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

GIVEN: 

$k x-2 y+6=0$

$4 x-3 y+9=0$

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,

$\frac{k}{4}=\frac{-2}{-3}=\frac{6}{9}$

$\frac{k}{4}=\frac{-2}{-3}$

$k=\frac{4 \times 2}{3}$

$k=\frac{8}{3}$

Hence for $k=\frac{8}{3}$ the system of equation have infinitely many solutions.