Infinitely Many Solutions of a Pair of Linear Equations

Video Explanation

Question

Determine the values of \(a\) and \(b\) so that the following system of linear equations has infinitely many solutions:

\[ (2a-1)x + 3y – 5 = 0, \qquad 3x + (b-1)y – 2 = 0 \]

Solution

Step 1: Identify Coefficients

From the given equations,

\[ a_1 = 2a – 1, \quad b_1 = 3, \quad c_1 = -5 \]

\[ a_2 = 3, \quad b_2 = b – 1, \quad c_2 = -2 \]

Step 2: 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 3: Apply the Condition

\[ \frac{c_1}{c_2} = \frac{-5}{-2} = \frac{5}{2} \]

So,

\[ \frac{2a-1}{3} = \frac{5}{2} \quad \text{and} \quad \frac{3}{\,b-1\,} = \frac{5}{2} \]

Step 4: Find the Value of a

\[ 2(2a-1) = 15 \]

\[ 4a – 2 = 15 \]

\[ 4a = 17 \]

\[ a = \frac{17}{4} \]

Step 5: Find the Value of b

\[ 2 \times 3 = 5(b-1) \]

\[ 6 = 5b – 5 \]

\[ 5b = 11 \]

\[ b = \frac{11}{5} \]

Conclusion

The given system of equations has infinitely many solutions for:

\[ \boxed{a = \frac{17}{4}, \quad b = \frac{11}{5}} \]

\[ \therefore \quad \text{Both equations represent the same straight line.} \]