Find the value of k for which each of the following system of equations have infinitely many solutions :

GIVEN: 

$2 x+3 y=2$

$(k+2) x+(2 x+1) y=2(k-1)$

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,

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

Consider the following for k

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

$2(2 k+1)=3(k+2)$

$4 k+2=3 k+6$

$4 k-3 k=6-2$

$\Rightarrow \quad k=4$

Now consider the following

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

$3 \times 2(k-1)=2(2 k+1)$

$6 k-6=4 k+2$

$6 k-4 k=6+2$

$2 k=8$

$\Rightarrow \quad k=4$

Hence for $k=4$ the system of equation have infinitely many solutions.