If the lines given by `3x+ 2ky =2 and 2x+ 5y+1= 0` are parallel, then the value of k is

Options

A. `(-5)/(4)`
B. `(2)/(5)`
C. `(15)/(4)`
D. `(3)/(2)`

Answer

To find the value of \( k \) for which the lines given by the equations \( 3x + 2ky = 2 \) and \( 2x + 5y + 1 = 0 \) are parallel, we can follow these steps:

### Step 1: Write the equations in standard form
The first equation is already in the form \( A_1x + B_1y = C_1 \):
\[ 3x + 2ky = 2 \]

The second equation can be rearranged to fit the standard form:
\[ 2x + 5y + 1 = 0 \]
This can be rewritten as:
\[ 2x + 5y = -1 \]
So, we have:
\[ A_2 = 2, \quad B_2 = 5, \quad C_2 = -1 \]

### Step 2: Identify coefficients
From the first equation \( 3x + 2ky = 2 \), we identify:
- \( A_1 = 3 \)
- \( B_1 = 2k \)
- \( C_1 = 2 \)

From the second equation \( 2x + 5y = -1 \), we have:
- \( A_2 = 2 \)
- \( B_2 = 5 \)
- \( C_2 = -1 \)

### Step 3: Set up the condition for parallel lines
For two lines to be parallel, the following condition must hold:
\[ \frac{A_1}{A_2} = \frac{B_1}{B_2} \quad \text{and} \quad \frac{A_1}{A_2} \neq \frac{C_1}{C_2} \]

### Step 4: Calculate \( \frac{A_1}{A_2} \) and \( \frac{B_1}{B_2} \)
Calculating \( \frac{A_1}{A_2} \):
\[ \frac{A_1}{A_2} = \frac{3}{2} \]

Calculating \( \frac{B_1}{B_2} \):
\[ \frac{B_1}{B_2} = \frac{2k}{5} \]

### Step 5: Set the ratios equal
Since the lines are parallel, we set the ratios equal:
\[ \frac{3}{2} = \frac{2k}{5} \]

### Step 6: Cross-multiply to solve for \( k \)
Cross-multiplying gives:
\[ 3 \cdot 5 = 2k \cdot 2 \]
\[ 15 = 4k \]

### Step 7: Solve for \( k \)
Dividing both sides by 4:
\[ k = \frac{15}{4} \]

### Conclusion
Thus, the value of \( k \) for which the lines are parallel is:
\[ \boxed{\frac{15}{4}} \]