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

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

$c x+2 y=3$

$12 x+c y=6$

Solution:

GIVEN:

$c x+3 y=3$

$12 x+c y=6$

To find: To determine for what value of c 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,

$\frac{c}{12}=\frac{3}{c} \neq \frac{3}{6}$

$\frac{c}{12}=\frac{3}{c}$

$c^{2}=12 \times 3$

$c^{2}=36$

$c=\pm 6$

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

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

$2 x+k y=11$

$5 x-7 y=5$

Solution:

GIVEN:

$2 x+k y=11$

$5 x-7 y=11$

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,

$\frac{2}{5}=\frac{k}{-7} \neq \frac{11}{5}$

$\frac{2}{5}=\frac{k}{-7}$

$k=-\frac{14}{5}$

Hence for $k=\frac{-14}{5}$ the system of equation has no solution.