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:

\[ 2x + ky = 11, \qquad 5x – 7y = 5 \]

Solution

Step 1: Write in Standard Form

\[ 2x + ky – 11 = 0 \quad (1) \]

\[ 5x – 7y – 5 = 0 \quad (2) \]

Step 2: Identify Coefficients

From equations (1) and (2),

\[ a_1 = 2, \quad b_1 = k, \quad c_1 = -11 \]

\[ a_2 = 5, \quad b_2 = -7, \quad c_2 = -5 \]

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

\[ \frac{a_1}{a_2} = \frac{2}{5} \]

So,

\[ \frac{k}{-7} = \frac{2}{5} \]

\[ k = -\frac{14}{5} \]

Step 5: Verify with Third Ratio

\[ \frac{c_1}{c_2} = \frac{-11}{-5} = \frac{11}{5} \]

Since

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

the system is inconsistent.

Conclusion

The given system of equations has no solution for:

\[ \boxed{k = -\dfrac{14}{5}} \]

\[ \therefore \quad 2x – \frac{14}{5}y = 11 \text{ and } 5x – 7y = 5 \text{ represent parallel lines.} \]