Condition for Infinitely Many Solutions of a Pair of Linear Equations

Video Explanation

Question

Find the value of \(k\) for which the following system of equations has infinitely many solutions:

\[ 2x + (k-2)y = k, \qquad 6x + (2k-1)y = 2k + 5 \]

Solution

Step 1: Write in Standard Form

\[ 2x + (k-2)y – k = 0 \quad (1) \]

\[ 6x + (2k-1)y – (2k+5) = 0 \quad (2) \]

Step 2: Identify Coefficients

From equations (1) and (2),

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

\[ a_2 = 6, \quad b_2 = 2k-1, \quad c_2 = -(2k+5) \]

Step 3: Condition for Infinitely Many Solutions

A pair of linear equations has infinitely many solutions if

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

Step 4: Apply the Condition

\[ \frac{a_1}{a_2} = \frac{2}{6} = \frac{1}{3} \]

Equate with the second ratio:

\[ \frac{k-2}{2k-1} = \frac{1}{3} \]

\[ 3(k-2) = 2k – 1 \]

\[ 3k – 6 = 2k – 1 \]

\[ k = 5 \]

Check with the third ratio:

\[ \frac{k}{2k+5} = \frac{5}{15} = \frac{1}{3} \]

Hence, the condition is satisfied.

Conclusion

The given system of equations has infinitely many solutions for:

\[ \boxed{k = 5} \]

\[ \therefore \quad 2x + 3y = 5 \text{ and } 6x + 9y = 15 \text{ represent the same line.} \]