine the value of \( k \) for which the pair of linear equations \( 3x + 5y = 3 \) and \( 6x + ky = 8 \) do not have a solution, we can follow these steps:

### Step 1: Write the equations in standard form
The equations are already in standard form:
1. \( 3x + 5y - 3 = 0 \) (Equation 1)
2. \( 6x + ky - 8 = 0 \) (Equation 2)

### Step 2: Identify coefficients
From the equations, we can identify the coefficients:
- For Equation 1:
- \( a_1 = 3 \)
- \( b_1 = 5 \)
- \( c_1 = -3 \)

- For Equation 2:
- \( a_2 = 6 \)
- \( b_2 = k \)
- \( c_2 = -8 \)

### Step 3: Use the condition for no solution
For the pair of linear equations to have no solution, the following condition must hold:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \]

### Step 4: Set up the ratios
Substituting the coefficients into the ratios:
\[ \frac{3}{6} = \frac{5}{k} \]

### Step 5: Solve for \( k \)
Cross-multiplying gives:
\[ 3k = 6 \cdot 5 \]
\[ 3k = 30 \]
Now, divide both sides by 3:
\[ k = \frac{30}{3} = 10 \]

### Step 6: Verify the condition for no solution
Now, we need to check the second part of the condition:
\[ \frac{c_1}{c_2} = \frac{-3}{-8} = \frac{3}{8} \]
Since \( \frac{3}{6} = \frac{1}{2} \) and \( \frac{1}{2} \neq \frac{3}{8} \), the condition for no solution is satisfied.

### Conclusion
Thus, the value of \( k \) for which the equations do not have a solution is:
\[ \boxed{10} \]