Question

The value of k for which the pair of linear equations \(kx - 3y = 5\), \(4x - 6y = 10\) has infinitely many solutions, is :

• A. 4
• B. 3
• C. 2
• D. 1

Hint:

Always convert the linear equations to the standard form \(ax + by + c = 0\) before extracting coefficients to prevent sign-related errors, particularly for the constant term.
Additionally, verify that the ratio of constants \(c_1/c_2\) is equal to the other ratios to guarantee infinitely many solutions rather than no solution (parallel lines).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given a pair of linear equations in two variables: \(kx - 3y = 5\) and \(4x - 6y = 10\). We need to determine the value of the constant \(k\) such that the system has infinitely many solutions.

Step 2: Key Formula or Approach:
For a pair of linear equations in standard form:
\[ a_1x + b_1y + c_1 = 0 \]
\[ a_2x + b_2y + c_2 = 0 \]
The condition for the system to have infinitely many solutions (coincident lines) is:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \]

Step 3: Detailed Explanation:
1. First, write both equations in the standard form:
\[ kx - 3y - 5 = 0 \implies a_1 = k, \quad b_1 = -3, \quad c_1 = -5 \]
\[ 4x - 6y - 10 = 0 \implies a_2 = 4, \quad b_2 = -6, \quad c_2 = -10 \]
2. Apply the condition for infinitely many solutions:
\[ \frac{k}{4} = \frac{-3}{-6} = \frac{-5}{-10} \]
3. Simplify the known ratios:
\[ \frac{-3}{-6} = \frac{1}{2} \]
\[ \frac{-5}{-10} = \frac{1}{2} \]
4. Since both ratios simplify to \(\frac{1}{2}\), set the ratio containing \(k\) equal to \(\frac{1}{2}\):
\[ \frac{k}{4} = \frac{1}{2} \]
5. Solve for \(k\) by cross-multiplying:
\[ 2k = 4 \implies k = 2 \]
This value ensures that both lines coincide and have infinitely many points of intersection.

Step 4: Final Answer:
The value of \(k\) for which the system has infinitely many solutions is 2, corresponding to option (C).