Infinitely Many Solutions of a Pair of Linear Equations

Video Explanation

Question

Find the values of \(a\) and \(b\) for which the following system of linear equations has infinitely many solutions:

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

Solution

Step 1: Write in Standard Form

\[ (2a-3)x – 3y – 5 = 0 \quad (1) \]

\[ 3x + (b-2)y – 3 = 0 \quad (2) \]

Step 2: Identify Coefficients

From equations (1) and (2),

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

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

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

\[ \frac{c_1}{c_2} = \frac{-5}{-3} = \frac{5}{3} \]

So,

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

Step 5: Find the Value of a

\[ 2a – 3 = 5 \]

\[ 2a = 8 \]

\[ a = 4 \]

Step 6: Find the Value of b

\[ -9 = 5(b-2) \]

\[ -9 = 5b – 10 \]

\[ 5b = 1 \]

\[ b = \frac{1}{5} \]

Conclusion

The given system of equations has infinitely many solutions for:

\[ \boxed{a = 4, \quad b = \frac{1}{5}} \]

\[ \therefore \quad 5x – 3y = 5 \text{ and } 3x – \frac{9}{5}y = 3 \text{ represent the same line.} \]