Condition for No Solution of a Pair of Linear Equations

Video Explanation

Question

Find the value of \(k\) for which the following system of equations has no solution:

\[ kx + 3y = k – 3, \qquad 12x + ky = 6 \]

Solution

Step 1: Write in Standard Form

\[ kx + 3y – (k-3) = 0 \quad (1) \]

\[ 12x + ky – 6 = 0 \quad (2) \]

Step 2: Identify Coefficients

From equations (1) and (2),

\[ a_1 = k, \quad b_1 = 3, \quad c_1 = -(k-3) \]

\[ a_2 = 12, \quad b_2 = k, \quad c_2 = -6 \]

Step 3: Condition for No Solution

A pair of linear equations has no solution if

\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \]

Step 4: Apply the Condition

\[ \frac{k}{12} = \frac{3}{k} \]

\[ k^2 = 36 \]

\[ k = \pm 6 \]

Step 5: Check the Third Ratio

For \(k = 6\):

\[ \frac{c_1}{c_2} = \frac{-3}{-6} = \frac{1}{2}, \qquad \frac{a_1}{a_2} = \frac{6}{12} = \frac{1}{2} \]

(All three ratios equal ⇒ infinitely many solutions, not acceptable)

For \(k = -6\):

\[ \frac{a_1}{a_2} = \frac{-6}{12} = -\frac{1}{2}, \qquad \frac{b_1}{b_2} = \frac{3}{-6} = -\frac{1}{2}, \qquad \frac{c_1}{c_2} = \frac{9}{-6} = -\frac{3}{2} \]

\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \]

Hence, the system is inconsistent.

Conclusion

The given system of equations has no solution for:

\[ \boxed{k = -6} \]

\[ \therefore \quad -6x + 3y = -9 \text{ and } 12x – 6y = 6 \text{ represent parallel lines.} \]