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:

\[ 2x – ky + 3 = 0, \qquad 3x + 2y – 1 = 0 \]

Solution

Step 1: Identify Coefficients

From the given equations,

\[ a_1 = 2, \quad b_1 = -k, \quad c_1 = 3 \]

\[ a_2 = 3, \quad b_2 = 2, \quad c_2 = -1 \]

Step 2: 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 3: Apply the Condition

\[ \frac{a_1}{a_2} = \frac{2}{3} \]

So,

\[ \frac{-k}{2} = \frac{2}{3} \]

\[ -3k = 4 \]

\[ k = -\frac{4}{3} \]

Step 4: Verify with Third Ratio

\[ \frac{c_1}{c_2} = \frac{3}{-1} = -3 \]

Since

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

the given system of equations is inconsistent.

Conclusion

The given system of equations has no solution for:

\[ \boxed{k = -\dfrac{4}{3}} \]

\[ \therefore \quad 2x + \frac{4}{3}y + 3 = 0 \text{ and } 3x + 2y – 1 = 0 \text{ represent parallel lines.} \]