Find the value of `k` for which pair of linear equations `3x + 2y = –5` and `x – ky = 2` has a unique solution.

Answer

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

### Step 1: Write the equations in standard form
The first equation is already in standard form:
\[ 3x + 2y + 5 = 0 \]
The second equation can be rewritten as:
\[ x - ky - 2 = 0 \]

### Step 2: Identify coefficients
From the equations, we can identify the coefficients:
- For the first equation \( 3x + 2y + 5 = 0 \):
- \( a_1 = 3 \)
- \( b_1 = 2 \)
- \( c_1 = 5 \)

- For the second equation \( x - ky - 2 = 0 \):
- \( a_2 = 1 \)
- \( b_2 = -k \)
- \( c_2 = -2 \)

### Step 3: Apply the condition for unique solutions
For the pair of linear equations to have a unique solution, the following condition must be satisfied:
\[ \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \]
Substituting the values we identified:
\[ \frac{3}{1} \neq \frac{2}{-k} \]
This simplifies to:
\[ 3 \neq \frac{2}{-k} \]

### Step 4: Cross-multiply to eliminate the fraction
Cross-multiplying gives:
\[ 3(-k) \neq 2 \]
This simplifies to:
\[ -3k \neq 2 \]

### Step 5: Solve for \( k \)
Rearranging the inequality gives:
\[ 3k \neq -2 \]
Dividing both sides by 3, we find:
\[ k \neq -\frac{2}{3} \]

### Conclusion
Thus, the value of \( k \) for which the pair of linear equations has a unique solution is any real number except \( -\frac{2}{3} \).

---