Infinitely Many Solutions of a Pair of Linear Equations

Video Explanation

Question

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

\[ 2x + 3y – 7 = 0, \qquad (a-1)x + (a+1)y = 3a – 1 \]

Solution

Step 1: Write Both Equations in Standard Form

\[ 2x + 3y – 7 = 0 \quad (1) \]

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

Step 2: Identify Coefficients

From equations (1) and (2),

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

\[ a_2 = a-1, \quad b_2 = a+1, \quad c_2 = -(3a-1) \]

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

Equate the first two ratios:

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

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

\[ 2a + 2 = 3a – 3 \]

\[ a = 5 \]

Step 5: Verify with the Third Ratio

Substitute \(a = 5\):

\[ \frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}, \quad \frac{c_1}{c_2} = \frac{7}{14} = \frac{1}{2} \]

All ratios are equal, hence the condition is satisfied.

Conclusion

The given system of equations has infinitely many solutions for:

\[ \boxed{a = 5} \]

\[ \therefore \quad 2x + 3y – 7 = 0 \text{ and } 4x + 6y – 14 = 0 \text{ represent the same line.} \]