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:

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

Solution

Step 1: Write in Standard Form

\[ 2x + 3y – k = 0 \quad (1) \]

\[ (k-1)x + (k+2)y – 3k = 0 \quad (2) \]

Step 2: Identify Coefficients

From equations (1) and (2),

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

\[ a_2 = k-1, \quad b_2 = k+2, \quad c_2 = -3k \]

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 equate the first two ratios:

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

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

\[ 2k + 4 = 3k – 3 \]

\[ k = 7 \]

Now check with the third ratio:

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

Hence, the condition is satisfied.

Conclusion

The given system of equations has infinitely many solutions for:

\[ \boxed{k = 7} \]

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