What is the value of k for which the pair of linear equations kx-2y=3 and 3x+y=5 has a unique solution.

Options

A. k=6
B. `kne-6`
C. k=-6
D. None of these

Answer

To find the value of \( k \) for which the pair of linear equations \( kx - 2y = 3 \) and \( 3x + y = 5 \) has a unique solution, we can follow these steps:

### Step 1: Write the equations in standard form
The given equations are:
1. \( kx - 2y = 3 \)
2. \( 3x + y = 5 \)

We can rewrite them in the standard form \( Ax + By + C = 0 \):
1. \( kx - 2y - 3 = 0 \) (or \( kx - 2y + (-3) = 0 \))
2. \( 3x + y - 5 = 0 \) (or \( 3x + y + (-5) = 0 \))

### Step 2: Identify coefficients
From the standard form, we can identify:
- For the first equation: \( A_1 = k \), \( B_1 = -2 \), \( C_1 = -3 \)
- For the second equation: \( A_2 = 3 \), \( B_2 = 1 \), \( C_2 = -5 \)

### Step 3: Use the condition for a unique solution
For the system of equations to have a unique solution, the condition is:
\[ \frac{A_1}{A_2} \neq \frac{B_1}{B_2} \]
Substituting the coefficients we identified:
\[ \frac{k}{3} \neq \frac{-2}{1} \]
This simplifies to:
\[ \frac{k}{3} \neq -2 \]

### Step 4: Solve for \( k \)
To find the value of \( k \) that does not satisfy this condition, we can set up the equation:
\[ \frac{k}{3} = -2 \]
Multiplying both sides by 3 gives:
\[ k = -6 \]

### Step 5: Conclusion
Thus, for the pair of equations to have a unique solution, \( k \) must not equal \(-6\). Therefore, the value of \( k \) for which the equations have a unique solution is:
\[ k \neq -6 \]