Infinitely Many Solutions of a Pair of Linear Equations

Video Explanation

Question

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

\[ 2x + 3y = 9, \qquad (p+q)x – (2p-q)y = 3(p+q+1) \]

Solution

Step 1: Write the Equations in Standard Form

\[ 2x + 3y – 9 = 0 \quad (1) \]

\[ (p+q)x – (2p-q)y – 3(p+q+1) = 0 \quad (2) \]

Step 2: Identify Coefficients

From equations (1) and (2),

\[ a_1 = 2, \quad b_1 = 3, \quad c_1 = -9 \]

\[ a_2 = (p+q), \quad b_2 = -(2p-q), \quad c_2 = -3(p+q+1) \]

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

First equate the first two ratios:

\[ \frac{2}{p+q} = \frac{3}{2p-q} \]

\[ 2(2p-q) = 3(p+q) \]

\[ 4p – 2q = 3p + 3q \]

\[ p = 5q \quad (i) \]

Now equate the first and third ratios:

\[ \frac{2}{p+q} = \frac{9}{3(p+q+1)} \]

\[ \frac{2}{p+q} = \frac{3}{p+q+1} \]

\[ 2(p+q+1) = 3(p+q) \]

\[ 2p + 2q + 2 = 3p + 3q \]

\[ p + q = 2 \quad (ii) \]

Step 5: Solve the System

From (i) and (ii),

\[ p = 5q \]

\[ 5q + q = 2 \]

\[ 6q = 2 \]

\[ q = \frac{1}{3} \]

\[ p = \frac{5}{3} \]

Conclusion

The given system of equations has infinitely many solutions for:

\[ \boxed{p = \frac{5}{3}, \quad q = \frac{1}{3}} \]

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