The value of k, so that the following system of equations `3x – y + 5 = 0` and `6x - 2y + k=0` has no solution, is

Options

A. `K ne 1 `
B. `k =1`
C. `k = 10 `
D. `k ne 10`

Answer

To determine the value of \( k \) such that the system of equations has no solution, we need to analyze the given equations:

1. The first equation is:
\[ 3x - y + 5 = 0 \]
Rearranging this gives:
\[ 3x - y = -5 \]

2. The second equation is:
\[ 6x - 2y + k = 0 \]
Rearranging this gives:
\[ 6x - 2y = -k \]

### Step 1: Identify coefficients
From the equations, we can identify:
- For the first equation \( 3x - y + 5 = 0 \):
- \( a = 3 \)
- \( b = -1 \)
- \( c = 5 \)

- For the second equation \( 6x - 2y + k = 0 \):
- \( p = 6 \)
- \( q = -2 \)
- \( r = k \)

### Step 2: Apply the condition for no solution
For the system of equations to have no solution, the following condition must be satisfied:
\[ \frac{a}{p} = \frac{b}{q} \quad \text{and} \quad \frac{a}{p} \neq \frac{c}{r} \]

### Step 3: Calculate \( \frac{a}{p} \) and \( \frac{b}{q} \)
Calculating \( \frac{a}{p} \):
\[ \frac{a}{p} = \frac{3}{6} = \frac{1}{2} \]

Calculating \( \frac{b}{q} \):
\[ \frac{b}{q} = \frac{-1}{-2} = \frac{1}{2} \]

Since both ratios are equal:
\[ \frac{a}{p} = \frac{b}{q} = \frac{1}{2} \]

### Step 4: Set up the inequality for \( c \) and \( r \)
Now we need to ensure:
\[ \frac{c}{r} \neq \frac{1}{2} \]
Substituting \( c = 5 \) and \( r = k \):
\[ \frac{5}{k} \neq \frac{1}{2} \]

### Step 5: Solve for \( k \)
Cross-multiplying gives:
\[ 5 \neq \frac{1}{2}k \]
Multiplying both sides by 2:
\[ 10 \neq k \]
This implies:
\[ k \neq 10 \]

### Conclusion
The value of \( k \) such that the system of equations has no solution is:
\[ k \neq 10 \]