Infinitely Many Solutions of a Pair of Linear Equations

Video Explanation

Question

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

\[ cx + 3y + 3 – c = 0, \qquad 12x + cy – c = 0 \]

Solution

Step 1: Write in Standard Form

\[ cx + 3y – (c – 3) = 0 \quad (1) \]

\[ 12x + cy – c = 0 \quad (2) \]

Step 2: Identify Coefficients

From equations (1) and (2),

\[ a_1 = c, \quad b_1 = 3, \quad c_1 = -(c – 3) \]

\[ a_2 = 12, \quad b_2 = c, \quad c_2 = -c \]

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

\[ \frac{a_1}{a_2} = \frac{c}{12} \]

Equate with the second ratio:

\[ \frac{3}{c} = \frac{c}{12} \]

\[ c^2 = 36 \]

\[ c = \pm 6 \]

Step 5: Verify with the Third Ratio

For \(c = 6\):

\[ \frac{c_1}{c_2} = \frac{-(6-3)}{-6} = \frac{3}{6} = \frac{1}{2}, \qquad \frac{a_1}{a_2} = \frac{6}{12} = \frac{1}{2} \]

Condition is satisfied.

For \(c = -6\):

\[ \frac{c_1}{c_2} = \frac{-(-9)}{6} = \frac{9}{6} \neq \frac{1}{2} \]

Condition is not satisfied.

Conclusion

The given system of equations has infinitely many solutions for:

\[ \boxed{c = 6} \]

\[ \therefore \quad 6x + 3y – 3 = 0 \text{ and } 12x + 6y – 6 = 0 \text{ represent the same line.} \]