Find the value(s) of k so that the pair of equations `x+ 2y =5 and 3x+ ky + 15=0` has a unique solution

Answer

To find the value(s) of \( k \) such that the pair of equations \( x + 2y = 5 \) and \( 3x + ky + 15 = 0 \) has a unique solution, we can follow these steps:

### Step 1: Rewrite the equations in standard form
The first equation is already in standard form:
\[ x + 2y - 5 = 0 \]
For the second equation, we can rewrite it as:
\[ 3x + ky + 15 = 0 \quad \text{(already in standard form)} \]

### Step 2: Identify coefficients
From the equations, we can identify the coefficients:
- For the first equation \( a_1 = 1, b_1 = 2, c_1 = -5 \)
- For the second equation \( a_2 = 3, b_2 = k, c_2 = 15 \)

### Step 3: Apply the condition for a unique solution
The condition for the pair of equations to have a unique solution is:
\[ \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \]
Substituting the values of the coefficients:
\[ \frac{1}{3} \neq \frac{2}{k} \]

### Step 4: Cross-multiply to eliminate the fractions
Cross-multiplying gives us:
\[ 1 \cdot k \neq 3 \cdot 2 \]
This simplifies to:
\[ k \neq 6 \]

### Step 5: Conclusion
Thus, the pair of equations will have a unique solution for all values of \( k \) except \( k = 6 \).

### Final Answer
The values of \( k \) for which the pair of equations has a unique solution are:
\[ k \in \mathbb{R} \text{ except } k = 6 \]

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