Find the value of $k$ for which the system of equations $kx+2y=5$, $3x+y=1$ has no solution. 

Given: Equations: $kx+2y=5$, $3x+y=1$

To do: To find the value of $k$.

If a pair of linear  equations is given by

$a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$ 

When a system of linear equations will represent two Parallel Lines there will be no point of intersection and consequently, there will be no pair of values of $x$ and $y$ which satisfy both equations. Thus, the system has no solution and such pair of linear equation are known as inconsistent pair of linear equations

For parallel lines(inconsistent)

$\frac{a_1}{a_2}=\frac{b_1}{b_2}\
eq\frac{c_1}{c_2}$

The given pair of linear equations are 

$5x+ky+7=0          ........( 1)$

$x+2y−3=0           .........( 2)$

On comparing with General form of a pair of linear equations  in two variables x & y is: 

we get

$a_1=5,\ b_1=k,\ c_1=7,\ a_2=1,\ b_2=2,\ c_2=−3$

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

$\Rightarrow k=5\times2$

$\Rightarrow k=10$

Thus for $k=10$ the given system of equations have no solutions.