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

GIVEN:

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

$6 x+(2 k-1) y=2 k+5$

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 

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

Here,

$\frac{2}{6}=\frac{(k-2)}{(2 k-1)}=\frac{k}{2 k+5}$

Consider the following relation to find k

$\frac{2}{6}=\frac{(k-2)}{(2 k-1)}$

$2(2 k-1)=6(k-2)$

$4 k-2=6 k-12$

$6 k-4 k=12-2$

$2 k=10$

$k=5$

Now again consider the following

$\frac{(k-2)}{(2 k-1)}=\frac{k}{2 k+5}$

$(2 k+5)(k-2)=k(2 k-1)$

$2 k^{2}-4 k+5 k-10=2 k^{2}-k$

$2 k^{2}-4 k+5 k-10=2 k^{2}-k$

$2 k=10$

$k=5$
Hence for $k=5$ the system of equation have infinitely many solutions