Question

The value of k for which the system of linear equations kx – y – 2 = 0 and 6x – 2y – 3 = 0 has infinitely many solutions, is (does)

• A. \(\frac{1}{2}\)
• B. 3
• C. 4
• D. Not exist

Hint:

Always check the constant term ratio first!
If \(\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\), the lines can never be coincident (which is the requirement for infinitely many solutions).
They can only be parallel (if \(k = 3\)) or intersecting. This immediately tells you that the answer is "Not exist"!

Answer 1

The Correct Option is D

Step 1: First find the $k$ that makes the lines parallel.
Using $\frac{a_1}{a_2}=\frac{b_1}{b_2}$: $\frac{k}{6}=\frac{-1}{-2}=\frac12$, which gives $k=3$.
Step 2: Check whether this candidate also satisfies the constant-term ratio.
For infinitely many solutions we also need $\frac{b_1}{b_2}=\frac{c_1}{c_2}$. Here $\frac{c_1}{c_2}=\frac{-2}{-3}=\frac23$.
Step 3: Compare the two ratios.
$\frac{b_1}{b_2}=\frac12$ but $\frac{c_1}{c_2}=\frac23$, and these are not equal. So even at $k=3$ the lines are only parallel and distinct, never coincident, and no other value of $k$ can fix this mismatch either.
Step 4: Conclude.
No value of $k$ gives infinitely many solutions, so the answer is "Not exist", matching option (D).
\[ \boxed{\text{Not exist}} \]