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: Understanding the Question:
The topic of this question is a Pair of Linear Equations in Two Variables.
A system of linear equations can have a unique solution, infinitely many solutions, or no solution.
We are given two linear equations and we need to determine the value of the constant \(k\) for which the system has infinitely many solutions.

Step 2: Key Formula or Approach:
For a pair of linear equations:
\[ a_1x + b_1y + c_1 = 0 \] \[ a_2x + b_2y + c_2 = 0 \] The algebraic condition for the system to have infinitely many solutions is:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \] We will identify the coefficients from both equations and check if there exists a value of \(k\) that satisfies this simultaneous ratio equality.

Step 3: Detailed Explanation:

• Identify the coefficients from both given equations:
First Equation: \(kx - y - 2 = 0 \implies a_1 = k, \quad b_1 = -1, \quad c_1 = -2\)
Second Equation: \(6x - 2y - 3 = 0 \implies a_2 = 6, \quad b_2 = -2, \quad c_2 = -3\)

• Apply the condition for infinitely many solutions:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \] Substitute the coefficient values:
\[ \frac{k}{6} = \frac{-1}{-2} = \frac{-2}{-3} \]

• Simplify the fractions:
\[ \frac{k}{6} = \frac{1}{2} = \frac{2}{3} \]

• Analyze the simplified ratios:
Notice that:
\[ \frac{1}{2} \neq \frac{2}{3} \] Since the second ratio \(\frac{b_1}{b_2} = \frac{1}{2}\) is not equal to the third ratio \(\frac{c_1}{c_2} = \frac{2}{3}\), the three ratios can never be equal to one another regardless of the value of \(k\).
Therefore, there is no value of \(k\) that can satisfy the condition for infinitely many solutions.


Step 4: Final Answer:
The value of \(k\) for which the system has infinitely many solutions does not exist.
Therefore, the correct option is (D).