Find the value of k for which each of the following system of equations have no solution :

GIVEN:

$3 x-4 y+7=0$

$k x+3 y-5=0$

To find: To determine for what value of k the system of equation has no solution 

We know that the system of equations

$a_{1} x+b_{1} y=c_{1}$

$a_{2} x+b_{2} y=c_{2}$

For no solution

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

Here,

$\frac{3}{k}=\frac{-4}{3} \neq \frac{-7}{5}$

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

$k=\frac{-9}{4}$

Hence for $k=\frac{-9}{4}$ the system of equation has no solution.