Question

Find the values of p and q for which the following system of linear equations has an infinite number of solutions.
$2x + 3y = 9,\left( {p + q} \right)x + \left( {2p - q} \right)y = 3\left( {p + q + 1} \right)$

Answer 1

Hint- Here, we will proceed by comparing the given pair of linear equations with any general pair of linear equations i.e., ${a_1}x + {b_1}y + {c_1} = 0$ and ${a_2}x + {b_2}y + {c_2} = 0$. Then using the condition for having infinite number of solutions i.e., $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}$.

Complete step-by-step solution -
The given system of linear equations are
2x + 3y = 9 \\\ \Rightarrow 2x + 3y - 9 = 0{\text{ }} \to {\text{(1)}} \\\
$\text{and}$
\left( {p + q} \right)x + \left( {2p - q} \right)y = 3\left( {p + q + 1} \right) \\\ \Rightarrow \left( {p + q} \right)x + \left( {2p - q} \right)y - 3\left( {p + q + 1} \right){\text{ = 0 }} \to {\text{(2)}} \\\
As we know that for any pair of linear equations ${a_1}x + {b_1}y + {c_1} = 0{\text{ }} \to {\text{(3)}}$ and ${a_2}x + {b_2}y + {c_2} = 0{\text{ }} \to {\text{(4)}}$ to have infinite number of solutions, the condition which must be satisfied is that the ratio of the coefficients of x should be equal to the ratio of the coefficients of y which further should be equal to the ratio of the constant terms in the pair of linear equations.
The condition is $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}{\text{ }} \to (5{\text{)}}$
By comparing equations (1) and (3), we get
${a_1} = 2,{b_1} = 3,{c_1} = - 9$
By comparing equations (2) and (4), we get
${a_2} = p + q,{b_2} = 2p - q,{c_2} = - 3\left( {p + q + 1} \right)$
For the given pair of linear equations to have infinite number of solutions, equation (5) must be satisfied
By equation (5), we can write

$$\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}} \\\ \Rightarrow \dfrac{2}{{p + q}} = \dfrac{3}{{2p - q}} = \dfrac{{ - 9}}{{ - 3\left( {p + q + 1} \right)}} \\\ \Rightarrow \dfrac{2}{{p + q}} = \dfrac{3}{{2p - q}} = \dfrac{3}{{p + q + 1}}{\text{ }} \to {\text{(6)}} \\\$$

By equation (6), we can write

$$\Rightarrow \dfrac{2}{{p + q}} = \dfrac{3}{{2p - q}} \\\ \Rightarrow 2\left( {2p - q} \right) = 3\left( {p + q} \right) \\\ \Rightarrow 4p - 2q = 3p + 3q \\\ \Rightarrow 4p - 3p = 3q + 2q \\\ \Rightarrow p = 5q{\text{ }} \to {\text{(7)}} \\\$$

By equation (6), we can write

$$\Rightarrow \dfrac{3}{{2p - q}} = \dfrac{3}{{p + q + 1}} \\\ \Rightarrow 3\left( {p + q + 1} \right) = 3\left( {2p - q} \right) \\\ \Rightarrow 3p + 3q + 3 = 6p - 3q \\\ \Rightarrow 3p + 3q + 3 - 6p + 3q = 0 \\\ \Rightarrow - 3p + 6q + 3 = 0 \\\$$

By substituting the value of p from equation (7) in the above equation, we get

$$\Rightarrow - 3\left( {5q} \right) + 6q + 3 = 0 \\\ \Rightarrow - 15q + 6q = - 3 \\\ \Rightarrow - 9q = - 3 \\\ \Rightarrow q = \dfrac{{ - 3}}{{ - 9}} = \dfrac{1}{3} \\\$$

By putting q = $\dfrac{1}{3}$ in equation (7), we get
$\Rightarrow p = 5\left( {\dfrac{1}{3}} \right) = \dfrac{5}{3}$
Therefore, the required values of p and q for which the given system of linear equations has infinite number of solutions are $\dfrac{5}{3}$ and $\dfrac{1}{3}$ respectively.

Note- Any general pair of linear equations which are given by ${a_1}x + {b_1}y + {c_1} = 0$ and ${a_2}x + {b_2}y + {c_2} = 0$ can also have unique solution (consistent solution) if the condition $\dfrac{{{a_1}}}{{{a_2}}} \ne \dfrac{{{b_1}}}{{{b_2}}}$ is satisfied. Also, for these pair of linear equations to have no solution, the condition $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} \ne \dfrac{{{c_1}}}{{{c_2}}}$ should always be satisfied.