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:

\[ 3x + 4y = 12, \qquad (a+b)x + 2(a-b)y = 5a – 1 \]

Solution

Step 1: Write the Equations in Standard Form

\[ 3x + 4y – 12 = 0 \quad (1) \]

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

Step 2: Identify Coefficients

From equations (1) and (2),

\[ a_1 = 3, \quad b_1 = 4, \quad c_1 = -12 \]

\[ a_2 = a+b, \quad b_2 = 2(a-b), \quad c_2 = -(5a-1) \]

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

\[ \frac{a_1}{a_2} = \frac{3}{a+b}, \qquad \frac{b_1}{b_2} = \frac{4}{2(a-b)} = \frac{2}{a-b} \]

Equating the first two ratios:

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

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

\[ 3a – 3b = 2a + 2b \]

\[ a = 5b \quad (i) \]

Now equate the first and third ratios:

\[ \frac{3}{a+b} = \frac{12}{5a-1} \]

\[ 3(5a-1) = 12(a+b) \]

\[ 15a – 3 = 12a + 12b \]

\[ 3a – 12b = 3 \]

\[ a – 4b = 1 \quad (ii) \]

Step 5: Solve the System

From (i): \(a = 5b\)

Substitute in (ii):

\[ 5b – 4b = 1 \]

\[ b = 1 \]

\[ a = 5 \]

Conclusion

The given system of equations has infinitely many solutions for:

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

\[ \therefore \quad 3x + 4y = 12 \text{ and } 6x + 8y = 24 \text{ represent the same line.} \]