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 equations has infinitely many solutions:

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

Solution

Step 1: Write the Equations in Standard Form

\[ x + 2y – 1 = 0 \quad (1) \]

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

Step 2: Identify Coefficients

From equations (1) and (2),

\[ a_1 = 1, \quad b_1 = 2, \quad c_1 = -1 \]

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

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

Equating the first two ratios:

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

\[ a + b = 2(a – b) \]

\[ a = 3b \quad (i) \]

Now equate the second and third ratios:

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

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

\[ a + b = 4 \quad (ii) \]

Step 5: Solve the System

From (i): \(a = 3b\)

Substitute in (ii):

\[ 3b + b = 4 \]

\[ 4b = 4 \]

\[ b = 1 \]

\[ a = 3 \]

Conclusion

The given system of equations has infinitely many solutions for:

\[ \boxed{a = 3, \quad b = 1} \]

\[ \therefore \quad x + 2y = 1 \text{ and } 2x + 4y = 2 \text{ represent the same line.} \]