Question

Find values of $p$ and $q$ for which the following system of linear equations has infinite number of solutions:
$2x+3y=9$ , $(p+q)x+(2p-q)y=3(p+q+1)$

Answer 1

Given equations:-

$2x+3y=9(p+q)x+(2p-q)y=3(p+q+1)$

Here, $\frac{a_1}{a_2}=\frac{2}{p+q},\frac{b_1}{b_2}=\frac{3}{2p-q},\frac{c_1}{c_2}=\frac{9}{3(p+q+1)}$

For a pair of linear equations to have infinitely many solutions:

$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

So, we need, $\frac{2}{p+q}=\frac{3}{2p-q}=\frac{9}{3(p+q+1)}$

$or,\frac{2}{p+q}=\frac{3}{2p-q}=>2(2p-q)=3(p+q)=>4p-2q=3p+3q=>p=5q....\left(i\right)Also,\frac{3}{2p-q}=\frac{9}{3(p+q+1)}=>9(p+q+1)=9(2p-q)=>p+q+1=2p-q=>2p-p=q+q+1=>p=2q+1Substituting\left(i\right),wehave,5q=2q+1=>q=\frac{1}{3}Also,p=5q=5\left(\frac{1}{3}\right)=\frac{5}{3}$

$\therefore p=\frac{5}{3}andq=\frac{1}{3}$