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:

\[ 2x + 3y = 7, \qquad (k+1)x + (2k-1)y = 4k + 1 \]

Solution

Step 1: Write in Standard Form

\[ 2x + 3y – 7 = 0 \quad (1) \]

\[ (k+1)x + (2k-1)y – (4k+1) = 0 \quad (2) \]

Step 2: Identify Coefficients

From equations (1) and (2),

\[ a_1 = 2, \quad b_1 = 3, \quad c_1 = -7 \]

\[ a_2 = k+1, \quad b_2 = 2k-1, \quad c_2 = -(4k+1) \]

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 first two ratios:

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

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

\[ 4k – 2 = 3k + 3 \]

\[ k = 5 \]

Now check with the third ratio:

\[ \frac{2}{6} = \frac{1}{3}, \qquad \frac{7}{21} = \frac{1}{3} \]

Hence, the condition is satisfied.

Conclusion

The given system of equations has infinitely many solutions for:

\[ \boxed{k = 5} \]

\[ \therefore \quad 2x + 3y = 7 \text{ and } 6x + 9y = 21 \text{ represent the same line.} \]