Find the values of $ k $ for which the system
$2 x+k y=1$
$ 3 x-5 y=7 $
will have no solution. Is there a value of $k$ for which the system has infinitely many solutions?

Given: 

The given system of equations is:

\(2 x+k y=1\)
\( 3 x-5 y=7 \)

To do: 

We have to find the value of $k$ for which the given system of equations has no solution and if there is a value of $k$ for which the given system of equations has infinitely many solutions.

Solution:

The given system of equations can be written as:

$2x+ky-1=0$

$3x-5y-7=0$

The standard form of system of equations of two variables is $a_{1} x+b_{1} y+c_{1}=0$ and $a_{2} x+b_{2} y-c_{2}=0$.

Comparing the given system of equations with the standard form of equations, we have,

$a_1=2, b_1=k, c_1=-1$ and $a_2=3, b_2=-5, c_2=-7$

The condition for which the above system of equations has no solution is

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

Therefore,

$\frac{2}{3}=\frac{k}{-5}≠\frac{-1}{-7}$

$\frac{2}{3}=\frac{k}{-5}≠\frac{1}{7}$

$\frac{2}{3}=\frac{k}{-5}$

$k=\frac{-5\times2}{3}$

$k=\frac{-10}{3}$

The condition for which the given system of equations has infinitely many solutions is

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

$\frac{a_{1}}{a_{2}}=\frac{2}{3}$

$\frac{c_1}{c_2}=\frac{-1}{-7}=\frac{1}{7}$

Here, 

$\frac{a_{1}}{a_{2}} \  ≠ \frac{c_{1}}{c_{2}} \ $

Therefore, there is no value of $k$ for which the given system of equations have infinite solutions.

The value of $k$ for which the given system of equations has no solution is $\frac{-10}{3}$.