Infinitely Many Solutions of a Pair of Linear Equations

Video Explanation

Question

Prove that there is a value of \(c\) (\(c \neq 0\)) for which the system

\[ 6x + 3y = c – 3, \qquad 12x + cy = c \]

has infinitely many solutions. Find this value.

Solution

Step 1: Write the Equations in Standard Form

\[ 6x + 3y – (c – 3) = 0 \quad (1) \]

\[ 12x + cy – c = 0 \quad (2) \]

Step 2: Identify Coefficients

From equations (1) and (2),

\[ a_1 = 6, \quad b_1 = 3, \quad c_1 = -(c – 3) \]

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

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

\[ \frac{a_1}{a_2} = \frac{6}{12} = \frac{1}{2} \]

Equate with the second ratio:

\[ \frac{3}{c} = \frac{1}{2} \]

\[ c = 6 \]

Step 5: Verify the Third Ratio

Substitute \(c = 6\):

\[ \frac{c_1}{c_2} = \frac{-(6-3)}{-6} = \frac{-3}{-6} = \frac{1}{2} \]

Thus,

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

Hence, the given system has infinitely many solutions.

Conclusion

There exists a non-zero value of \(c\) for which the given system has infinitely many solutions.

\[ \boxed{c = 6} \]

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