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:

\[ 2x – 3y = 7, \qquad (a+b)x – (a+b-3)y = 4a + b \]

Solution

Step 1: Write in Standard Form

\[ 2x – 3y – 7 = 0 \quad (1) \]

\[ (a+b)x – (a+b-3)y – (4a+b) = 0 \quad (2) \]

Step 2: Identify Coefficients

From equations (1) and (2),

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

\[ a_2 = a+b, \quad b_2 = -(a+b-3), \quad c_2 = -(4a+b) \]

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

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

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

\[ 2a + 2b – 6 = 3a + 3b \]

\[ a + b = -6 \quad (3) \]

Now equate the first and third ratios:

\[ \frac{2}{a+b} = \frac{7}{4a+b} \]

Substitute \(a+b=-6\):

\[ \frac{2}{-6} = \frac{7}{4a+b} \]

\[ -\frac{1}{3} = \frac{7}{4a+b} \]

\[ 4a + b = -21 \quad (4) \]

Step 5: Solve the System

From (3) and (4):

\[ a + b = -6 \]

\[ 4a + b = -21 \]

Subtracting,

\[ 3a = -15 \]

\[ a = -5 \]

\[ b = -1 \]

Conclusion

The given system of equations has infinitely many solutions for:

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

\[ \therefore \quad 2x – 3y = 7 \text{ and } -6x + 9y = -21 \text{ represent the same line.} \]