Condition for No Solution of a Pair of Linear Equations

Video Explanation

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

\[ 3x – 4y + 7 = 0, \qquad kx + 3y – 5 = 0 \]

From the given equations,

\[ a_1 = 3, \quad b_1 = -4, \quad c_1 = 7 \]

\[ a_2 = k, \quad b_2 = 3, \quad c_2 = -5 \]

A pair of linear equations has no solution if

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

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

\[ 9 = -4k \]

\[ k = -\frac{9}{4} \]

Now check the third ratio:

\[ \frac{c_1}{c_2} = \frac{7}{-5} \neq \frac{3}{k} \]

Hence, the condition for no solution is satisfied.

The given system of equations has no solution for:

\[ \boxed{k = -\dfrac{9}{4}} \]

\[ \therefore \quad 3x – 4y + 7 = 0 \text{ and } -\frac{9}{4}x + 3y – 5 = 0 \text{ represent parallel lines.} \]

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