For what value of k the equations `2x+ky=5 and 3x-3y=6` have unique solution?

Options

A. `k=2`
B. `k=-2`
C. `k ne2`
D. `kne-2`

Answer

To determine the value of \( k \) for which the equations \( 2x + ky = 5 \) and \( 3x - 3y = 6 \) have a unique solution, we will use the condition for a unique solution in a system of linear equations.

### Step-by-Step Solution:

1. **Identify the equations**:
We have the two equations:
\[ 2x + ky = 5 \quad \text{(Equation 1)} \]
\[ 3x - 3y = 6 \quad \text{(Equation 2)} \]

2. **Rearranging the equations**:
We can rewrite these equations in the standard form \( Ax + By + C = 0 \):
- For Equation 1:
\[ 2x + ky - 5 = 0 \quad \Rightarrow \quad 2x + ky + (-5) = 0 \]
Here, \( a_1 = 2 \), \( b_1 = k \), and \( c_1 = -5 \).
- For Equation 2:
\[ 3x - 3y - 6 = 0 \quad \Rightarrow \quad 3x - 3y + (-6) = 0 \]
Here, \( a_2 = 3 \), \( b_2 = -3 \), and \( c_2 = -6 \).

3. **Condition for unique solution**:
The condition for the system of equations to have a unique solution is:
\[ \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \]
Substituting the values:
\[ \frac{2}{3} \neq \frac{k}{-3} \]

4. **Cross-multiplying**:
To eliminate the fractions, we cross-multiply:
\[ 2 \cdot (-3) \neq 3 \cdot k \]
This simplifies to:
\[ -6 \neq 3k \]

5. **Solving for \( k \)**:
Dividing both sides by 3 gives:
\[ -2 \neq k \]

6. **Conclusion**:
Therefore, for the equations to have a unique solution, \( k \) must not equal -2.

### Final Answer:
The value of \( k \) for which the equations have a unique solution is:
\[ k \neq -2 \]