Find the value of $k$ for which the following system of equations having infinitely many solution:
$kx\ +\ 3y\ =\ 2k\ +\ 1$
$2(k\ +\ 1)x\ +\ 9y\ =\ 7k\ +\ 1$

Given: The system of equation is$kx\ +\ 3y\ =\ 2k\ +\ 1$; $2(k\ +\ 1)x\ +\ 9y\ =\ 7k\ +\ 1$

 

To do: Find the value of $k$ for which the system of the equation has infinitely many solutions

Solution: 

The given system of the equation can be written as:

$kx\ +\ 3y\ =\ 2k\ +\ 1$

$2(k\ +\ 1)x\ +\ 9y\ =\ 7k\ +\ 1$

The given system of equation is in the form

$a_1x+b_1y+c_1=0$

$a_2x+b_2y+c_2=0$

Here, $a_1=k,b_1=3 ,c_1=-(2k+1) ; a_2=2(k+1),b_2=9,c_2=-(7k+1)$

For unique solution we must have: 

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

$\frac{k}{2(k+1)}=\frac{3}{9}=\frac{-(2k+1)}{-(7k+1)}$

$\frac{k}{2(k+1)}=\frac{1}{3}=\frac{2k+1}{7k+1}$

$\frac{k}{2k+2)}=\frac{1}{3}$ and $\frac{1}{3}=\frac{2k+1}{7k+1}$

$3k=2k+2$ and $7k+1=6k+3$

$3k-2k=2$and $7k-6k=3-1$

$k=2$and  $k=2$

K=2 satisfies both the conditions

Therefore, given system of the equation has infinitely many solutions at $k=2$