Find the values of p and q for which the following system of linear equations has infinite number of solutions:

Question:

Find the values of p and q for which the following system of linear equations has infinite number of solutions:

$2 x+3 y=9$

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

Solution:

GIVEN:

$2 x+3 y=9$

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

To find: To determine for what value of k the system of equation has infinitely many solutions

We know that the system of equations

$a_{1} x+b_{1} y=c_{1}$

$a_{2} x+b_{2} y=c_{2}$

For infinitely many solution

$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

Here

$\frac{2}{(p+q)}=\frac{3}{(2 p-q)}=\frac{3}{(p+q+1)}$

$\frac{3}{(2 p-q)}=\frac{3}{(p+q+1)}$

$3(p+q+1)=3(2 p-q)$

$3 n+3 a+3=6 n-3 a$

$3 p+3 q+3=6 p-3 q$

$3 p-6 q-3=0 \ldots \ldots$ (1)

Again consider

$\frac{2}{(p+q)}=\frac{3}{(2 p-q)}$

$3(p+q)=2(2 p-q)$

$3 p+3 q=4 p-2 q$

$p-5 q=0$...(2)

Multiplying eq. (2) by 3 and subtracting from eq. (1)

$3 p-6 q-3-3 p+15 q=0$

$9 q=3$

$q=\frac{1}{3}$

Putting the value of q in eq. (2)

$p-\frac{5}{3}=0$

$p=\frac{5}{3}$

Hence for $p=\frac{5}{3}$ and $q=\frac{1}{3}$ the system of equation has infinitely many solution.