Question

The pair of linear equations \(kx + 3y = k - 3\) and \(12x + ky = k\) has NO solution for the value of \(k\) equal to :

• A. 6
• B. -6
• C. \(\pm 6\)
• D. 0

Hint:

Be extremely careful not to select option (C) \(\pm 6\) in a hurry.
The value \(k = 6\) makes the equations identical, giving infinite solutions.
Only \(k = -6\) makes them parallel with no solution. Always verify your candidate values against the final inequality ratio!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
This question is from the topic of Linear Equations in Two Variables.
We need to find the value of the parameter \(k\) for which the system of equations has no solution.

Step 2: Key Formula or Approach:
For two linear equations \(a_1 x + b_1 y = c_1\) and \(a_2 x + b_2 y = c_2\), the condition for NO solution (parallel lines) is:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \]

Step 3: Detailed Explanation:
Identify the coefficients from the given equations:
Equation 1: \(kx + 3y = k - 3 \implies a_1 = k, \quad b_1 = 3, \quad c_1 = k - 3\)
Equation 2: \(12x + ky = k \implies a_2 = 12, \quad b_2 = k, \quad c_2 = k\)
Now, set up the ratio relations:
\[ \frac{k}{12} = \frac{3}{k} \neq \frac{k-3}{k} \]
First, solve the equality:
\[ \frac{k}{12} = \frac{3}{k} \]
\[ k^2 = 36 \implies k = 6 \quad \text{or} \quad k = -6 \]
Now, we must check which of these values satisfies the inequality \(\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\).
Case 1: If \(k = 6\):
The ratios are:
\[ \frac{a_1}{a_2} = \frac{6}{12} = \frac{1}{2} \]
\[ \frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2} \]
\[ \frac{c_1}{c_2} = \frac{6-3}{6} = \frac{3}{6} = \frac{1}{2} \]
Since all three ratios are equal (\(\frac{1}{2} = \frac{1}{2} = \frac{1}{2}\)), the system has infinitely many solutions. Thus, \(k = 6\) is incorrect.
Case 2: If \(k = -6\):
The ratios are:
\[ \frac{a_1}{a_2} = \frac{-6}{12} = -\frac{1}{2} \]
\[ \frac{b_1}{b_2} = \frac{3}{-6} = -\frac{1}{2} \]
\[ \frac{c_1}{c_2} = \frac{-6-3}{-6} = \frac{-9}{-6} = \frac{3}{2} \]
Here, we have:
\[ -\frac{1}{2} = -\frac{1}{2} \neq \frac{3}{2} \]
This satisfies the condition for no solution perfectly.

Step 4: Final Answer:
The value of \(k\) for which the system has no solution is \(-6\).