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:

\[ kx + 3y = 2k + 1, \qquad 2(k+1)x + 9y = 7k + 1 \]

Solution

Step 1: Write in Standard Form

\[ kx + 3y – (2k+1) = 0 \quad (1) \]

\[ 2(k+1)x + 9y – (7k+1) = 0 \quad (2) \]

Step 2: Identify Coefficients

From equations (1) and (2),

\[ a_1 = k, \quad b_1 = 3, \quad c_1 = -(2k+1) \]

\[ a_2 = 2(k+1), \quad b_2 = 9, \quad c_2 = -(7k+1) \]

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

First compare the second ratio:

\[ \frac{b_1}{b_2} = \frac{3}{9} = \frac{1}{3} \]

Now equate with the first ratio:

\[ \frac{k}{2(k+1)} = \frac{1}{3} \]

\[ 3k = 2(k+1) \]

\[ 3k = 2k + 2 \]

\[ k = 2 \]

Check with the third ratio:

\[ \frac{2k+1}{7k+1} = \frac{5}{15} = \frac{1}{3} \quad \text{(for } k=2\text{)} \]

Hence, the condition is satisfied.

Conclusion

The given system of equations has infinitely many solutions for:

\[ \boxed{k = 2} \]

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