Question

Find the value of ${}^{\prime }{k}^{\prime }$ for which each of the following systems of equations has no solution: $8x+5y=9,kx+10y=15.$

Answer 1

The given system of equations:
$8x+5y=9$
$8x+5y-9=0\dots \dots \left(i\right)$
$kx+10y=15$
$kx+10y-15=0\dots \dots \left(ii\right)$

These equations are of the following form:

$a_{1}x+b_{1}y+c_1=0,a_{2}x+b_{2}y+c_2=0$
Here, $a_1=8,b_1=5,c_1=-9$ and $a_2=k,b_2=10,c_2=-15$
In order that the given system has no solution, we must have:

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

i.e. $\frac{8}{k}=\frac{5}{10}\ne \frac{-9}{-15}$

or, $\frac{8}{k}=\frac{1}{2}\ne \frac{3}{5}$

or, $\frac{8}{k}=\frac{1}{2}$ and $\frac{8}{k}\ne \frac{3}{5}$

$\Rightarrow k=16$ and $k\ne \frac{40}{3}$

Hence, the given system of equations has no solution when ${}^{\prime }{k}^{\prime }$ is equal to $16.$