Find the value(s) of k for which the pair of equations `{{:(kx + 2y = 3) , (3x + 6y = 10):}` has a unique solution.

Answer

To find the value(s) of \( k \) for which the pair of equations

\[ \begin{align*} 1. & \quad kx + 2y = 3 \\ 2. & \quad 3x + 6y = 10 \end{align*} \]

has a unique solution, we will follow these steps:

### Step 1: Write the equations in standard form
We can rewrite the equations in the standard form \( Ax + By + C = 0 \):

1. \( kx + 2y - 3 = 0 \) (Equation 1)
2. \( 3x + 6y - 10 = 0 \) (Equation 2)

### Step 2: Identify coefficients
From the equations, we identify the coefficients:

- For Equation 1: \( A_1 = k, B_1 = 2, C_1 = -3 \)
- For Equation 2: \( A_2 = 3, B_2 = 6, C_2 = -10 \)

### Step 3: Determine the condition for a unique solution
For a pair of linear equations to have a unique solution, the determinant of the coefficients must be non-zero. The determinant \( D \) is given by:

\[ D = A_1B_2 - A_2B_1 \]

Substituting the values:

\[ D = k \cdot 6 - 3 \cdot 2 \]
\[ D = 6k - 6 \]

### Step 4: Set the determinant not equal to zero
For the equations to have a unique solution, we need:

\[ 6k - 6 \neq 0 \]

### Step 5: Solve the inequality
Solving the inequality:

\[ 6k - 6 \neq 0 \]
\[ 6k \neq 6 \]
\[ k \neq 1 \]

### Conclusion
The pair of equations has a unique solution for all values of \( k \) except \( k = 1 \).

Thus, the final answer is:

\[ \text{The value of } k \text{ for which the equations have a unique solution is } k \neq 1. \]

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