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

Given: 

The given system of equations is:

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

To do: 

We have to find the values of $p$ and $q$ for which the following system of linear equations has infinite number of solutions.

Solution:

The given system of equations can be written as:

$2x+3y-9=0$
$(p+q)x+(2p-q)y-3(p+q+1)=0$

The standard form of system of equations of two variables is $a_{1} x+b_{1} y+c_{1}=0$ and $a_{2} x+b_{2} y-c_{2}=0$.

Comparing the given system of equations with the standard form of equations, we have,

$a_1=2, b_1=3, c_1=-9$ and $a_2=(p+q), b_2=(2p-q), c_2=-3(p+q+1)$

The condition for which the given system of equations has infinitely many solutions is

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

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

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

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

$(2p-q)\times2=3\times(p+q)$ and $(p+q+1)\times2=3\times(p+q)$

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

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

$p=5q$ and $p+q=2$

Using $p=5q$ in $p+q=2$, we get,

$5q+q=2$

$6q=2$

$q=\frac{2}{6}=\frac{1}{3}$

This implies,

$p=5q=5\times\frac{1}{3}=\frac{5}{3}$

The values of $p$ and $q$ for which the given system of equations has infinitely many solutions is $\frac{5}{3}$ and $\frac{1}{3}$ respectively.