Question

For what value of k, the system of equations
$x+2y=5$
$3x+ky-15=0$
has no solution?

Answer 1

The given system of equations can be written as
$x+2y=5$
$3x+ky=15$

This is of the form $a_{1}x+b_{1}y=c_1$
$a_{2}x+b_{2}y=c_2$,

where, $a_1=1,b_1=2,c_1=5$

and $a_1=3,b_2=k,c_2=15$

For no solution, we must have
$\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$

The given system of equations will have no solution, if
$\frac{1}{3}=\frac{2}{k}\ne \frac{5}{15}$

$\Rightarrow \frac{1}{3}=\frac{2}{k}$ and $\frac{2}{k}\ne \left(\frac{1}{3}\right)$

$\Rightarrow k=6$ and $k\ne 6$, which is not possible.

Hence, there is no value of $k$ for which the given system for equations has no solution.