7. If the lines given by \(3x + 2ky = 2\) and \(2x + 5y + 1 = 0\) are parallel, then the value of \(k\) is
\(-\dfrac{5}{2}\)
\(\dfrac{15}{3}\)
\(\dfrac{15}{4}\)
\(\dfrac{4}{5}\)
Step 1: Recall the condition for parallel lines.
Two lines \(a_1x + b_1y + c_1 = 0\) and \(a_2x + b_2y + c_2 = 0\) are parallel if
\(\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2}\) but \(\dfrac{c_1}{c_2}\) is different.
Step 2: Write equations in standard form.
First line: \(3x + 2ky = 2\)
Rewrite: \(3x + 2ky - 2 = 0\)
So, \(a_1 = 3, \; b_1 = 2k, \; c_1 = -2\).
Second line: \(2x + 5y + 1 = 0\)
So, \(a_2 = 2, \; b_2 = 5, \; c_2 = 1\).
Step 3: Apply parallel condition.
Set \(\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2}\).
That is: \(\dfrac{3}{2} = \dfrac{2k}{5}\).
Step 4: Solve for \(k\).
Cross multiply: \(3 \times 5 = 2 \times 2k\).
\(15 = 4k\).
\(k = \dfrac{15}{4}\).
Step 5: Conclusion.
Therefore, the value of \(k\) is \(\dfrac{15}{4}\).
Correct Option: (C)