Find the value of k for which the system has (i) a unique solution, and (ii) no solution.

Solution:

GIVEN:

$k x+2 y=5$

$3 x+y=1$

To find: To determine for what value of k the system of equation has

(1) Unique solution

(2) No solution

We know that the system of equations

$a_{1} x+b_{1} y=c_{1}$

$a_{2} x+b_{2} y=c_{2}$

(1) For Unique solution

$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$

Here,

$\frac{k}{3} \neq \frac{2}{1}$

$k \neq 6$

Hence for $k \neq 6$ the system of equation has unique solution.

(2) For no solution

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

$\frac{k}{3}=\frac{2}{1} \neq \frac{5}{1}$

$k=6$

Hence for $k=6$ the system of equation has no solution