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

GIVEN: 

$2 x-3 y=7$

$(k+2) x-(2 k+1) y=3(2 k-1)$

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

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

Consider the following

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

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

$4 k+2=3 k+6$

$4 k-3 k=6-2$

$k=4$

Now consider the following

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

$-3 \times 3(2 k-1)=-7(2 k+1)$

$18 k-9=14 k+7$

$4 k=16$

$\Rightarrow \quad k=4$

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